Đáp án:
a, $\left \{ {{x\neq0} \atop {(x\neq1}} \right.$
b, B = $\frac{x^2}{x-1}$
c, x = $\frac{1}{2}$
Giải thích các bước giải:
a, ĐKXĐ: $\left \{ {{x\neq0} \atop {(x-1)\neq0}} \right.$ ⇔ $\left \{ {{x\neq0} \atop {(x\neq1}} \right.$
b, B = $\frac{x^2+x}{x^2-2x+1}$ : ($\frac{x+1}{x}$ + $\frac{1}{x-1}$ + $\frac{2-x^2}{x^2-x}$)
= $\frac{x.(x+1)}{(x-1)^2}$ : ($\frac{(x+1)(x-1)}{x(x-1)}$ + $\frac{x}{x(x-1)}$ + $\frac{2-x^2}{x(x-1)}$)
= $\frac{x.(x+1)}{(x-1)^2}$ : $\frac{x^2-1+x+2-x^2}{x(x-1)}$
= $\frac{x.(x+1)}{(x-1)^2}$.$\frac{x(x-1)}{x+1}$
= $\frac{x^2}{x-1}$
c, B = $\frac{-1}{2}$
⇔ $\frac{x^2}{x-1}$ = $\frac{-1}{2}$
⇔ 2$x^2$ = 1 - x
⇔ 2$x^2$ + x - 1 = 0
⇔ \(\left[ \begin{array}{l}x=\frac{1}{2}(nhận)\\x=-1(loại)\end{array} \right.\)
d, B = $\frac{x^2}{x-1}$
