Đáp án+Giải thích các bước giải:

 `1)`

`\frac{(2-a)^{2}}{2ab}.\frac{b}{2-a}+\frac{1}{2}`

`(đk:a\ne0;b\ne0;a\ne2)`

`=\frac{(2-a)^{2}.b}{2ab.(2-a)}+1/2`

`=\frac{(2-a).b}{2ab}+1/2`

`=\frac{2-a}{2a}+1/2`

`=\frac{2-a}{2a}+\frac{a}{2a}`

`=\frac{2-a+a}{2a}`

`=\frac{2}{2a}`

`=1/a`

`2)`

`\frac{5a}{3a}-\frac{5x-1}{(x+1)^{2}}.\frac{3x+3}{9}`

`(đk:a\ne0;x\ne-1)`

`=\frac{5}{3}-\frac{(5x-1).3.(x+1)}{(x+1)^{2}.3.3}`

`=5/3-\frac{(5x-1).3}{(x+1).3.3}`

`=5/3-\frac{5x-1}{3.(x+1)}`

`=\frac{5.(x+1)}{3.(x+1)}-\frac{5x-1}{3.(x+1)}`

`=\frac{5x+5-5x+1}{3.(x+1)}`

`=\frac{6}{3.(x+1)}`

`=\frac{2}{x+1}`