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/r/n13/r/n一元一次不等式及不等式組/r/n中考要求/r/n內(nèi)容/r/n基本要求/r/n略高要求/r/n較高要求/r/n不等式/r/n(/r/n組/r/n)/r/n能根據(jù)具體問題中的大小關(guān)系了解不等式的意義./r/n能根據(jù)具體問題中的數(shù)量關(guān)系列出不等式/r/n(/r/n組/r/n)/r/n./r/n不等式的性質(zhì)/r/n理解不等式的基本性質(zhì)./r/n會利用不等式的性質(zhì)比較兩個實數(shù)的大小./r/n解一元一次不/r/n等式/r/n(/r/n組/r/n)/r/n了解一元一次不等式/r/n(/r/n組/r/n)/r/n的解的意義,會在數(shù)軸上表示/r/n(/r/n確定/r/n)/r/n其解集./r/n會解一元一次不等式和由兩個一元一次不等式組成的不等式組,并會根據(jù)條件求整數(shù)解./r/n能根據(jù)具體問題中的數(shù)量關(guān)系列出一元一次不等式解決簡單問題./r/n例題精講/r/n板塊一、一元一次不等式及不等式組的解法/r/n【例/r/n1/r/n】求不等式/r/nx/r/n+/r/n3/r/n凹〉/r/n1/r/n-口/r/n的解集./r/n82/r/n【解析】對本例,首先應(yīng)去分母,化成標準形式求解./r/n去分母,得/r/n8/r/nx/r/n+/r/n3/r/n(/r/nx/r/n+/r/n1/r/n)/r/n>/r/n8/r/n一/r/n4/r/n(/r/nx/r/n-/r/n5/r/n)/r/n,/r/n去括號,得/r/n8x/r/n+/r/n3x/r/n+/r/n3/r/n〉/r/n8/r/n一/r/n4/r/nx/r/n+/r/n20/r/n移項,得/r/n8x/r/n+/r/n3x/r/n+/r/n4/r/nx/r/n〉/r/n8/r/n+/r/n20/r/n-/r/n3/r/n,/r/n合并同類項,得/r/n15x/r/n〉/r/n25/r/n系數(shù)化為/r/n1/r/n,得/r/nx/r/n>/r/n5/r/n3/r/n【答案】/r/nx/r/n>/r/n5/r/n3/r/n【鞏固】/r/n當/r/nx/r/n為何值時,代數(shù)式凹/r/n-/r/n1/r/n的值不小于的值?/r/n34/r/n【解析】解決此類問題首先應(yīng)理解/r/n“/r/n不小于/r/n”/r/n的意思,進而再列出不等式,按照解一元一次不等式方法求解/r/n2/r/nx/r/n+/r/n1/r/n3/r/n+/r/n5/r/nx/r/n依題意,得土/r/n1/r/n-/r/n1/r/n三/r/n3/r/n^/r/n5/r/nx/r/n3/r/n4/r/n???/r/n4/r/n(/r/n???/r/n4/r/n(/r/n2/r/nx/r/n+/r/n1/r/n)/r/n-/r/n12/r/n三/r/n3/r/n(/r/n3/r/n+/r/n5/r/nx/r/n)/r/n,/r/n8/r/nx/r/n-/r/n15x/r/n2/r/n9/r/n+/r/n12/r/n-/r/n4/r/n,/r/n-/r/n7x/r/n2/r/n17/r/n???x/r/nW/r/n17/r/n7/r/n所以,當/r/nx/r/n時,代數(shù)式竽/r/n一/r/n1/r/n的值不小于罟的值/r/n?/r/n17/r/n【答案】/r/nx/r/nW/r/n-/r/n17/r/n【例/r/n2/r/n】求不等式</r/n1/r/n的正整數(shù)解./r/n12/r/n解析】對于求不等式的正整數(shù)解,應(yīng)先不考慮這一限制條件,按解一元一次不等式的方法求解后,再研/r/n
/r/n究限制條件,便可達到目的./r/n去分母,/r/n得/r/n4/r/nx/r/n-/r/n5/r/n</r/n12/r/n移項,合并,得/r/n4/r/nx/r/n</r/n17/r/n17/r/n系數(shù)化為/r/n1/r/n,得/r/nx/r/n</r/n17/r/n4/r/n??/r/n?求原不等式正整數(shù)解./r/n???x/r/n=/r/n1/r/n,/r/n2/r/n,/r/n3/r/n,/r/n4/r/n為原不等式正整解./r/n【答案】/r/nx/r/n=/r/n1,2,3,4/r/n【鞏固】/r/n不等式/r/nx/r/n+/r/n3/r/n3/r/n2/r/nx/r/n的負整數(shù)解是/r/n一/r/n__/r/n答案】/r/n-/r/n5/r/n、/r/n-/r/n4/r/n、/r/n-/r/n3/r/n、/r/n-/r/n2/r/n、/r/n-/r/n1/r/n丨/r/n3/r/nx/r/n—/r/n1/r/n〉一/r/n4/r/n【例/r/n3/r/n】解不等式組/r/n]/r/n4/r/n,并把它的解集表示在數(shù)軸上./r/n[/r/n2/r/nx/r/n</r/nx/r/n+/r/n2/r/n【解析】/r/nF/r/nx/r/n一/r/n1/r/n〉一/r/n4/r/nnF/r/n>-/r/n1/r/nn-/r/nl/r/n</r/nx/r/n</r/n2./r/n二/r/n[/r/n2/r/nx/r/n</r/nx/r/n+/r/n2/r/n[/r/nx/r/n</r/n2/r/n_3-2/r/n-1/r/no1/r/n23/r/n?/r/n原不等式組的解集是/r/n一/r/n1/r/n</r/nx/r/n</r/n2./r/n在數(shù)軸上表示為:答案】/r/n一/r/n1/r/n</r/nx/r/n</r/n2/r/n2/r/nx/r/n一/r/n3/r/n1/r/n【鞏固】/r/n解不等式:亠/r/n3/r/n</r/n2/r/n</r/n-/r/nx/r/n+/r/n1/r/n42/r/n解析】原不等式相當于:/r/n解得/r/n解析】原不等式相當于:/r/n解得/r/n2/r/n</r/nx/r/n</r/n丄/r/n【答案】/r/n2/r/n</r/nx/r/n</r/n11/r/n鞏固】/r/n解不等式組:/r/n1/r/nx/r/n鞏固】/r/n解不等式組:/r/n1/r/nx/r/n-/r/n10/r/n〉/r/n4/r/n+/r/nx/r/n-/r/n10/r/n【解/r/n析】原方程組的解為/r/nF/r/n>/r/n'/r/n且'^/r/n=/r/n10/r/n,綜合得/r/nx/r/n>/r/n8/r/n且/r/nx/r/n=/r/n10;/r/n[/r/nx/r/n>/r/n8/r/n【答案】/r/nx/r/n>/r/n8/r/n且/r/nx/r/n=/r/n10/r/n【鞏固】/r/n如果/r/n2m/r/n、/r/nm/r/n、/r/n1/r/n-/r/nm/r/n這三個實數(shù)在數(shù)軸上所對應(yīng)的點從左到右依次排列,求/r/nm/r/n的取值范圍./r/n{m/r/n</r/n1/r/n—/r/nm/r/n,解得/r/nm/r/n</r/n0/r/n./r/n2/r/nm/r/n</r/nm/r/n答案】/r/nm/r/n</r/n0/r/n
/r/n【例/r/n4/r/n】已知方程組爲/r/ny/r/n=/r/n2/r/n+/r/n3/r/nm/r/n的解滿足/r/nx/r/n>/r/n0/r/n,/r/ny/r/n>/r/n0/r/n,試求/r/nm/r/n的取值范圍/r/n【答案】解方程組得/r/nJ/r/n1/r/n+/r/n3/r/nm/r/n【答案】解方程組得/r/nJ/r/n1/r/n+/r/n3/r/nm/r/nx/r/n=/r/n一/r/n2/r/n1/r/n—/r/n3/r/nmy/r/n=/r/n2/r/n->/r/n0/r/n1/r/n2/r/n,解得/r/nm/r/n>—/r/n1/r/n—/r/n3/r/nm/r/n3/r/n>/r/n0/r/n2/r/n【鞏固】求使方程組/r/n]/r/n4/r/n;/r/n+打=+/r/n:/r/n+/r/n3/r/n的解'/r/nx/r/n、/r/ny/r/n都是正數(shù)的/r/nm/r/n的取值范圍?/r/n【答案】解方程組得/r/nJ/r/nx/r/n=/r/n8/r/n—/r/nm/r/n,/r/n1/r/ny/r/n=/r/n2/r/nm/r/n—/r/n6/r/n?/r/n?x/r/n、/r/ny/r/n都是正數(shù)/r/n,/r/n【鞏固】求使方程組/r/n]/r/n4/r/n;/r/n+打=+/r/n:/r/n+/r/n3/r/n的解'/r/nx/r/n、/r/ny/r/n都是正數(shù)的/r/nm/r/n的取值范圍?/r/n【答案】解方程組得/r/nJ/r/nx/r/n=/r/n8/r/n—/r/nm/r/n,/r/n1/r/ny/r/n=/r/n2/r/nm/r/n—/r/n6/r/n?/r/n?x/r/n、/r/ny/r/n都是正數(shù)/r/n,/r/n???/r/nJ/r/n8/r/n—/r/nm/r/n>/r/n0/r/n,解得/r/n3/r/n</r/nm/r/n</r/n8/r/n1/r/n2/r/nm/r/n—/r/n6/r/n>/r/n0/r/n【鞏固】在方程組/r/nJ"/r/n+/r/ny/r/n=/r/n1/r/n—/r/nm/r/n中,若未知數(shù)/r/nx/r/n、/r/ny/r/n滿足/r/nx/r/n+/r/ny/r/n>/r/n0/r/n1/r/nx/r/n+/r/n2/r/ny/r/n=/r/n2/r/n,則/r/nm/r/n的取值范圍為./r/n答案】/r/n2/r/nx/r/n+/r/ny/r/n=/r/n1/r/n—/r/nm/r/n①/r/n,/r/n①/r/n+/r/n②/r/n得,/r/n3(/r/nx/r/n+/r/ny/r/n)/r/n=/r/n3/r/n-/r/nm,x/r/n+/r/n2/r/ny/r/n=/r/n2/r/n3/r/n—/r/nm/r/n??/r/n丁>/r/n0/r/n'解得/r/nm/r/n</r/n3/r/n例/r/n5/r/n】/r/n已知/r/nx/r/n、/r/ny/r/n、/r/nz/r/n為三個非負有理數(shù),且滿足/r/n3/r/nx/r/n+/r/n2/r/ny/r/n+/r/nz/r/n=/r/n5/r/n,/r/nx/r/n+/r/ny/r/n—/r/nz/r/n=/r/n2/r/n,若/r/nS/r/n=/r/n2/r/nx/r/n+/r/ny/r/n—/r/nz/r/n的最大值和最小值之和是多少?/r/n解析】/r/n將/r/nx/r/n、/r/ny/r/n、/r/nz/r/n中的一個字母看做常數(shù),解方程,然后將結(jié)果代入/r/nS/r/n2/r/nx/r/n+/r/ny/r/n-/r/nz/r/n進行消元/r/n答案】/r/n方法一、由嚴/r/n+/r/n2y/r/n+/r/nz/r/n=/r/n5/r/n解得,/r/nJ/r/nx/r/n=/r/n1/r/n-/r/n3z/r/n,/r/n1/r/nx/r/n+/r/ny/r/n—/r/nz/r/n=/r/n2/r/n1/r/ny/r/n=/r/n4/r/nz/r/n+/r/n1/r/n1/r/n-/r/n3/r/nz/r/n>/r/n0/r/nx/r/n、/r/ny/r/n、/r/nz/r/n為三個非負有理數(shù),/r/n???/r/n^/r/n4/r/nz/r/n+/r/n1/r/n>/r/n0/r/n,解得/r/n0/r/n</r/nz<T/r/n將/r/nJ/r/nx/r/n=/r/n1/r/n-/r/n3z/r/n1/r/ny/r/n=/r/n4/r/nz/r/n+/r/n1/r/n代入/r/nS/r/n=/r/n2/r/nx/r/n+/r/ny/r/n—/r/nz/r/n得,/r/nS/r/n=/r/n3/r/n—/r/n3/r/nz/r/n?/r/n?/r/n?/r/n2/r/n</r/nS/r/n</r/n3/r/n,/r/n?/r/n?/r/n?/r/nS/r/n的最大值與最小值之和為/r/n5/r/n方法二、根據(jù)題意得/r/nx/r/n=/r/nS/r/n—/r/n2/r/nS/r/n—/r/n2/r/n>/r/n0/r/n3/r/nx/r/n+/r/n2/r/ny/r/n+/r/nz/r/n=/r/n5/r/nJ/r/nx/r/n+/r/ny/r/n—/r/nz/r/n=/r/n2/r/n,解得/r/nJ/r/n2/r/nx/r/n+/r/ny/r/n—/r/nz/r/n=/r/nS/r/n15/r/n—/r/n4/r/nS/r/n,/r/ny/r/n=/r/n3/r/n3/r/n—/r/nS/r/nz=/r/n—/r/n3/r/n?*x/r/n、/r/ny/r/n、/r/nz/r/n都是非負數(shù),?/r/n:/r/n◎/r/n>/r/n0/r/n
/r/n3/r/n3/r/n—/r/nS/r/n>/r/n0/r/n1/r/n3/r/nS/r/n>/r/n2/r/n???」/r/nS/r/n</r/n15/r/n?/r/n?/r/n?/r/n2/r/n</r/nS/r/n</r/n3/r/n,/r/n?/r/n?/r/n?/r/nS/r/n的最大值與最小值之和為/r/n5/r/n4/r/nS/r/n</r/n3/r/n板塊二、解含有參數(shù)的不等式/r/n【例/r/n6/r/n】解關(guān)于/r/nx/r/n的不等式/r/n2/r/nmx/r/n+/r/n3/r/nV/r/n3/r/nx/r/n+/r/nn/r/n【答案】由原不等式,得:/r/n(2/r/nm/r/n-/r/n3)/r/nx/r/nV/r/nn/r/n-/r/n3/r/n3/r/n./r/nn/r/n一/r/n3/r/n當/r/n2/r/nm/r/n-/r/n3/r/n>/r/n0/r/n,即/r/nm/r/n〉一/r/n時,其解集為/r/nx/r/n</r/n2/r/n2/r/nm/r/n-/r/n3/r/n當/r/n2/r/nm/r/n-/r/n3/r/n</r/n0/r/n,即/r/nm/r/n<—/r/n時,其解集為/r/nx/r/n>/r/n——/r/n2/r/nm/r/n-/r/n3/r/n3/r/n當/r/n2/r/nm/r/n-/r/n3/r/n=/r/n0/r/n,即/r/nm=—/r/n時,/r/n2/r/n若/r/nn/r/n-/r/n3/r/n>/r/n0/r/n,即/r/nn/r/n>/r/n3/r/n,解集為所有數(shù);若/r/nn/r/n-/r/n3/r/n</r/n0/r/n,即/r/nn/r/n</r/n3/r/n,原不等式無解./r/n【鞏固】/r/n解關(guān)于/r/nx/r/n的不等式:/r/na/r/n(/r/nx/r/n-/r/na/r/n)/r/n>/r/nb/r/n(/r/nx/r/n-/r/nb/r/n)/r/n【答案】由原不等式得:/r/n(/r/na/r/n-/r/nb/r/n)/r/nx/r/n>/r/n(/r/na/r/n-/r/nb/r/n)(/r/na/r/n+/r/nb/r/n)/r/n當/r/na/r/n-/r/nb/r/n>/r/n0/r/n,即得不等式解集為/r/nx/r/n>/r/na/r/n+/r/nb/r/n;/r/n當/r/na/r/n-/r/nb/r/n=/r/n0/r/n,即得/r/n0/r/n>/r/n0/r/n,不等式無解;/r/n當/r/na/r/n-/r/nb/r/n</r/n0/r/n,即得不等式解集為/r/nx/r/n</r/na/r/n+/r/nb/r/n./r/n【例/r/n7/r/n】已知關(guān)于/r/nx/r/n的不等式/r/n3/r/n(/r/nx/r/n-/r/na/r/n)/r/n</r/n4/r/n(/r/nx/r/n-/r/n1/r/n)/r/n+/r/n9/r/n的解集是/r/nx/r/n>/r/n1/r/n,求/r/na/r/n的值。/r/n解析】解這個不等式:/r/n3/r/nx/r/n-/r/n3/r/na/r/n</r/n4/r/nx/r/n-/r/n4/r/n+/r/n9,3/r/nx/r/n-/r/n4/r/nx/r/n</r/n3/r/na/r/n-/r/n4/r/n+/r/n9,/r/nx/r/n>-/r/n3/r/na/r/n-/r/n5/r/n解集是/r/nx/r/n>/r/n1/r/n,/r/n-/r/n3/r/na/r/n-/r/n5/r/n=/r/n1/r/n,解得/r/na/r/n=/r/n-/r/n2/r/n。/r/n答案】/r/na/r/n=-/r/n2/r/n【例/r/n8/r/n】已知/r/nx/r/n=/r/n3/r/n是關(guān)于/r/nx/r/n的不等式/r/n3/r/nx/r/n-/r/n竺竺/r/n>/r/n弐的解,求/r/na/r/n的取值范圍。/r/n23/r/n【解/r/n析】將/r/nx/r/n=/r/n3/r/n代入不等式,得/r/n9/r/n-/r/n丸竺/r/n>/r/n2/r/n。解這個不等式,得/r/na/r/n</r/n4/r/n。/r/n2/r/n答案】/r/na/r/n</r/n4/r/n【例/r/n9/r/n】已知關(guān)于/r/nx/r/n的不等式/r/n(4/r/na/r/n-/r/n3/r/nb/r/n)/r/nx/r/n>/r/n2/r/nb/r/n-/r/na/r/n的解集為/r/nx/r/n</r/n|/r/n,求/r/nax/r/n>/r/nb/r/n的解集./r/n解析】根據(jù)題意可得:/r/n2/r/nb/r/n-/r/na/r/n4/r/n5/r/n4a/r/n一/r/n3b/r/n</r/n0/r/n且/r/n4/r/na/r/n-/r/n3/r/nb/r/n=/r/n9/r/n,可得/r/n6b/r/n=/r/n5a/r/n,/r/na/r/n</r/n0/r/n,/r/nax/r/n>/r/nb/r/n的解集為/r/nx/r/n</r/n6/r/n-/r/n【答案】/r/nx/r/n</r/n6/r/n解不等式/r/n3/r/nax/r/n+/r/n5/r/nb/r/n>/r/n0/r/n./r/n【鞏固】/r/n已知關(guān)于/r/nx/r/n的不等式/r/n(2/r/na/r/n-/r/nb/r/n)/r/nx/r/n+/r/na/r/n解不等式/r/n3/r/nax/r/n+/r/n5/r/nb/r/n>/r/n0/r/n./r/n【解析】/r/nV/r/n(2/r/na/r/n-/r/nb/r/n)/r/nx/r/n>/r/n5/r/nb/r/n-/r/na/r/n的解集為/r/nx/r/n</r/n10/r/n,可得/r/n2/r/na/r/n-/r/nb/r/n</r/n0/r/n,且/r/nx/r/n</r/n—/r/n,/r/n?/r/n:—/r/n=/r/n10/r/n,解得/r/nb/r/n=/r/n3/r/na/r/n,/r/nTOC\o"1-5"\h\z/r/n7/r/n2/r/na/r/n-/r/nb/r/n2/r/na/r/n-/r/nb/r/n7/r/n5/r/n2/r/na/r/n-/r/nb/r/n=/r/n2/r/na/r/n-/r/n3/r/na/r/n=/r/n—/r/n</r/n0/r/n,即/r/na/r/n</r/n0/r/n./r/n不等式/r/n3/r/nax/r/n+/r/n5/r/nb/r/n>/r/n0/r/n的解集為/r/nx/r/n</r/n-—=/r/n-/r/n1/r/n./r/n5/r/n5/r/n3/r/na/r/n答案】/r/nx/r/n</r/n-/r/n1/r/nx/r/n-/r/na/r/n</r/n0/r/n①/r/n【例/r/n10/r/n】求關(guān)于/r/nx/r/n的不等式組/r/nL/r/n-/r/n1/r/nx/r/n+/r/n2/r/n①的解集。/r/n+/r/n</r/nx/r/n②/r/n〔/r/n2/r/n3/r/n【答案】解/r/n①/r/n得/r/nx/r/n</r/na/r/n,由/r/n②/r/n得/r/nx/r/n>/r/n1/r/n。/r/n應(yīng)分情況討論:/r/n(⑴當/r/na/r/nW/r/n1/r/n時,原不等式組無解。/r/n⑵/r/n當/r/na/r/n>/r/n1/r/n時,原不等式組的解集為/r/n1/r/n</r/nx/r/n</r/na/r/n。/r/nxax/r/n+/r/n2/r/n>/r/n【鞏固】/r/n解關(guān)于/r/nx/r/n的不等式組:/r/n</r/n3/r/n2/r/n6/r/n2(/r/nx/r/n+/r/n1)/r/n>/r/n11/r/n—/r/nx/r/n【答案】原不等式組可化為/r/nI/r/nx/r/n>/r/n3a/r/n+/r/n2/r/n,/r/n[/r/nx/r/n>/r/n3/r/n當/r/n3/r/na/r/n+/r/n2/r/n>/r/n3/r/n,即/r/na/r/n>/r/n-時,不等式組的解集為/r/nx/r/n>/r/n3/r/na/r/n+/r/n2/r/n;/r/n3/r/n當/r/n3/r/na/r/n+/r/n2/r/n</r/n3/r/n,即/r/na/r/n</r/n-時,不等式組的解集為/r/nx/r/n>/r/n3/r/n./r/n3/r/nI/r/nx/r/n+/r/n9/r/n</r/n5/r/nx/r/n+/r/n1/r/n【例/r/n11/r/n】不等式組/r/nI/r/nx/r/n的解集是/r/nx/r/n>/r/n2/r/n,求/r/nm/r/n的取值范圍./r/nI/r/nx/r/n>/r/nm/r/n+/r/n1/r/n【答案】/r/na/r/n</r/n-/r/n1/r/n【例/r/n12/r/n】試確定/r/nc/r/n的范圍,使不等式組/r/nx/r/n-/r/n■_/r/n->3/r/n-/r/n(/r/n2/r/nx/r/n+/r/n5/r/n)/r/n5/r/n</r/n11/r/n1.5c/r/n—/r/n(x/r/n+/r/n1/r/n)/r/n>/r/n—/r/n(c/r/n-/r/nx/r/n)/r/n+/r/n0.5/r/n(/r/n2x/r/n-/r/n1/r/n)/r/nI/r/n2/r/n2/r/n⑴只有一個整數(shù)解;/r/n⑵沒有整數(shù)解./r/n【解/r/n析】/r/n⑴/r/n解不等式/r/n①/r/n得/r/nx/r/n>/r/n1.7/r/n,解不等式/r/n②/r/n得/r/nx/r/n</r/nc/r/n./r/n(1)/r/n要使不等式組只有一個整數(shù)解,則不等式的解/r/n集為/r/n-/r/n1.7/r/n</r/nx/r/n</r/nc/r/n,且這個惟一的整數(shù)必為/r/n-/r/n1/r/n,故/r/n-/r/n1/r/n</r/nc/r/nW/r/n0/r/n./r/n⑵/r/n要使不等式組沒有整數(shù)解,則/r/nc/r/nW/r/n-/r/n1/r/n./r/n【答案】/r/n⑴/r/n-/r/n1/r/n</r/nc/r/nW/r/n0/r/n;/r/n⑵/r/nc/r/nW/r/n-/r/n1/r/n板塊三、不等式及不等式組的應(yīng)用/r/n【例/r/n13/r/n】若干名學(xué)生合影留念,需交照像費/r/n20/r/n元/r/n(/r/n有兩張照片/r/n)/r/n,如果另外加洗一張照片,又需收費/r/n1.5/r/n元,要使每人平均出錢不超過/r/n4/r/n元錢,并都分到一張照片,至少應(yīng)有幾名同學(xué)參加照像?/r/n【解析】設(shè)有/r/nx/r/n位同學(xué)參加照像,根據(jù)題意得:/r/n20/r/n+/r/n1.5(/r/nx/r/n-/r/n2)/r/nW/r/n4/r/nx/r/n,解得/r/nx/r/n2/r/n6.8/r/n,所以至少應(yīng)有/r/n7/r/n名同學(xué)參加照像./r/n【答案】/r/n7/r/n【例/r/n14/r/n】商業(yè)大廈購進某種商品/r/n1000/r/n件,售價定為進價的/r/n125/r/n%.現(xiàn)計劃節(jié)日期間按原售價讓利/r/n10/r/n%,至多售出/r/n100/r/n件商品;而在銷售淡季按原定價的/r/n60/r/n%大甩賣.為使全部商品售完后贏利,在節(jié)日和淡季之外要按原定價銷售出至少多少件商品?/r/n【解析】設(shè)進價為/r/na/r/n元,按原定價售出/r/nx/r/n件,節(jié)日讓利售出/r/ny/r/n件/r/n(0/r/n</r/ny/r/n</r/n100)/r/n./r/n依題意有/r/na/r/n-/r/nx/r/n-/r/n125%/r/n+/r/na/r/n-/r/ny/r/n?/r/n125%/r/n-/r/n(1/r/n-/r/n10%)/r/n+/r/n(1000/r/n-/r/nx/r/n-/r/ny/r/n)/r/n-/r/na/r/n-/r/n125%/r/n-/r/n60%/r/n>/r/n1000/r/na/r/n,/r/n
/r/n整理得/r/n4/r/nx/r/n+/r/n3/r/ny/r/n>/r/n2000/r/n,由于/r/n0/r/n</r/ny/r/n</r/n100/r/n,所以/r/nx/r/n>/r/n425/r/n,因此按原定價至少銷售/r/n426/r/n件./r/n【答案】/r/n426/r/n件/r/n【例/r/n15/r/n】某高速公路收費站有/r/nm/r/n(/r/nm/r/n>/r/n0/r/n)輛汽車排隊等候通過,假設(shè)通過收費站得車流量保持不變,每個收費窗口的收費檢票的速度也是不變的,若開放一個收費窗口,則需/r/n20/r/nmin/r/n才能將原來排隊等候的汽車以及后來到的汽車全部收費通過。若同時開放兩個收費窗口,只需/r/n8/r/nmin/r/n就可以將原來排隊等候的汽車以及后來到的汽車全部收費通過,若要求在/r/n3/r/nmin/r/n內(nèi)將排隊等候收費的汽車全部通過,并使后來到站的汽車也隨到隨時收費通過,請問至少要同時開放幾個收費窗口?/r/n【答案】解:設(shè)一個收費窗口每分鐘可通過/r/nx/r/n輛汽車,車流量為/r/ny/r/n輛/r/n//r/nmin/r/n,需開放/r/nn/r/n個收費窗口才能在/r/n3/r/nmin/r/n內(nèi)將排隊等候及后來到的汽車全部通過,根據(jù)題意得/r/nm/r/n+/r/n20/r/ny/r/n=/r/n20/r/nx/r/n①/r/n,/r/n最小正整數(shù)/r/nn/r/n=/r/n5/r/n即至少需要開/r/nm/r/n+/r/n8/r/ny/r/n=/r/n2/r/nx/r/n8/r/nx/r/n②/r/n,由/r/n①②/r/n得/r/nj/r/n歹,代入/r/n③/r/n得/r/nn/r/n最小正整數(shù)/r/nn/r/n=/r/n5/r/n即至少需要開/r/n—/r/nI/r/nm/r/n=/r/n40/r/ny/r/n9/r/nm/r/n+/r/n3/r/ny/r/n</r/nn/r/n?/r/n3/r/nx/r/n③/r/n
/r/n放/r/n5/r/n個收費窗口/r/n例/r/n16/r/n】為加強公民的節(jié)水意識,某市制定了以下用水收費標準:每戶每月用水未超過/r/n7/r/n立方米時,每立方米收費/r/n1.0/r/n元并加收/r/n0.2/r/n元的城市污水處理費,超過/r/n7/r/n立方米的部分每立方米收費/r/n1/r/n./r/n5/r/n元并加收/r/n0/r/n./r/n4/r/n元的城市污水處理費,如果某單元共有用戶/r/n50/r/n戶,某月共交水費/r/n541.6/r/n元,且每戶的用水量均未超過/r/n10/r/n立方米,求這個月用水未超過/r/n7/r/n立方米的最多有多少戶?/r/n【解/r/n析】若月用水量恰為/r/n7/r/n立方米,則應(yīng)交費/r/n7/r/nx/r/n(1/r/n+/r/n0.2)/r/n=/r/n8.4(/r/n元/r/n)/r/n,因為/r/n〉/r/n8.4/r/n,即每戶平均用水超/r/n過/r/n7/r/n立方米,所以設(shè)有/r/nx/r/n戶用水未超過/r/n7/r/n立方米,要使/r/nx/r/n取值最大,則需另夕卜/r/n(50/r/n-/r/nx/r/n)/r/n戶用水量盡/r/n可能大,故若這/r/nx/r/n戶每戶用水均為/r/n7/r/n立方米及另外/r/n(50/r/n-/r/nx/r/n)/r/n戶每戶用水為/r/n10/r/n立方米,則總水費應(yīng)不少于/r/n541.6/r/n,/r/n2/r/n即:/r/n8.4/r/nx/r/n+/r/n(50/r/n-/r/nx/r/n)/r/nx/r/n[8.4/r/n+/r/n(10/r/n-/r/n7)/r/nx/r/n(1.5/r/n+/r/n0.4)]/r/n>/r/n541.6/r/n,解得:/r/nx/r/n</r/n28—/r/n,所以這個月用水未超過/r/n7/r/n3/r/n立方米的最多有/r/n28/r/n戶./r/n答案】/r/n28/r/n戶/r/n例/r/n17/r/n】為了加強學(xué)生的交通安全意識,某中學(xué)和交警大隊聯(lián)合舉行了/r/n“/r/n我當一日小警察/r/n”/r/n活動,星期天選派部分學(xué)生到交通路口執(zhí)勤,協(xié)助交警維護交通秩序。若每個路口安排/r/n4/r/n人,那么還剩下/r/n78/r/n人;若每個路口安排/r/n8/r/n人,那么最后一個路口不足/r/n8/r/n人,但不少于/r/n4/r/n人,求這個中學(xué)共選派執(zhí)勤學(xué)生多少人,在幾個交通路口安排執(zhí)勤/r/n【答案】設(shè)選派執(zhí)勤學(xué)生/r/nx/r/n人,在/r/ny/r/n個交通路口執(zhí)勤,則根據(jù)題意得/r/nx/r/n-/r/n4/r/ny/r/nx/r/n-/r/n4/r/ny/r/n=/r/n78/r/n4/r/n</r/nx/r/n-/r/n8(/r/ny/r/n-/r/n1)/r/n</r/n8/r/n解得/r/n19.5/r/n</r/ny/r/n</r/n20.5/r/n。/r/n???/r/n整數(shù)/r/ny/r/n=/r/n20/r/n,/r/nx/r/n=/r/n158/r/n,/r/n?/r/n學(xué)校選派了/r/n158/r/n名學(xué)生在/r/n20/r/n個路口安排執(zhí)勤/r/n例/r/n18/r/n】/r/n2008/r/n年/r/n8/r/n月,北京奧運會帆船比賽將在青島國際帆船中心舉行.觀看帆船比賽的船票分為兩種/r/nA/r/n種船票/r/n600/r/n元/r/n//r/n張,/r/nB/r/n種船票/r/n120/r/n元/r/n//r/n張.某旅行社要為一個旅行團代購部分船票,在購票費不超過/r/n5000/r/n元的情況下,購買/r/nA,B/r/n兩種船票共/r/n15/r/n張,要求/r/nA/r/n種船票的數(shù)量不少于/r/nB/r/n種船票數(shù)量的一半.若設(shè)購買/r/nA/r/n種船票/r/nx/r/n張,請你解答下列問題:/r/n共有幾種符合題意的購票方案?寫出解答過程;/r/n根據(jù)計算判斷:哪種購票方案更省錢?/r/n解析】/r/n(1)/r/n由題意:/r/n600/r/nx/r/n+/r/n120/r/n(/r/n15/r/n-/r/nx/r/n)/r/nW/r/n5000/r/nx/r/n三丄/r/n(/r/n15/r/n-/r/nx/r/n)/r/n〔/r/n2/r/n
/r/n方案一:/r/nA/r/n種船票/r/n5/r/n張,/r/nB/r/n種船票/r/n10/r/n張/r/n;/r/n方案二:/r/nA/r/n種船票/r/n6/r/n張,/r/nB/r/n種船票/r/n9/r/n張/r/n(2)/r/n因為/r/nB/r/n種船票價格便宜,因此/r/nB/r/n種船票越多,總購票費用少./r/n??.第一種方案省錢,為/r/n5/r/nx/r/n600/r/n+/r/n120/r/nx/r/n10/r/n=/r/n4200/r/n(/r/n元/r/n)/r/n答案】(/r/n1/r/n)共兩種購票方案:/r/n方案一:/r/nA/r/n種船票/r/n5/r/n張,/r/nB/r/n種船票/r/n10/r/n張/r/n;/r/n方案二:/r/nA/r/n種船票/r/n6/r/n張,/r/nB/r/n種船票/r/n9/r/n張(/r/n2/r/n)第一種方案省錢/r/n例/r/n19/r/n】/r/n2007/r/n年我市某縣籌備/r/n20/r/n周年縣慶,園林部門決定利用現(xiàn)有的/r/n3490/r/n盆甲種花卉和/r/n2950/r/n盆乙種花卉搭配/r/nA/r/n,/r/nB/r/n兩種園藝造型共/r/n50/r/n個擺放在迎賓大道兩側(cè),已知搭配一個/r/nA/r/n種造型需甲種花卉/r/n80/r/n盆,乙種花卉/r/n40/r/n盆,搭配一個/r/nB/r/n種造型需甲種花卉/r/n50/r/n盆,乙種花卉/r/n90/r/n盆./r/n⑴某校九年級⑴班課外活動小組承接了這個園藝造型搭配方案的設(shè)計,問符合題意的搭配方案有幾種?請你幫助設(shè)計出來./r/n⑵若搭配一個/r/nA/r/n種造型的成本是/r/n800/r/n元,搭配一個/r/nB/r/n種造型的成本是/r/n960/r/n元,試說明/r/n(1)/r/n中哪種方案成本最低?最低成本是多少元?/r/n【解析】/r/n⑴/r/n設(shè)搭配/r/nA/r/n種造型/r/nx/r/n個,則/r/nB/r/n種造型為/r/n(50/r/n-/r/nx/r/n)/r/n個,/r/n依題意,得:/r/n80/r/nx/r/n+/r/n50(50/r/n依題意,得:/r/n80/r/nx/r/n+/r/n50(50/r/n-/r/nx/r/n)/r/n</r/n3490/r/n40/r/nx/r/n+/r/n90(50/r/n-/r/nx/r/n)/r/n</r/n2950/r/n,解得:/r/nx/r/n</r/n33/r/nx/r/n>/r/n31/r/n?/r/n31/r/n</r/nx/r/n</r/n33/r/n???x/r/n是整數(shù),/r/n???/r/nx/r/n可取/r/n31/r/n,/r/n32/r/n,/r/n33/r/n,??.可設(shè)計三種搭配方案:/r/nA/r/n種園藝造型/r/n31/r/n個,/r/nB/r/n種園藝造型/r/n19/r/n個;/r/nA/r/n種園藝造型/r/n32/r/n個,/r/nB/r/n種園藝造型/r/n18/r/n個;/r/nA/r/n種園藝造型/r/n33/r/n個,/r/nB/r/n種園藝造型/r/n17/r/n個./r/n(2)(/r/n法/r/n1)/r/n:由于/r/nB/r/n種造型的造價成本高于/r/nA/r/n種造型成本.所以/r/nB/r/n種造型越少,成本越低,故應(yīng)選擇方案/r/n③/r/n,成本最低,最低成本為:/r/n33/r/nx/r/n800/r/n+/r/n17/r/nx/r/n960/r/n=/r/n42720/r/n(/r/n元/r/n)/r/n(/r/n法/r/n2)/r/n:方案/r/n①/r/n需成本:/r/n31/r/nx/r/n800/r/n+/r/n19/r/nx/r/n960/r/n=/r/n43040/r/n(/r/n元/r/n)/r/n方案/r/n②/r/n需成本:/r/n32/r/nx/r/n800/r/n+/r/n18/r/nx/r/n960/r/n=/r/n42880/r/n(/r/n元/r/n)/r/n方案/r/n③/r/n需成本:/r/n33/r/nx/r/n800/r/n+/r/n17/r/nx/r/n960/r/n=/r/n42720/r/n(/r/n元/r/n)/r/n答案】(/r/n1/r/n)可設(shè)計三種搭配方案:/r/nA/r/n種園藝造型/r/n31/r/n個,/r/nB/r/n種園藝造型/r/n19/r/n個;/r/nA/r/n種園藝造型/r/n32/r/n個,/r/nB/r/n種園藝造型/r/n18/r/n個;/r/nA/r/n種園藝造型/r/n33/r/n個,/r/nB/r/n種園藝造型/r/n17/r/n個./r/n(/r/n2/r/n)方案/r/n③/r/n成本最低,最低成本為:/r/n42720/r/n(/r/n元/r/n)/r/n例/r/n20/r/n】某園林的門票每張/r/n10/r/n元,一次使用,考慮到人們的不同需求,也為了吸引更多的游客,該園林除保留原來的售票方法外,還推出了一種/r/n“/r/n購買個人年票/r/n”/r/n的售票方法(個人年票從購買日起,可供持票者使用一年)。年票分/r/nA/r/n、/r/nB/r/n、/r/nC/r/n三類:/r/nA/r/n類年票每張/r/n120/r/n元,持票者進入園林時,無需再購買門票;/r/nB/r/n類年票每張/r/n60/r/n元,/r/n持票者進入園林時,需購買門票,每次/r/n2/r/n元;/r/nC/r/n類門票每張/r/n40/r/n元,持票者進入該園林時,需在購買門票,每次/r/n3/r/n元/r/n
/r/n⑴如果你只選擇一種購買門票的方式,并且你計劃在一年中用/r/n80/r/n元花在該園林的門票上,試通過/r/n計算,找出可使進入該園林的次數(shù)最多的購票方式/r/n⑵求一年中進入該園林至少超過多少次時,購買/r/nA/r/n類年票比較合算/r/n【答案】/r/n⑴/r/n不可能選/r/nA/r/n類年票/r/n若選/r/nB/r/n類年票,/r/n80/r/n二/r/n60/r/n=/r/n10/r/n次/r/n2/r/n若選/r/nC/r/n類年票,則/r/n80/r/n-/r/n40/r/n=/r/n13-/r/n次/r/n33/r/n若不購買年票,則夕/r/n=/r/n8/r/n次/r/n10/r/n所以計劃用/r/n80/r/n元花在該園林的門票上時,選擇購買/r/nC/r/n類年票的方法進入園林的次數(shù)最多,為/r/n13/r/n次/r/nx/r/n>/r/n30/r/n2/r/n解得/r/n</r/nx/r/n>/r/nx/r/n>/r/n30/r/n2/r/n解得/r/n</r/nx/r/n>/r/n26—/r/n3/r/nx/r/n>/r/n12/r/n60/r/n+/r/n2/r/nx/r/n>/r/n120/r/n則/r/n”0/r/n+/r/n3/r/nx/r/n>/r/n120/r/n、/r/n10/r/nx/r/n>/r/n120/r/n所以,一年中進入該園林至少超過/r/n30/r/n次時,購買/r/nA/r/n類年票比較合算/r/n課堂檢測/r/n討論/r/nax/r/n</r/nb/r/n的解集./r/nb/r/n【答案】當/r/na/r/n>/r/n0/r/n時,解集為/r/nx/r/n</r/n-/r/n;/r/na/r/nb/r/n當/r/na/r/n</r/n0/r/n時,解集為/r/nx/r/n>—/r/n;/r/na/r/n當/r/na/r/n=/r/n0/r/n時/r/n若/r/nb/r/n>/r/n0/r/n,則解集為所有數(shù);/r/n若/r/n—/r/n</r/n0/r/n,不等式無解./r/n已知/r/nx/r/n、/r/ny/r/n同時滿足三個條件:①/r/n3/r/nx/r/n-/r/n2/r/ny/r/n=/r/n4/r/n-/r/np/r/n:②/r/n4/r/nx/r/n-/r/n3/r/ny/r/n=/r/n2/r/n+/r/np/r/n:③/r/nx/r/n>/r/ny/r/n則/r/np/r/n的取值范圍/r/n是/r/n【答案】/r/n②/r/n—/r/n①/r/n得/r/nx/r/n—/r/ny/r/n=/r/n2/r/np/r/n—/r/n2/r/n>/r/n0/r/n,?/r/n:/r/np/r/n>/r/n1/r/n已知非負數(shù)/r/na/r/n、/r/n—/r/n、/r/nc/r/n滿足條件:/r/n3a/r/n+/r/n2—/r/n+/r/nc/r/n=/r/n4/r/n,/r/n2a/r/n+/r/n—/r/n+/r/n3c/r/n=/r/n5/r/n,設(shè)/r/nS/r/n=/r/n5a/r/n+/r/n4—/r/n+/r/n7c/r/n的最小值為/r/nm/r/n,最大值為/r/nn/r/n,求/r/nm/r/n—/r/nn/r/n的值/r/n【答案】/r/n丄/r/n2/r/n開學(xué)初,小芳和小亮去學(xué)校商店購買學(xué)習(xí)用品,小芳用/r/n18/r/n元錢買了/r/n1/r/n支鋼筆和/r/n3/r/n本筆記本;小亮用/r/n
/r/n31/r/n元買了同樣的鋼筆/r/n2/r/n支和筆記本/r/n5/r/n本./r/n⑴求每支鋼筆和每本筆記本的價格;/r/n⑵校運會后,班主任拿出/r/n200/r/n元學(xué)校獎勵基金交給班長,購買上述價格的鋼筆和筆記本共/r/n48/r/n件作為獎品,獎給校運會中表現(xiàn)突出的同學(xué),要求筆記本數(shù)不少于鋼筆數(shù),共有多少種購買方案?請你一一寫出./r/n【解/r/n析】/r/n⑴/r/n設(shè)每支鋼筆/r/nx/r/n元,每支筆記本/r/ny/r/n本./r/nx/r/n+/r/n3/r/ny/r/n=/r/n18/r/n./r/n\/r/nx/r/n=/r/n3/r/n2/r/nx/r/n+/r/n5/r/ny/r/n=/r/n31/r/n,/r/n??/r/nj/r/ny/r/n=/r/n5/r/n⑵/r/n設(shè)購買鋼筆/r/na/r/n支,筆記本/r/nb/r/n個./r/na/r/n+/r/nb/r/n=/r/n48/r/n</r/n3/r/na/r/n+/r/n5/r/nb/r/nW/r/n200,/r/nb/r/n三/r/na/r/n+/r/nb/r/n=/r/n48/r/n</r/n3/r/na/r/n+/r/n5/r/nb/r/nW/r/n200,/r/nb/r/n三/r/na/r/na/r/n三/r/n20/r/nb/r/nW/r/n28/r/n則共有五種購買方案/r/na/r/n=/r/n20,21,22,23,24/r/nb/r/n=/r/n28,27,26,25,24/r/n答案】(/r/n1/r/n)每支鋼筆/r/n3/r/n元,每支筆記本/r/n5/r/n本./r/n5/r/n)五種方案:/r/na/r/n=/r/n20,21,22,23,24/r/nb/r/n=/r/n28,27,26,25,24/r/n課后作業(yè)/r/n若不等式/r/n(/r/na/r/n+/r/nb/r/n)/r/nx/r/n+/r/n(2/r/na/r/n-/r/n3/r/nb/r/n)/r/n</r/n0/r/n的解集為/r/nx/r/n>-/r/n1/r/n,求不等式/r/n(/r/na/r/n-/r/n3/r/nb/r/n)/r/nx/r/n+/r/n(/r/nb/r/n-/r/n2/r/na/r/n)/r/n>/r/n0/r/n的解集./r/n【解析】原的解集為/r/nx/r/n>-/r
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