2018-2019河南省信陽市平橋區(qū)明港鎮(zhèn)七年級(jí)數(shù)學(xué)期末答案

七年級(jí)數(shù)學(xué)參考答案選擇題:（每小題3分，共30分）-5、ACBBD6--10、DCDBA二、填空題：（每小題3分，共15分）11、-2；12、；13、4；14、甲；15、三、解答題：（75分）16.（1）原式=+﹣3.......................................................2分=﹣1................................................................4分（2）原式=﹣1+24÷4﹣9×..............................................2分=﹣1+6﹣1.......................................................3分=4．..................................................................4分17.解：（1）根據(jù)題意得：（a2﹣4ab+4b2）+（a2﹣4b2）............2分=a2﹣4ab+4b2+a2﹣4b2.....................4分=2a2﹣4ab；...................................5分當(dāng)a=﹣1，b=時(shí).................................................6分原式=2+2=4．............................................8分18.（1）解:去括號(hào)得：...........................................................2分移項(xiàng)得：..................................................................3分合并同類項(xiàng)得：.....................................................................4分把系數(shù)化為1得：

........................................................5分

（2）解:去分母.................................................1分去括號(hào)得：.....................................................2分移項(xiàng)得：.......................................................3分合并同類項(xiàng)得：...........................................................4分把系數(shù)化為1得：...........................................................5分19.解：∵AB=4cm，BC=2AB∴BC=8cm..............................................................2分∴AC=AB+BC=4+8=12cm.............................................4分∵M(jìn)是線段AC中點(diǎn)∴MC=AM=AC=6cm.................................................6分∴BM=AM﹣AB=6﹣4=2cm........................................8分20.解：原式=3a2﹣4ab﹣2a2﹣4ab......................................3分=a2﹣8ab.........................................................5分∵|a+1|+（2﹣b）2=0．∴a+1=0，2﹣b=0，即a=﹣1，b=2...............................................8分當(dāng)a=﹣1，b=2時(shí)原式=（﹣1）2﹣8×（﹣1）×2=17．........................................10分21.解：（1）∵∠AOC=40°∴∠COB=180°﹣∠AOC=180°﹣40°=140°.....................2分∵OE平分∠COB∴∠COE=∠COB=70°................................................3分又∵∠COD=90°∴∠EOD=∠COD﹣∠COE=20°....................................4分（2）∠DOE=..........................................................7分（3）∠DOE=∠AOC.................................................10分22.解：（1）設(shè)這天該經(jīng)營戶批發(fā)了番茄x千克，豆角（45﹣x）千克..................................................1分由題意得：2.4x+3.2（45﹣x）=120.........................................5分解得，x=30∴45﹣x=15...........................................................7分即這天該經(jīng)營戶批發(fā)了番茄30千克，豆角15千克..............8分（2）由題意可得：（3.6﹣2.4）×30+（5.0﹣3.2）×15=63（元），即當(dāng)天賣完這些番茄和豆角能盈利63元．.........................10分23.解：（1）②設(shè)∠BOC=α，則∠BOD=3α，若射線OD在∠AOB外部，如圖2：則∠COD=∠BOD﹣∠BOC=2α...................1分∵∠AOD=∠AOC∴∠AOD=∠COD=...............................2分∴∠AOB=∠BOD﹣∠AOD=3α﹣==70°...........3分∴α=30°即：∠BOC=30°.....................4分（2）當(dāng)射線OC在∠AOB外部時(shí)，射線OD的位置也只有兩種可能.①若射線OD在∠AOB內(nèi)部，如圖3所示設(shè)∠BOC=α，則∠COD=4α∵∠BOC＜45°，∠AOD=∠AOC∴∠AOB=∠BOD+∠AOD=3α+4α=7α=70°∴α=10°即：∠BOC=10°.........................7分②若射線OD在∠AOB外部，如圖4所示則∠COD=∠BOC+∠BOD=