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七年级数学题目,求解答要过程~好的给分值~计算:-8m²-【4m-2m²-(3m-2m²-7)】3x²-【7x-(4x-3)-2X²】1/2(a²-b)+1/3(a-b²)+1/6(a²+b²)5ab²-3【2a²-2(a²b-2a
更新时间:2024-03-29 22:57:47
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问题描述:

七年级数学题目,求解答要过程~好的给分值~

计算:

-8m²-【4m-2m²-(3m-2m²-7)】

3x²-【7x-(4x-3)-2X²】

1/2(a²-b)+1/3(a-b²)+1/6(a²+b²)

5ab²-3【2a²-2(a²b-2ab²)】

先化解,再求值

1/4(-4x²+2x-8)-(1/2x-1),其中X=1/2

(4a²-3a)-(2a²+a-1)+(2-a²+4a),其中a=-2

a-2(2a+b)+3(a-b),其中a=-2

4x²-(2x²+x-1)+(2-x²-3x),其中X=-1/2

5(3a²b-ab²)-(ab²+3a²b),其中a=1/2b=-1

已知:A=4a²+5b,B=-3a²-2b,求2A-B的值,其中a=-2,b=1

已知:(x+1)²+|y-1|=0,求2(xy-5xy²)-(3xy²-xy)的值

已知:A=x³+2x²y+2y³-1,B=3+y³+2x²y+2x³若A+B+C=0,求C

要过程,好的,给分值,多少都无所谓~

贺毅岳回答:
  计算:   -8m²-[4m-2m²-(3m-2m²-7)]   =-8m²-(4m-2m²-3m+2m²+7)   =-8m²-4m+2m²+3m-2m²-7   =(-8m²+2m²-2m²)-4m-7   =-8m²-4m-7   3x²-[7x-(4x-3)-2x²]   =3x²-(7x-4x+3-2x²)   =3x²-7x+4x-3+2x²   =(3x²+2x²)+(-7x+4x)-3   =5x²-3x-3   1/2(a²-b)+1/3(a-b²)+1/6(a²+b²)   =1/2a²-1/2b+1/3a-1/3b²+1/6a²+1/6b²   =(1/2a²+1/6a²)-1/2b+1/3a+(-1/3b²+1/6b²)   =2/3a²-1/2b+1/3a-1/6b²   5ab²-3[2a²-2(a²b-2ab²)]   =5ab²-3(2a²-2a²b+4ab²)   =5ab²-6a²+6a²b-12ab²   =(5ab²-12ab²)-6a²+6a²b   =-7ab²-6a²+6a²b   先化简,再求值   1/4(-4x²+2x-8)-(1/2x-1),其中x=1/2   =-x²+1/2x-2-1/2x+1   =-x²-1   =-(1/2)²-1   =-1/4-1   =-5/4   (4a²-3a)-(2a²+a-1)+(2-a²+4a),其中a=-2   =4a²-3a-2a²-a+1+2-a²+4a   =(4a²-2a²-a²)+(-3a-a+4a)+(1+2)   =a²+3   =(-2)²+3   =4+3   =7   a-2(2a+b)+3(a-b),其中a=-2   =a-4a-2b+3a-3b   =(a-4a+3a)-(2b+3b)   =-5b   =-5×(-2)   =10   4x²-(2x²+x-1)+(2-x²-3x),其中x=-1/2   =4x²-2x²-x+1+2-x²-3x   =(4x²-2x²-x²)-(x+3x)+(1+2)   =x²-4x+3   =(-1/2)²-4×(-1/2)+3   =1/4+2+3   =21/4   5(3a²b-ab²)-(ab²+3a²b)其中a=1/2,b=-1   =15a²b-5ab²-ab²-3a²b   =(15a²b-3a²b)-(5ab²+ab²)   =12a²b-6ab²   =12×(1/2)²×(-1)-6×1/2×(-1)²   =12×1/4×(-1)-6×1/2×1   =-3-3   =-6   已知A=4a²+5b,B=-3a²-2b,求2A-B的值,其中a=-2,b=1   2A-B   =2(4a²+5b)-(-3a²-2b)   =8a²+10b+3a²+2b   =11a²+12b   =11×(-2)²+12×1   =11×4+12   =56   已知(x+1)²+|y-1|=0,求2(xy-5xy²)-(3xy²-xy)的值   ∵(x+1)²+|y-1|=0   ∴x+1=0且y-1=0   ∴x=-1,y=1   2(xy-5xy²)-(3xy²-xy)   =2xy-10xy²-3xy²+xy   =3xy-13xy²   =3×(-1)×1-13×(-1)×1²   =-3+13   =10   已知A=x³+2x²y+2y³-1,B=3+y³+2x²y+2x³,若A+B+C=0,求C   A+B+C=0   C=-(A+B)   C=-(x³+2x²y+2y³-1+3+y³+2x²y+2x³)   =-x³-2x²y-2y³+1-3-y³-2x²y-2x³   =-(x³+2x³)-(2x²y+2x²y)-(2y³+y³)+(1-3)   =-3x³-4x²y-3y³-2
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