初一数学化简求值答案
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1、已知A,B是方程x^2+2x-5=0的握逗两个实数根,
求(A^2+2AB+2A)(B^2+2AB+2B)的值.
由A,B是方程x^2+2x-5=0的段轮卖两个实数根得:
AB=-5,A+B=-2
A^2+2AB+2A)(B^2+2AB+2B)
=AB(A+2B+2)(B+2A+2)
=-5(-2+B+2)(-2+A+2)
=-5AB
=25
2、1/2(x+y+z)方+1/2(x-y-z)(x-y+z)-z(x+y),其中x-y=6,xy=21.要详细步骤
化简得:
1/2(x+y+z)方+1/2(x-y-z)(x-y+z)-z(x+y)=
1/2[(x+y)方+2z(x+y)+z方]+1/2[(x-y)方-z方]-z(x+y)=
1/2(x+y)方+1/2(x-y)方=x方+y方
由x-y=6,xy=21得,x方+y方=(x-y)方+2xy=78
3、a^2-ab+2b^2=3 求2ab-2a^2-4b^2-7的值
2ab-2a^2-4b^2-7
=2(ab-a^2-2b^2)-7
=-2(a^2-ab+2b^2)-7
=(-2)*3-7
=-6-7=-13
4、若A=2x^2+3xy-2x-3,B=-x^2+xy+2,且3A+6B的值与x无关,求y的桐耐值
3A+6B=6x^2+9xy-6x-9-6x^2+6xy+12
=15xy-6x+3
=x(15y-6)+3
5、9x+6x^2 -3(x-2/3x^2).其中x=-2
9x+6x² -3(x-2/3x²)
=9x+6x²-3x+2x²
=8x²+6x
=8×(-2)²+6×(-2)
=32-12
=20
求(A^2+2AB+2A)(B^2+2AB+2B)的值.
由A,B是方程x^2+2x-5=0的段轮卖两个实数根得:
AB=-5,A+B=-2
A^2+2AB+2A)(B^2+2AB+2B)
=AB(A+2B+2)(B+2A+2)
=-5(-2+B+2)(-2+A+2)
=-5AB
=25
2、1/2(x+y+z)方+1/2(x-y-z)(x-y+z)-z(x+y),其中x-y=6,xy=21.要详细步骤
化简得:
1/2(x+y+z)方+1/2(x-y-z)(x-y+z)-z(x+y)=
1/2[(x+y)方+2z(x+y)+z方]+1/2[(x-y)方-z方]-z(x+y)=
1/2(x+y)方+1/2(x-y)方=x方+y方
由x-y=6,xy=21得,x方+y方=(x-y)方+2xy=78
3、a^2-ab+2b^2=3 求2ab-2a^2-4b^2-7的值
2ab-2a^2-4b^2-7
=2(ab-a^2-2b^2)-7
=-2(a^2-ab+2b^2)-7
=(-2)*3-7
=-6-7=-13
4、若A=2x^2+3xy-2x-3,B=-x^2+xy+2,且3A+6B的值与x无关,求y的桐耐值
3A+6B=6x^2+9xy-6x-9-6x^2+6xy+12
=15xy-6x+3
=x(15y-6)+3
5、9x+6x^2 -3(x-2/3x^2).其中x=-2
9x+6x² -3(x-2/3x²)
=9x+6x²-3x+2x²
=8x²+6x
=8×(-2)²+6×(-2)
=32-12
=20
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