2021年廣東梅州市中考數(shù)學(xué)試卷及解析

年州市初中畢生學(xué)業(yè)考試

說(shuō)明本卷頁(yè)，小題滿(mǎn)分．試時(shí)分．注事：1．答題前，考生務(wù)必在答題卡用黑色字跡的鋼筆或簽字筆填寫(xiě)準(zhǔn)考證號(hào)、姓名、試室號(hào)、座位號(hào)，再用鉛把試室號(hào)、座位號(hào)的對(duì)應(yīng)數(shù)字涂黑．．選擇題每小題選出答案后，用2B鉛筆把答題卡上對(duì)應(yīng)答案選項(xiàng)涂黑，如需改動(dòng)，用橡皮擦干凈后，再重新選涂其他答案，答案不能答在試卷上．．非選擇題必須用黑色字跡鋼筆或簽字筆作答，答案必須寫(xiě)在答題卡各題目指定區(qū)域內(nèi)相應(yīng)位置上；如需改動(dòng)，先劃掉原來(lái)的答案，然后再寫(xiě)上新的答案；不準(zhǔn)使用鉛筆和涂改液．不按以上要求作答的答案無(wú)效．．考生必須保持答題卡的整潔．考試結(jié)束后，將試卷和答題卡一并交回．．本試卷不用裝訂，考完后統(tǒng)一交縣招生辦（中招辦）封存．參公：次函數(shù)

yax

的對(duì)稱(chēng)軸是直線(xiàn)

=

b2a

，頂點(diǎn)坐標(biāo)是（

b2，24一、選擇題：每小題分，共分．每小題給出四個(gè)案，其中只有一個(gè)是正確的．．下各組數(shù)中，互為相反數(shù)的是（）A2和

11B．-和－22

．-2-D．2和

12．如圖1的何體的俯視圖是（）圖

A

B．

C．

D．．列事件中，必然事件是（）Ａ．任意擲一枚均勻的硬幣，正面朝上Ｂ．黑暗中從一串不同的鑰匙中隨意摸出一把，用它打開(kāi)了門(mén)Ｃ．通常情況下，水往低處流Ｄ．上學(xué)的路上一定能遇到同班同學(xué)．如圖2所，圓O的AB垂平半徑．則四邊形（）A是正形是長(zhǎng)形C．是形D．上答案都不對(duì)

圖．一列貨運(yùn)火車(chē)從梅州站出發(fā)，勻加速行駛一段時(shí)間后開(kāi)始勻速行駛，過(guò)了一段時(shí)，火車(chē)到達(dá)下一個(gè)車(chē)站停下裝完貨后火車(chē)又勻加速行駛一時(shí)間后再次開(kāi)始勻速行駛，那么可以近似地刻畫(huà)出火車(chē)在這段時(shí)間內(nèi)的速度變化情況的是（）

二、填空題：每小題分，共分．．算：

1()2

．7．

如圖，要測(cè)量A、B兩間距離，在O點(diǎn)樁，取OA的點(diǎn)，

圖OB的點(diǎn)D，得CD=30米則AB米．．如，點(diǎn)到∠兩的距離相等，若=30°則∠度．圖．如，AB是O的徑，=70°，則∠A度．

·10函數(shù)

y

1x

的自變量

的取值范圍是_．．某校年級(jí)二班名生的年齡情況如下表所示：年齡人數(shù)

歲

歲

歲

歲

圖5則該班學(xué)生年齡的中位數(shù)________從該班隨機(jī)地抽取一人，抽到學(xué)生的年齡恰好是歲的概率等于_______．已知線(xiàn)mx與曲線(xiàn)y

kx

的一個(gè)交點(diǎn)的標(biāo)為（-1-2；

=____；們的另一個(gè)交點(diǎn)坐標(biāo)．13觀察下列等式：①-1=4×2②

-2

=4×3③

-3

=4×4④（）

-（）=（）（……則第個(gè)等式_．第n個(gè)式為_(kāi)____是整數(shù)）三解下各：題10小，81分．解應(yīng)出字明推過(guò)或算驟．題分7分如圖，已知

△

：

1111（1的等于．（2）將△ABC向平移個(gè)單位到______；

△

，則點(diǎn)的對(duì)應(yīng)點(diǎn)（3若將△ABC繞按時(shí)針?lè)较蛐D(zhuǎn)90后得到

AB

，則A點(diǎn)應(yīng)點(diǎn)

的坐標(biāo)是_．．題分7分右圖是我國(guó)運(yùn)動(dòng)員在年、年年三屆奧運(yùn)會(huì)上獲得獎(jiǎng)牌數(shù)的統(tǒng)計(jì)圖．請(qǐng)你根據(jù)統(tǒng)計(jì)圖提供的信息，回答下列問(wèn)題：（1在1996年年、這三屆奧運(yùn)會(huì)上，我國(guó)運(yùn)動(dòng)員獲得獎(jiǎng)牌總數(shù)最多的一屆奧運(yùn)會(huì)________年．（2在1996年、2000年這三屆奧運(yùn)會(huì)上運(yùn)員共獲獎(jiǎng)_枚．（3）據(jù)以上統(tǒng)計(jì)，預(yù)測(cè)我國(guó)運(yùn)動(dòng)員在年運(yùn)會(huì)上能獲得的獎(jiǎng)牌總數(shù)大約為枚．．題分7分解分式方程：

．．題分7分如圖所，在長(zhǎng)和寬分別是

a

、

的矩形紙片的四個(gè)角都剪去一個(gè)邊長(zhǎng)為的方形．用，b表示紙片剩部分的面積；當(dāng)=6，b且剪去部分的面積等于剩余部分的面積時(shí)，求正方形的邊長(zhǎng)．

圖

EF

．題分8分如圖，四邊形ABCD是平行四邊形．O是角線(xiàn)AC的中點(diǎn)，過(guò)點(diǎn)O的線(xiàn)分別交ABDC于點(diǎn)E、，與、延長(zhǎng)線(xiàn)分別交于點(diǎn)G、．（1寫(xiě)出圖中不全等的兩個(gè)相似三角形（不要求證明（2除=，AD=，OAOC這三對(duì)相等的線(xiàn)段外，圖中還有多相等的線(xiàn)段，請(qǐng)選出其中一對(duì)加以證明．圖．題分8分如圖所示，直線(xiàn)與坐標(biāo)軸的交點(diǎn)坐標(biāo)分別是（-，0，O是標(biāo)系原點(diǎn)．（1求直線(xiàn)L所應(yīng)的函數(shù)的達(dá)式；（2若以O(shè)為心，半徑為的圓與直線(xiàn)L切，求的．．題分8分已知關(guān)于的一元二次方程x-．…①若x是方程①的一個(gè)根，求的和方程①的另一根；對(duì)任意實(shí)數(shù)，判斷方程①的根的情況，并明理由．．題分8分

如圖10所，E是方形ABCD的AB的動(dòng)點(diǎn)，EF⊥DE交BC于F（1求證：∽；（2設(shè)方形的邊長(zhǎng)為，AE=

，BF

y

．當(dāng)

取什么值時(shí)，

y

有最大求出這個(gè)最大值．．題分10．“一方有難，八方支援．在抗“．汶川特大地震災(zāi)害中，某市組織輛汽車(chē)裝運(yùn)食品、藥品、生活用品三種救災(zāi)物資共噸到災(zāi)民安置點(diǎn)．按計(jì)劃20輛汽車(chē)都要裝運(yùn)每輛汽車(chē)只能裝同一種救災(zāi)物資且必須裝滿(mǎn)據(jù)表提供的信息，解答下列問(wèn)題：

物資種類(lèi)食藥生用品每輛汽車(chē)運(yùn)載量（噸）64每噸所需運(yùn)費(fèi)（/噸）（1設(shè)裝運(yùn)食品的車(chē)輛數(shù)為

，裝運(yùn)藥品的車(chē)輛數(shù)為

y

．求

y

與

的函數(shù)關(guān)系式；（2如裝運(yùn)食品的車(chē)輛數(shù)不少于5輛裝藥品的車(chē)輛數(shù)不少于4輛，那么車(chē)輛的安排有幾種方案寫(xiě)出每種安排方案；（3在）的條件下，若要求總運(yùn)費(fèi)最少，應(yīng)采用哪種安排方并求出最少總費(fèi)．．題分11分如圖11所示，在梯形ABCD中，已知∥，AD⊥DBAD=CB=4以所直線(xiàn)為

軸，過(guò)D且垂直于AB的線(xiàn)為y軸建立平面直角坐標(biāo)系．（1求的數(shù)及、D、C三的坐標(biāo)（2求ADC三點(diǎn)的拋物線(xiàn)的解析式及其對(duì)稱(chēng)軸．（3P是拋物線(xiàn)的對(duì)稱(chēng)軸L上點(diǎn)使PDB為等腰三角形的點(diǎn)P有幾個(gè)（必求點(diǎn)的標(biāo)，只需說(shuō)明理由）

111111參考答案與評(píng)分意見(jiàn)一、選擇題：每小題分，共15分．每小題出四個(gè)答案，其中只有一個(gè)是正確的．．C；2．A；3C；4；5B．二、填空題：每小題分，共分．．2．

．

．60

．

．x>1

11歲12m=2k=2（12分1362

-4

（1分22

=4×（）（分三解答下列各題本題有10小題共81分解答應(yīng)寫(xiě)出文字說(shuō)推過(guò)程或演算步驟．．題分7分如圖，已知ABC：（1的等于_．（2）若將ABC向右平移2個(gè)單位得到△

，則

A

點(diǎn)的對(duì)應(yīng)點(diǎn)

A

的坐標(biāo)______；（3將△繞點(diǎn)按順時(shí)針?lè)较蛐D(zhuǎn)

0后得到

C

，則A點(diǎn)對(duì)應(yīng)點(diǎn)A的坐是．解110．···································分（2，····················································································

（3，·························································································7分．題分7分右圖是我國(guó)運(yùn)動(dòng)員在年、2000年年屆奧運(yùn)會(huì)上獲得獎(jiǎng)牌數(shù)的統(tǒng)計(jì)圖．請(qǐng)你根據(jù)統(tǒng)計(jì)圖提供的信息，回答下列問(wèn)題：（1在年、、2004這三屆奧運(yùn)會(huì)上，我國(guó)運(yùn)動(dòng)員獲得獎(jiǎng)牌總數(shù)最多的一屆奧運(yùn)會(huì)_年（2年2000年年這三屆奧運(yùn)會(huì)上，我國(guó)運(yùn)動(dòng)員共獲獎(jiǎng)牌枚（3根以上統(tǒng)計(jì)預(yù)我國(guó)運(yùn)動(dòng)員在年運(yùn)會(huì)上能獲得的獎(jiǎng)牌總數(shù)大約為_(kāi)________．解1年；··················2（2172；························································································4分（3．························································································7分

，，（注：預(yù)數(shù)字在64～83的得3分，84得2，～103得分大于或小于的得0分）．題分7分解分式方程：

．解方程兩邊同乘以

-2，得1-

+2（

）=1········································2分即x+2-4=1，··············································································解得x=4·······················································································經(jīng)檢驗(yàn)，是方程的根．····························································．本滿(mǎn)分7分如圖所示長(zhǎng)和寬分別是

a

b

的矩形紙片的四個(gè)角都剪去一個(gè)邊長(zhǎng)為

的正方形．··············································································································用a，b，表示紙片剩余部分的面積；當(dāng)=6，b且剪去部分的面積等于剩余部分的面積時(shí)，求正方形的邊長(zhǎng)．解1－；·····································2分（2依題意有：－4將=6，b，代入上式，得

2分2=3，·········6分解得

x舍去)

．·················7分即正方形的邊長(zhǎng)為

．

圖．題分8分如圖，四邊形

是平行四邊形．O是對(duì)角線(xiàn)

的中點(diǎn)，過(guò)點(diǎn)

O

的直線(xiàn)

EF

分別交、于

E

、

F

，與CBAD延長(zhǎng)線(xiàn)分別交于點(diǎn)、．（1寫(xiě)出圖中不全等的兩個(gè)相似三角形（不要求證明（2除=，AD=，OAOC這三對(duì)相等的線(xiàn)段外，圖中還有多相等的線(xiàn)段，請(qǐng)選出其中一對(duì)加以證明．解1AEH與DFH·······························（或

與

，或

與

，或

DFH與

）（2）=．·················································3分證：∵四邊形是平行四邊形，AB∥，·····························EAOAOE

，··································，···································6分∴

≌

COF

，·······························OE．············································8分（意此有種法選外對(duì)，此準(zhǔn)

圖分．題分8分如圖所示，直線(xiàn)L與坐標(biāo)軸的交點(diǎn)坐標(biāo)分別是A（-3，0（04是標(biāo)原點(diǎn)．

1212（1求直線(xiàn)L所應(yīng)的函數(shù)的達(dá)式；（2若以O(shè)為心，半徑為的與直線(xiàn)L相切，求R的．解1設(shè)所求為

y

=

kx

+

．·································································1分將A-3，0（0）的坐標(biāo)代入，得

······································2分解得

=4，

=

43

．·································3所求為

y

=

43

．··································分（2設(shè)切點(diǎn)為P，OP則⊥AB，=Rt中，OA，OB，得AB=5，························································6分因?yàn)椋?/p>

122

得·································································=

125

．·································································································8分（本題可用相似三角形求解）．題分8分已知關(guān)于的一元二次方程x2-2=0①．若

是這個(gè)方程的一個(gè)根，求

的值和方程①的另一根；對(duì)任意的實(shí)數(shù)

，判斷方程①的根的情況，并說(shuō)明理由．解1

是方程①的一個(gè)根，所以1+

-2=0，····································1分解得

=1．·····················································································方程為2，解得，x，x．所以方程的另一根為．····································································4分（2

=m2+8，····································································5分因?yàn)閷?duì)于任意實(shí)數(shù)

，

20，·····························································6分所以

2

，····················································································所以對(duì)于任意的實(shí)數(shù)

，方程①有兩個(gè)不相等的實(shí)數(shù)根．·························8分．題分8分如圖10所，E是方形ABCD的AB的動(dòng)點(diǎn)，EF⊥DE交BC于點(diǎn)．（1求證：

∽

；（2設(shè)正方形的邊長(zhǎng)為4AE=，BFy當(dāng)x取么值時(shí)，y最大?并出這個(gè)最大值．證：（1因?yàn)槭钦叫?，所以∠?

，

所以∠ADE+∠DEA=

，·······················1分又EF⊥，所以AED+=，························································2所以∠ADE=∠FEB，················································································3分所以

ADE∽

BEF···············································································（2解：由（）

ADE∽

，=4，BE

，得4

，得························································································5分y

=

14

11()[2)(x4

，·································6分所以當(dāng)x=2時(shí)有大值，···································································7分y

的最大值為1····················································································．題分10．“一方有難，八方支援．在抗“．汶川特大地震災(zāi)害中，某市組織輛汽車(chē)裝運(yùn)食品、藥品、生活用品三種救災(zāi)物資共噸到災(zāi)民安置點(diǎn)．按計(jì)劃20輛汽車(chē)都要裝運(yùn)每輛汽車(chē)只能裝同一種救災(zāi)物資且必須裝滿(mǎn)據(jù)表提供的信息，解答下列問(wèn)題：

物資種類(lèi)食藥品生活用品每輛汽車(chē)運(yùn)載量（噸）654每噸所需運(yùn)費(fèi)（/噸）120100（1設(shè)裝運(yùn)食品的車(chē)輛數(shù)為

，裝運(yùn)藥品的車(chē)輛數(shù)為

y

．求

y

與

的函數(shù)關(guān)系式；（2如裝運(yùn)食品的車(chē)輛數(shù)不少于5輛裝藥品的車(chē)輛數(shù)不少于4輛，那么車(chē)輛的安排有幾種方案寫(xiě)出每種安排方案；（3在）的條件下，若要求總運(yùn)費(fèi)最少，應(yīng)采用哪種安排方并求出最少總費(fèi)．解1根據(jù)題意，裝運(yùn)食品的車(chē)輛數(shù)為裝運(yùn)藥品的車(chē)輛數(shù)為，那么裝運(yùn)生活用品的車(chē)輛數(shù)為

(

．··················································1分則有

4(20)100

，····························································2分整理得，

20x

．··········································································（2由（）知，裝運(yùn)食品，藥品，生活品三種物資的車(chē)輛數(shù)分別為

，

，由題意，得

x，20x≥

··········································································4分解這個(gè)不等式組，得

·······························································。5分因?yàn)?/p>

為整數(shù)，所以

的值為5，78所以安排方案有種：····················方案一：裝運(yùn)食品輛、藥品10輛生用品5輛·································。5分方案二：裝運(yùn)食品輛、藥品輛生活用品6輛·······································6分方案三：裝運(yùn)食品輛、藥品輛生活用品7輛··································6。5分方案四：裝運(yùn)食品輛、藥品輛生活用品8輛······································（3）設(shè)總運(yùn)費(fèi)為W（元?jiǎng)t

W

=6

×120+5

）×160+4

．·························8

最小11223234最小1122323445因?yàn)?，以W值隨x的大而減?。ぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁ?。5分要使總運(yùn)費(fèi)最少，需W小，則．··················································9故選方案4···············································································。5分

=16000-480×8=12160元．·······················································10分最少總運(yùn)費(fèi)為12160元．題分11分如圖11所，在梯形中已知CD，ADDB，AD=DC=CBAB．以AB所直線(xiàn)為軸，過(guò)且直于的直線(xiàn)為y軸立平面角坐標(biāo)系．（1求的數(shù)及、D、C三的坐標(biāo)（2求過(guò)A、、C三的拋物線(xiàn)的解析式及其對(duì)稱(chēng)軸L．（3P是拋物線(xiàn)的對(duì)稱(chēng)軸L上點(diǎn)么必求點(diǎn)P的坐標(biāo)，只需說(shuō)明理由）

為等腰三角形的點(diǎn)有幾不解（）

DCAB，=DC=，

∠CDB=∠，···············0。5∠DAB=∠CBA∠=2∠，∠∠DBA，

∠DAB=60，·······。5分∠DBA，，DC=AD=2

·········2分t

AOD=1，OD=

3

，·····················2。分

A（-，00

3

（2，

3

（2據(jù)拋物線(xiàn)和等腰梯形的對(duì)稱(chēng)性知滿(mǎn)條件的拋物線(xiàn)必過(guò)點(diǎn)（－1，（3故可設(shè)所求為

y=a（x-3）·····················································將點(diǎn)（，

3

）的坐標(biāo)代入上式得，

a

=

33

．所求拋物線(xiàn)的解析式為

33

(xx

····································其對(duì)稱(chēng)軸L為線(xiàn)

=1．··········································································（3為等腰三角形，有以下三種情況：①因直線(xiàn)L與DB不平行DB的直平分線(xiàn)與L有一個(gè)交點(diǎn)，PD=，

PDB等腰三角形；···········································································②因?yàn)橐訢為圓心，DB為徑的圓與直線(xiàn)L有個(gè)點(diǎn)、，DB=，DP，

PDB，

PDB等腰三角形；③與②同理L上有兩個(gè)點(diǎn)P、P，得=BP，BDBP．···················10分由于以上各點(diǎn)互不重合，所以在直線(xiàn)上使

為等腰三角形的點(diǎn)P5個(gè)．

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