Tính:
a, $\frac{x}{x+1}$ + $\frac{2x}{x-1}$ – $\frac{3x^{2}+1}{x^{2}-1}$
b, ($\frac{x}{x-1}$ + $\frac{x}{x^{2}-x}$) : $\frac{x+1}{3}$
Giúp với mik đang cần gấp
Tính:
a, $\frac{x}{x+1}$ + $\frac{2x}{x-1}$ – $\frac{3x^{2}+1}{x^{2}-1}$
b, ($\frac{x}{x-1}$ + $\frac{x}{x^{2}-x}$) : $\frac{x+1}{3}$
Giúp với mik đang cần gấp
`a) x/(x+1)+(2x)/(x-1)-(3x^2+1)/(x^2-1)` (ĐKXĐ: `x\ne +-1`)
`=x/(x+1)+(2x)/(x-1)-(3x^2+1)/((x-1)(x+1))`
`=(x(x-1)+2x(x+1)-(3x^2+1))/((x-1)(x+1))`
`=(x^2-x+2x^2+2x-3x^2-1)/((x-1)(x+1))`
`=(x-1)/((x-1)(x+1))`
`=1/(x+1)`
`b) (x/(x-1)+x/(x^2-x)):(x+1)/3` (ĐKXĐ: `x\ne 0;+1`)
`=(x/(x-1)+x/(x(x-1)).3/(x+1)`
`=(x.x+x)/(x(x-1)). 3/(x+1)`
`=(x^2+x)/(x(x-1)). 3/(x+1)`
`=(x(x+1)/(x(x-1)). 3/(x+1)`
`=(x+1)/(x-1). 3/(x+1)`
`=3/(x-1)`
Đáp án+giải thích các bước giải:
a)
$\dfrac{x}{x+1}+\dfrac{2x}{x-1}-\dfrac{3x^2+1}{x^2-1}$
$=\dfrac{x(x-1)}{(x-1)(x+1)}+\dfrac{2x(x+1)}{(x-1)(x+1)}-\dfrac{3x^2+1}{(x-1)(x+1)}$
$=\dfrac{x^2-x+2x^2+2x-3x^2-1}{(x-1)(x+1)}$
$=\dfrac{x-1}{(x-1)(x+1)}$
$=\dfrac{1}{x+1}$
b)
$(\dfrac{x}{x-1}+\dfrac{x}{x^2-x}):\dfrac{x+1}{3}$
$(=\dfrac{x}{x-1}+\dfrac{x}{x(x-1)}):\dfrac{x+1}{3}$
$(=\dfrac{x^2}{x(x-1)}+\dfrac{x}{x(x-1)}):\dfrac{x+1}{3}$
$=\dfrac{x^2+x}{x(x-1)}.\dfrac{3}{x+1}$
$=\dfrac{3x(x+1)}{x(x-1)(x+1)}$
$=\dfrac{3}{x-1}$