一、计算：1.y[y-3(x-z)]+y[3z-(y-3x)]2.(2x+3)(x-1)-(3x+2)(2x-3)二、先化简再求值：（5x+7)(2x-3)-(10x+1)(x-2),其中x=-½三、解方程：4（x-1）=5(2x+1)+6(1+x)四、已知：x=-5,y=1/5,求x²·x^2n·（y^n+1）²的值

Question

一、计算：1.y[y-3(x-z)]+y[3z-(y-3x)]
2.(2x+3)(x-1)-(3x+2)(2x-3)
二、先化简再求值：（5x+7)(2x-3)-(10x+1)(x-2),其中x=-½
三、解方程：4（x-1）=5(2x+1)+6(1+x)
四、已知：x=-5,y=1/5,求x²·x^2n·（y^n+1）²的值（n为正整数）

Answer 1

1、
=y(y-3z+3z)+y(3z-y+3x)
=y(y-3z+3x+3z-y+3x)
=y*6x
=6xy
2、
原式=(10x^2-x-21)-(10x^2-19x-2)
=10x^2-x-21-10x^2+19x+2
=18x-19
=18*(-1/2)-19
=-28
3、
===> 4x-4=10x+5+6+6x
===> 4x-4=16x+11
===> 12x=-15
===> x=-5/4
4、
x^2*2^2n*(y^n+1)^2
=x^(2n+2)*y^(2n+2)
=[x*y]^(2n+2)
=[-5*(1/5)]^(2n+2)
=(-1)^(2n+2)
=1