培優(yōu) 易錯 難題一元二次方程輔導專題訓練及詳細答案

一、一元二次方程真題與模擬題分類匯編（難題易錯題）/r/n隨著經(jīng)濟收入的不斷提高以及汽車業(yè)的快速發(fā)展，家用汽車已越來越多地進入普通家庭，汽車消費成為新亮點.抽樣調(diào)查顯示，截止/r/n2008/r/n年底全市汽車擁有量為/r/n14.4/r/n萬輛.已知/r/n2006/r/n年底全市汽車擁有量為/r/n10/r/n萬輛./r/n（1）/r/n求/r/n2006/r/n年底至/r/n2008/r/n年底我市汽車擁有量的年平均增長率；/r/n（/r/n2）/r/n為保護城市環(huán)境，要求我市到/r/n2010/r/n年底汽車擁有量不超過/r/n15.46/r/n4/r/n萬輛，據(jù)估計從/r/n2008/r/n年底起，此后每年報廢的汽車數(shù)量是上年底汽車擁有量的/r/n10%,/r/n那么每年新增汽車數(shù)量最多不超過多少輛？（假定每年新增汽車數(shù)量相同）/r/n【答案】詳見解析/r/n【解析】/r/n試題分析：/r/n（/r/n1/r/n）/r/n主要考查增長率問題，一般用增長后的量二增長前的量/r/nx/r/n（1+/r/n增長率）解決問題；/r/n（/r/n2）/r/n參照增長率問題的一般規(guī)律，表示出/r/n2010/r/n年的汽車擁有量，然后根據(jù)關(guān)鍵語列出不等式來判斷正確的解./r/n試題解析：/r/n（/r/n1）/r/n設年平均增長率為/r/nX,/r/n根據(jù)題意得：/r/n10/r/n（1+x）2=14.4,/r/n解得/r/nx=/r/n-/r/n2.2/r/n（不合題意舍去）/r/nx=0.2,/r/n答：年平均增長率為/r/n20%/r/n：/r/n（/r/n2）/r/n設每年新增汽車數(shù)量最多不超過/r/ny/r/n萬輛，根據(jù)題意得：/r/n2009/r/n年底汽車數(shù)量為/r/n14.4x90%+y,/r/n2010/r/n年底汽車數(shù)量為/r/n（/r/n14.4x90%+y）x90%+y,/r/n（14.4x90%+y）x90%+y<15.464,/r/ny<2./r/n答：每年新增汽車數(shù)量最多不超過/r/n2/r/n萬輛./r/n考點：一元二次方程一增長率的問題/r/n/.ki=l.k/r/n2/r/n=—3./r/n1/r/n*?*kw/r/n—/r/n,/r/nk/r/n=/r/n—/r/n3./r/n2/r/n解方/r/nS/r/n：/r/n（x+l）（x-3）=-1./r/n【答案】/r/nX!=l+73?X2=l/r/n-/r/nV3/r/n【解析】/r/n試題分析：根據(jù)方程的特點，先化為一般式，然后利用配方法求解即可./r/n試題解析：整理得：/r/nx/r/n2/r/n-2x=2,/r/n配方得：/r/nx/r/n2/r/n-/r/n2x/r/n+/r/nl=3,/r/n即/r/n（/r/nx/r/n?/r/nl）/r/n2=3,/r/n解得：/r/nxE/r/n+/r/n若，/r/nx/r/n2/r/n=l/r/n-/r/n"父母恩深重，恩憐無歇時"，每年/r/n5/r/n月的第二個星期口即為母親節(jié)，節(jié)口前夕巴蜀中學學生會計劃采購一批鮮花禮盒贈送給媽媽們./r/n（/r/n1/r/n）/r/n經(jīng)過和花店賣家議價，可在原標價的基礎上打八折購進，若在花店購買/r/n80/r/n個禮盒最多花費/r/n7680/r/n元，請求出每個禮盒在花店的最高標價；（用不等式解答）/r/n（/r/n2）/r/n后來學生會了解到通過"人眾點評"或"美團"同城配送會在/r/n（/r/n1）/r/n中花店最高售價的基/r/n礎上降價/r/n25%,/r/n學生會計劃在這兩個網(wǎng)站上分別購買相同數(shù)量的禮盒，但實際購買過程/r/n中，"人眾點評"網(wǎng)上的購買價格比原有價格上漲/r/n-m%,/r/n購買數(shù)量和原計劃一樣："美團"網(wǎng)/r/n2/r/n9/r/n上的購買價格比原有價格下降了三/元，購買數(shù)量在原計劃基礎上增加/r/n15/%,/r/n最終，在/r/n20/r/n兩個網(wǎng)站的實際消費總額比原計劃的預算總額增加了芽〃％，求出/r/nm/r/n的值./r/n2/r/n【答案】/r/n（1）120/r/n；/r/n（2）/r/n20./r/n【解析】/r/n試題分析：/r/n（/r/n1/r/n）/r/n本題介紹兩種解法：/r/n解法一：設標價為/r/nx/r/n元，列不等式為/r/n0.8x.80<7680,/r/n解出即可;/r/n解法二：根據(jù)單價二總價一數(shù)量先求出/r/n1/r/n個禮盒最多花費，再除以折扣可求出每個禮盒在花店的最高標價；/r/n（/r/n2）/r/n先假設學生會計劃在這兩個網(wǎng)站上分別購買的禮盒數(shù)為/r/na/r/n個禮盒，表示在“人眾點評"/r/n一/r/n網(wǎng)上的購買實際消費總額：/r/n120a/r/n（1-25%）/r/n（l/r/n+/r/n-m%）/r/n,/r/n在“美團〃網(wǎng)上的購買實際消費/r/n2/r/n9/r/n總額：/r/ncr[120/r/n（1-25%）/r/n-/r/n—m]/r/n（l+15m%）/r/n；/r/n根據(jù)"在兩個網(wǎng)站的實際消費總額比原計劃/r/n■/r/n一/r/n的預算總額增加了芋列方程解出即可./r/n2/r/n試題解析：/r/n（/r/n1）/r/n解：解法一：設標價為/r/nx/r/n元，列不等式為/r/n0.8x.80<7680,x<120/r/n；/r/n解法二：/r/n7680^80^0.8=96^0.8=120/r/n（元）?/r/n答：每個禮盒在花店的最高標價是/r/n120/r/n元；/r/n(/r/n2)/r/n解：假設學生會計劃在這兩個網(wǎng)站上分別購買的禮盒數(shù)為。個禮盒，由題意得：/r/nTOC\o"1-5"\h\z/r/n5/r/n9/r/n120x0.8/r/n。/r/n25%)/r/n(l+—m%)+a[120x0.8/r/n(1-25%)-/r/n一/r/nm]/r/n(l+15m%)=120x0.8a/r/n2/r/n20/r/n15/r/nnn/r/n5/r/n9/r/n(1-/r/n25%)/r/nx2(1+/r/n—m%)/r/n,/r/n即/r/n72a(1+/r/n-m%)/r/n+a(72/r/n-/r/n一/r/nm)(l+15m%)=144a2220/r/n(1+/r/n—m%)/r/n,/r/n整理得：/r/n0/r/n?/r/n0675m/r/n2/r/n-/r/n1.35m=0,/r/nm/r/n2/r/n-/r/n20m=0,/r/n解得：/r/nm/r/n訐/r/n0/r/n(舍)/r/n,/r/n2/r/n“2=20./r/n答：/r/nm/r/n的值是/r/n20./r/n點睛：本題是一元二次方程的應用，第二問有難度，正確表示出"人眾點評"或"美團"實際消費總額是解題關(guān)鍵./r/n5./r/n將/r/nm/r/n看作已知量，分別寫出當/r/n0<x<m/r/n和/r/nx>m/r/n時，/r/nF/r/n與/r/nX/r/n之間的函數(shù)關(guān)系式:/r/n小王經(jīng)營的網(wǎng)店專門銷售某種品牌的一種保溫杯，成本為/r/n30/r/n元/只，每天銷售量/r/ny/r/n(只)與銷售單價/r/nx/r/n(元)之間的關(guān)系式為/r/ny=-10x+700(40<x<55)/r/n,/r/n求當銷售單價為多少元時，每天獲得的利潤最大？最大利潤是多少元？/r/n【答案】當銷售單價為/r/n50/r/n元時，每天獲得的利潤最人，利潤的最犬值為/r/n4000/r/n元/r/n【解析】/r/n【分析】/r/n表示出一件的利潤為/r/n(x/r/n-30)/r/n,根據(jù)總利潤=單件利潤乘以銷倍數(shù)量，整理成頂點式即可解題./r/n【詳解】/r/n設每天獲得的利潤為/r/nw/r/n元，/r/n根據(jù)題意得：/r/nw=(x/r/n-30)/r/ny=(x-30)/r/n(-/r/nlOx+700)/r/n=-/r/n10x/r/n2/r/n+1000x/r/n-21000=-10/r/n(x/r/n-50)/r/n2+4000./r/n???a=/r/n?/r/n10<0,/r/n./r/n?.當/r/nx=50/r/n時，/r/nw/r/n取最人值，最人值為/r/n4000./r/n答：當銷售單價為/r/n50/r/n元時，每天獲得的利潤最人，利潤的最人值為/r/n4000/r/n元./r/n【點睛】/r/n本題考查了一元二次函數(shù)的實際應用，中等難度，熟悉函數(shù)的性質(zhì)是解題關(guān)鍵./r/n(問題)如圖①，在/r/naxbxc/r/n(長/r/nX/r/n寬/r/nx/r/n高，其中/r/na,b,c/r/n為正整數(shù))個小立方塊組成的長方體中，長方體的個數(shù)是多少？/r/n(探究)/r/n

/r/n□二「/r/nI/r/n探究一：/r/nBA/r/n圉②/r/n圖③/r/nB/r/nA/r/n□二「/r/nI/r/n探究一：/r/nBA/r/n圉②/r/n圖③/r/nB/r/nA/r/n圖④/r/n2x3/r/n（/r/n1/r/n）/r/n如圖/r/n②，/r/n在/r/n2x1x1/r/n個小立方塊組成的長方體中，棱/r/nAB/r/n上共有/r/n1+2=/r/n^-=3/r/n條線段,/r/n棱/r/nAC,AD/r/n上分別只有/r/n1/r/n條線段，則圖中長方體的個數(shù)為/r/n3xlxl=3./r/n__/r/n3x4/r/n（/r/n2）/r/n如圖③，在/r/n3x1x1/r/n個小立方塊組成的長方體中，棱/r/nAB/r/n上共有/r/n1+2+3=/r/n-^-=6/r/n條線段，棱/r/nAC,AD/r/n上分別只有/r/n1/r/n條線段，則圖中長方體的個數(shù)為/r/n6xlxl=6./r/n（/r/n3/r/n）/r/n依此類推，如圖④，在/r/naxlxl/r/n個小立方塊組成的長方體中，棱/r/nAB±/r/n共有/r/na（a+l）/r/nl+2+...+a/r/n二——線段，棱/r/nAC,AD±/r/n分別只有/r/n1/r/n條線段，則圖中長方體的個數(shù)為/r/n探究二：/r/n（/r/n4）/r/n如圖⑤，在/r/nax2“/r/n個小立方塊組成的長方體中，棱/r/nAB/r/n上有業(yè)巴條線段，棱/r/nAC/r/n2x3/r/n上有/r/n1/r/n+/r/n2=—=3/r/n條線段，棱/r/nAD±/r/n只有/r/n1/r/n條線段，則圖中長方體的個數(shù)為/r/n2/r/n炷/r/nU/r/n心/r/n如/r/n21）/r/n./r/n22/r/n（/r/n5）/r/n如圖⑥，在/r/nax3xl/r/n個小立方塊組成的長方體中，棱/r/nAB/r/n上有也也條線段，棱/r/nAC/r/n3x4/r/n上有/r/n1/r/n+2+/r/n3=-—=6/r/n條線段，棱/r/nAD±/r/n只有/r/n1/r/n條線段，則圖中長方體的個數(shù)為/r/n2/r/n（/r/n6）/r/n依此類推，如圖⑦，在/r/naxbxl/r/n個小立方塊組成的長方體中，長方體的個數(shù)為/r/n圖⑤/r/n圖⑥/r/n圖⑤/r/n圖⑥/r/n圉①/r/n探究三：/r/n（/r/n7）/r/n如圖⑧，在以/r/naxbx2/r/n個小立方塊組成的長方體中，棱/r/nAB/r/n上有也巴條線段，棱/r/n

/r/n7x3/r/n條線段，棱/r/nAD/r/n上有心〒必線段，則圖中長方體的個數(shù)為/r/n3a(a+l)b(b+l)3ab(a+l)(b+l)/r/nTOC\o"1-5"\h\z/r/nx/r/nx3/r/n=/r/n2/r/n2/r/n4/r/n（/r/n8）/r/n如圖⑨，在/r/naxbx3/r/n個小立方塊組成的長方體中，棱/r/nAB/r/n上有”/r/n2/r/n條線段，棱/r/nAC/r/n3x4/r/n上有/r/nW/r/n〒"條線段，則圖中長方體的個數(shù)為/r/n上有響條線段，棱/r/nAD/r/n（結(jié)論）如圖①，在/r/naxbxc/r/n個小立方塊組成的長方體中，長方體的個數(shù)為/r/n./r/n上有響條線段，棱/r/nAD/r/n（應用）在/r/n2x3x4/r/n個小立方塊組成的長方體中，長方體的個數(shù)為/r/n./r/n（拓展）/r/n如果在若干個小立方塊組成的正方體中共有/r/n1000/r/n個長方體，那么組成這個正方體的小立方塊的個數(shù)是多少？請通過計算說明你的結(jié)論./r/n…/r/nk/r/na（a+l）/r/nab（a+l）（b+l）/r/n2/r/n4/r/n【結(jié)論】/r/n【解析】/r/n【分析】/r/n（3）/r/n(/r/n5)/r/n(6)/r/n根據(jù)規(guī)律,根據(jù)規(guī)律,根據(jù)規(guī)律,根據(jù)規(guī)律,/r/n求出棱/r/nAB,AC,/r/n求出棱/r/nAB,AC,/r/n求出棱/r/nAB,AC,/r/n求出棱/r/nAB,AC,/r/nAD/r/n【結(jié)論】/r/n【解析】/r/n【分析】/r/n（3）/r/n(/r/n5)/r/n(6)/r/n根據(jù)規(guī)律,根據(jù)規(guī)律,根據(jù)規(guī)律,根據(jù)規(guī)律,/r/n求出棱/r/nAB,AC,/r/n求出棱/r/nAB,AC,/r/n求出棱/r/nAB,AC,/r/n求出棱/r/nAB,AC,/r/nAD/r/n上的線段條數(shù),/r/nAD/r/n上的線段條數(shù),/r/nAD/r/n上的線段條數(shù),/r/nAD/r/n上的線段條數(shù),/r/nAD/r/n上的線段條數(shù),/r/n即可得出結(jié)論;即可得出結(jié)論;即可得出結(jié)論;即可得出結(jié)論;即町得出結(jié)論;/r/n（8）/r/n（結(jié)論）根據(jù)規(guī)律，求出棱/r/nAB,AC,/r/n（應用/r/n）a=2,b=3,c=4/r/n代入（結(jié)論）中得出的結(jié)果，即可得出結(jié)論:/r/n（拓展）根據(jù)（結(jié)論）中得出的結(jié)果，建立方程求解，即可得出結(jié)論./r/n8/r/n8/r/n【詳解】/r/n解：探究一、/r/n(/r/n3)/r/n棱/r/nAB±/r/n共有”/r/n1/r/n線段，棱/r/nAC,AD/r/n上分別只有/r/n1/r/n條線段，/r/n2/r/n則圖中長方體的個數(shù)為世二/r/nU/r/n,/r/n22/r/n故答案為也；/r/n2/r/n探究二：/r/n(5)/r/n棱/r/nAB/r/n上有/r/na/r/n(/r/na/r/n")/r/n條線段,棱/r/nAC/r/n上有/r/n6/r/n條線段，棱/r/nAD±/r/n只有/r/n1/r/n條線/r/n2/r/n段，/r/n則圖中長方體的個數(shù)為士二耳/r/nx6xl=3a(a+1)/r/n,/r/n2/r/n故答案為/r/n3a(a+1)/r/n:/r/n(/r/n6)/r/n棱/r/nAB/r/n上有”條線段，棱/r/nAC/r/n上有/r/nb/r/n(/r/nZ1/r/n)/r/n條線段,棱/r/nAD/r/n±/r/n只有/r/n1/r/n條線段，/r/n22/r/n則圖中長方體的個數(shù)為心/r/nx/r/n迪/r/nkJ%")"/r/n】),/r/n2/r/n2/r/n4/r/nab(a+l)(b+l)/r/n故答案為———-——/r/nL./r/n4/r/n探究三：/r/n(8)/r/n棱/r/nAB/r/n上有/r/na/r/n(/r/na/r/n+"/r/n條線段，棱/r/nAC/r/n上有/r/n》卩~/r/n1/r/n)/r/n條線段,棱/r/nAD/r/n上有/r/n6/r/n條/r/n22/r/n線段，/r/n“.心亠2亠亠川八迤、./r/na(a+l)b(b+l)3ab(a+l)(b+l)/r/n則圖中長方體的個數(shù)為——-/r/nx/r/n——/r/nx6=/r/n一——/r/n——I,/r/n222/r/nz/r/n枷忙/r/n7/r/n3ab(a+l)(b+l)/r/n故答案為一-——-——/r/n；/r/n2/r/n(結(jié)論)棱/r/nAB/r/n上有/r/n+/r/n條線段,棱/r/nAC/r/n上有/r/n》/r/n(/r/nb*l/r/n)/r/n條線段,棱/r/nAD/r/n上有土二/r/n11/r/n條線/r/n222/r/n段，/r/n則圖中長方體的個數(shù)為心/r/nx/r/n叫已/r/nx3/r/njbc(W)(b+l)(c+l)/r/n,/r/n2228/r/nabc(a+l)(b+l)(c+l)/r/n故答案為一/r/n'/r/n——————/r/n'/r/n；/r/n8/r/n…/r/nabc(a+l)(b+l)(c+l)/r/n(應用)由(結(jié)論)知，一/r/n'/r/n——————/r/n■/r/n8/r/n???/r/n在/r/n2x3x4/r/n個小立方塊組成的長方體中，長方體的個數(shù)為/r/n2x3x4x/r/n(/r/n2+l)x/r/n(/r/n3+l)x/r/n(/r/n4+l)_/r/n=180,/r/n

/r/n故答案為為/r/n180/r/n：/r/n拓展：設正方體的每條棱上都有/r/nX/r/n個小立方體，即/r/na=b=c=x,/r/n由題意得/r/nm+l)‘/r/n8/r/nm+l)‘/r/n8/r/n=1000./r/n?/r/n??[x(x+1)/r/n]/r/n3/r/n=20/r/n3/r/n,/r/n/.x(x+1)/r/n=20,/r/nxi=4,/r/nx/r/n2/r/n=-5/r/n(不合題意，舍去)/r/n4x4x4=64/r/n所以組成這個正方體的小立方塊的個數(shù)是/r/n64./r/n【點睛】/r/n解此題的關(guān)鍵在于根據(jù)已知得出規(guī)律，題目較好，但有一定的難度，是一道比較容易出錯的題目./r/n&/r/n山西特產(chǎn)專賣店銷售核桃，其進價為每下克/r/n40/r/n元，按每「克/r/n60/r/n元出售，平均每天可售出/r/n100/r/nT'/r/n克，后來經(jīng)過市場調(diào)查發(fā)現(xiàn)，單價每降低/r/n2/r/n元，則平均每天的銷售口/r/nI/r/n增加/r/n20/r/n千克，若該專賣店銷售這種核桃要想平均每天獲利/r/n2240/r/n元，請回答：/r/n（/r/n1/r/n）/r/n每千克核桃應降價多少元？/r/n（/r/n2）/r/n在平均每天獲利不變的情況卞，為盡可能讓利于顧客，贏得市場，該店應按原售價的幾折出售？/r/n【答案】/r/n（/r/n1）4/r/n元或/r/n6/r/n元；/r/n（/r/n2）/r/n九折./r/n【解析】/r/n【詳解】/r/n解：/r/n（/r/n1）/r/n設每千克核桃應降價/r/nx/r/n元./r/n根據(jù)題意，得/r/n（60-X-40）/r/n（100+—x20）/r/n=2240,/r/n2/r/n化簡，得/r/nX/r/n2/r/n-/r/n10x+24=0,/r/n解得/r/nXi=4,/r/nx/r/n2/r/n=6./r/n答：每/r/nT/r/n?克核桃應降價/r/n4/r/n元或/r/n6/r/n元./r/n（2）/r/n由/r/n（/r/n1）/r/n可知每/r/nT/r/n?克核桃可降價/r/n4/r/n元或/r/n6/r/n元./r/n???/r/n要盡可能讓利于顧客，.??每/r/nT/r/n?克核桃應降價/r/n6/r/n元./r/n54/r/n此時，售價為：/r/n60-6=54/r/n（元）,/r/n—xl00%=90%./r/n60/r/n答：該店應按原售價的九折出售./r/n9./r/n如圖，在/r/nAABC/r/n中，/r/nZ/r/n8=90°,>48=6/r/ncm,/r/nBC=/r/n8/r/ncm,/r/n若點/r/nP/r/n從點人沿邊向/r/nB/r/n點以/r/nlcm/s/r/n的速度移動，點/r/nQ/r/n從/r/nB/r/n點沿/r/nBC/r/n邊向點/r/nC/r/n以/r/n2/r/ncm/s/r/n的速度移動，兩點同時出發(fā)./r/n（1）/r/n問幾秒后，/r/n△/r/nPBQ/r/n的面枳為/r/n8cm/r/n2/r/n?/r/n（2）/r/n出發(fā)幾秒后，線段/r/nPQ/r/n的長為/r/n4/r/nJ/r/n亍/r/ncm?/r/n（3/r/n）/r/nA/r/nPBQ/r/n的面積能否為/r/n10cm/r/n2/r/n?/r/n若能，求出時間；若不能，請說明理由./r/n【答案】⑴/r/n2/r/n或/r/n4/r/n秒/r/n；/r/n(2)4jJcm/r/n；/r/n⑶見解析./r/n【解析】/r/n【分析】/r/n由題意，可設/r/nP/r/n、/r/nQ/r/n經(jīng)過/r/nt/r/n秒，使/r/nZiPBQ/r/n的面積為/r/n8cm2,/r/n則/r/nPB=6-t,BQ=2t,/r/n根據(jù)三/r/n角形面枳的計算公式，/r/nSAPBO/r/n丄/r/nBPxBQ,/r/n列出表達式，解答出即可；/r/n2/r/n設經(jīng)過/r/nx/r/n秒后線段/r/nPQ/r/n的長為/r/n472/r/ncm,/r/n依題意得/r/nAP=x,BP=6-x,BQ=2x,/r/n利用勾股定理列方程求解；/r/n將/r/nAPBCl/r/n的面枳表示出來，根據(jù)/r/n△/r/n"2-4ac/r/n來判斷./r/n【詳解】/r/n設/r/nP/r/n，/r/nQ/r/n經(jīng)過/r/nt/r/n秒時，/r/n2XPBQ/r/n的面枳為/r/n8cm/r/n2/r/n,/r/n則/r/npB=6—t,BQ=2t,/r/nZB=90°,/r/n1/r/n—/r/n(6/r/n—/r/nt)x2t=8,/r/n解得/r/nti=2,t2=4,/r/n?/r/n??當/r/nP,Q/r/n經(jīng)過/r/n2/r/n或/r/n4/r/n秒時，/r/n2XPBQ/r/n的面枳為/r/n8cm/r/n2/r/n；/r/n設/r/nx/r/n秒后，/r/nPQ=4JTcm,/r/n由題意,得/r/n(6—x)/r/n2/r/n+4x/r/n2/r/n=32,/r/n2/r/n解得/r/nX1=—/r/n,/r/nX2=2,/r/n2_/r/n故經(jīng)過/r/ng/r/n秒或/r/n2/r/n秒后，線段/r/nPQ/r/n的長為/r/n4^2/r/ncm/r/n；/r/n⑶設經(jīng)過/r/ny/r/n秒，/r/nAPBCl/r/n的面積等于/r/n10cm/r/n2/r/n,/r/n1/r/nS/r/nA/r/npbq=/r/n—/r/nx(6—y)x2y=10,/r/n即/r/ny/r/n2/r/n-6y+10=0,/r/n/\=b/r/n2/r/n-4ac=36-4x10=-4<0,/r/n△/r/nPBQ/r/n的面積不會等于/r/n10cm/r/n2/r/n./r/n【點睛】/r/n本題考查了一元二次方程的應用，熟練的掌握一元二次方程的應用是本題解題的關(guān)鍵./r/n10./r/n我市茶葉專賣店銷售某品牌茶葉，其進價為每/r/nT/r/n?克/r/n240/r/n元，按每/r/nT/r/n?克/r/n400/r/n元出售，平均每周可售出/r/n200/r/nT-/r/n克/r/n,/r/n后來經(jīng)過市場調(diào)查發(fā)現(xiàn)，單價每降低/r/n10/r/n元，則平均每周的銷售量可增加/r/n40/r/nT/r/n?克，若該專賣店銷售這種品牌茶葉要想平均每周獲利/r/n41600/r/n元，請回答：/r/n（1）/r/n每/r/nT/r/n?克茶葉應降價多少元？/r/n（/r/n2）/r/n在平均每周獲利不變的情況下，為盡可能讓利于顧客，贏得市場，該店應按原售價的幾折出售？/r/n【答案】/r/n（/r/n1/r/n）/r/n每/r/nT/r/n?克茶葉應降價/r/n30/r/n元或/r/n80/r/n元；/r/n（2）/r/n該店應按原售價的/r/n8/r/n折出售./r/n【解析】/r/n【分析】/r/n（/r/n1/r/n）/r/n設每千克茶葉應降價/r/nx/r/n元，利用銷售量/r/nX/r/n每件利潤/r/n=416