Question

2x(x+1)+(2x+1)(2x-1)(x-1)的平方?

Answer 1

1.
(1/x+1)-(x+3/X^2-1)×(x^2-2x+1)/(x^2+4x+3)
=1/(x+1)-{(x+3)/[(x+1)(x-1)]}*{(x-1)^2/[(x+1)(x+3)]}
=1/(x+1)-(x-1)/[(x+1)^2]
=[(x+1)-(x-1)]/[(x+1)^2]
=2/(x^2+2x+1);
2.
因为(2x-1)^2+(x-2y)^2=0,
又因为
(2x-1)^2≥0,(x-2y)^2≥0,
所以(2x-1)^2=0,(x-2y)^2=0,
所以2x-1=0,x-2y=0,
所以x=1/2,y=1/4,
代入[1-{1/xy+x+y+1}(x^2-1/x1+x^2+x)]÷{(1/x)-(1/y)}即可;(因为式子有点乱,所以没法算)
3.把x=2,y=1代入是方程组ax-3y=1,x+by=5得
2a-3=1,2+b=5,
所以a=2,b=3
所以{(a/a-b)-(a^2/a^2-2ab+b^2)}÷{(a/a-b)+(a^2/b^2-a^2)
=[a/(a-b)-a^2/(a-b)^2]/[a/(a-b)+a^2/(b-a)(b+a)]
={[a(a-b)-a^2]/(a-b)^2}/{[a(a+b)-a^2]/(a-b)(a+b)}
=[-ab/(a-b)^2]*(a-b)(a+b)/ab
=-(a+b)/(a-b)
=-5/(-1)
=5;
4.因为(1/a)+(1/b)=(a+b)/(ab)=(5/a+b),
所以(a+b)^2=5ab,
所以a^2+b^2+2ab=5ab,
所以a^2+b^2=3ab,
所以(b/a)+(a/b)
=(a^2+b^2)/(ab)
=3ab/(ab)
=3.