七年级的计算题和答案100道
计算：-9÷27/4×(-7)+4-4×(+4/3)
解：原式= 9×4/27×7+4-4×4/3
=4/3×7-4×4/3+4
=(7-4)×4/3+4
=4+4
=8
22/7×(22/7-22/3)×7/22÷22/21
解：原式=22/7×(22/7-22/3)×7/22×21/22
=22/7×7/22×(22/7-22/3)×21/22
=22/7×21/22-22/3×21/22
=3-7
=-4
(-1)^2011÷1/15+(-1/5)×(-5)^3
解：原式=(-1)÷1/15+(-1/5)×(-125)
=(-1)×15+1/5×125
= -15+25
=10
-12^56×[(-1)^1999-22/3×(9/22-6/11)]
解：原式=-12^56×[(-1)^1999-(22/3×9/22-22/3×6/11)]
=-12^56×[(-1)^1999-(3-4)]
=-12^56×(-1+1)
=0
1、3ab-5ab2+3a2b-4ab+2ab2-3ab
解：原式=(3ab-3ab)-4ab+(-5ab2+2ab2)+3a2b
=-4ab+(-5+2)ab2+3a2b
=-4ab+(-3)ab2+3a2b
=-4ab-3ab2+3a2b
2、2(2x-y)-3(2x-y)2+2x-y-5(2x-y)2
解：原式=2(2x-y)-3(2x-y)2+(2x-y)-5(2x-y)2
=[2(2x-y)+(2x-y)]+[-3(2x-y)2-5(2x-y)2]
=(2+1)(2x-y)+(-3-5)(2x-y)2
=3(2x-y)-8(2x-y)2
1、化简
(1) (2a+b)-(a-2b)
解：原式=2a+b-a+2b
=2a-a+b+2b
=(2-1)a+(1+2)b
=a+3b
(2) -3(a-2b)-2(2a-b)
解：原式=-3a+(-3)(-2b)+(-2)(2a)+(-2)(-b)
=-3a+6b-4a+2b
=(-3-4)a+(6+2)b
=-7a+8b
(3) -1/4(-4a+4b/3)-1/3(3a-2b)
解：原式=-[1/4•(-4a)+1/4•4b/3]-(1/3•3a-1/3•2b)
=-(-a+b/3)-(a-2b/3)
=a-b/3-a+2b/3
=(a-a)+(-1/3+2/3)b
=b/3
(4) 3x2y-2[3x2y-3(3xyz+2x2z)+x2z]-3xyz
解：原式=3x2y-2[3x2y-(9xyz+6x2z)+x2z]-3xyz
=3x2y-2(3x2y-9xyz-6x2z+x2z)-3xyz
=3x2y-2[3x2y-9xyz+(-6x2z+x2z)]-3xyz
=3x2y-2(3x2y-9xyz-5x2z)-3xyz
=3x2y-(6x2y-18xyz-10x2z)-3xyz
=3x2y-6x2y+18xyz+10x2z-3xyz
=-3 x2y+15xyz+10x2z
2、若A=3x2y+4xy2，B=2xy2-x2y，求-2A-3B
解：-2A-3B=-2(3x2y+4xy2)-3(2xy2-x2y)
=-(6x2y+8xy2)-(6xy2-3x2y)
=-6x2y-8xy2-6xy2+3x2y
=(-6x2y+3x2y)+(-8xy2-6xy2)
=-3x2y-14xy2
3、先化简再求值：若｜x+1｜+(y-1/3)2=0，求8xy-2{xy+3[3x4y-2(x4y+2xy)+5xy]-2x4y}的值
解：化简：原式=8xy-2{xy+3[3x4y-(2x4y+4xy)+5xy]-2x4y}
=8xy-2[xy+3(3x4y-2x4y-4xy+5xy)-2x4y]
=8xy-2[xy+3(x4y+xy)-2x4y]
=8xy-2(xy+3x4y+3xy-2x4y)
=8xy-2(4xy+x4y)
=8xy-8xy-2x4y
=-2x4y
由｜x+1｜+(y-1/3)2=0，得
x=-1，y=1/3
将x=-1，y=1/3代入-2x4y，得
-2×(-1)4×1/3=-2/3
所以，原式=-2/3
4、有一道题：求两个多项式的差。小刚错把2a2b-3ab2当作被减数，解得差为-a2b，求正确的差。
解：设A=2a2b-3ab2，C=-a2b，另外一个多项式为B
小刚列的式子为：A-B=C，所以B=A-C
B=(2a2b-3ab2)-(-a2b)
=2a2b-3ab2+a2b
=3a2b-3ab2
正确的差应为：
B-A=(3a2b-3ab2)-(2a2b-3ab2)
=3a2b-3ab2-2a2b+3ab2
= a2b
(别解：因为a-b与b-a是相反数，小刚将被减数和减数搞反了，所以得到的差是正确的差的相反数，因此，正确的差为-(-a2b)= a2b)
标签：七年级,计算题
