tính: $\dfrac{x+1}{2x-2}+\dfrac{-2x}{x^2-1}$ 11/07/2021 Bởi Jade tính: $\dfrac{x+1}{2x-2}+\dfrac{-2x}{x^2-1}$
Đáp án + Giải thích các bước giải: `{x+1}/{2x-2}+{-2x}/{x^2-1}` `={x+1}/{2(x-1)}+{-2x}/{x^2-1^2}` `={(x+1)^2}/{2(x-1)(x+1)}+{2(-2x)}/{2(x-1)(x+1)}` `={x^2+2x+1}/{2(x-1)(x+1)}+{-4x}/{2(x-1)(x+1)}` `={x^2+2x+1-4x}/{2(x-1)(x+1)}` `={x^2-2x+1}/{2(x-1)(x+1)}` `={(x-1)^2}/[2(x-1)(x+1)}` `=[x-1]/[2(x+1)]` Bình luận
Đáp án+Giải thích các bước giải: $\left \{ {{2x-2=2(x-1)} \atop {x^2-1=(x-1)(x+1)}} \right.$ $⇔MTC=2(x-1)(x+1)$ $*)\dfrac{x+1}{2x-2}+ \dfrac{-2x}{x^2-1}$ $=\dfrac{x+1}{2(x-1)}+\dfrac{-2x}{(x-1)(x+1)}$ $=\dfrac{(x+1)(x+1)}{2(x-1)(x+1)}+\dfrac{2(-2x)}{2(x-1)(x+1)}$ $=\dfrac{(x+1)^2-4x}{2(x-1)(x+1)}$ $=\dfrac{x^2-2x+1}{2(x-1)(x+1)}$ $=\dfrac{(x-1)^2}{2(x-1)(x+1)}$ $=\dfrac{x-1}{2(x+1)}$ Bình luận
Đáp án + Giải thích các bước giải:
`{x+1}/{2x-2}+{-2x}/{x^2-1}`
`={x+1}/{2(x-1)}+{-2x}/{x^2-1^2}`
`={(x+1)^2}/{2(x-1)(x+1)}+{2(-2x)}/{2(x-1)(x+1)}`
`={x^2+2x+1}/{2(x-1)(x+1)}+{-4x}/{2(x-1)(x+1)}`
`={x^2+2x+1-4x}/{2(x-1)(x+1)}`
`={x^2-2x+1}/{2(x-1)(x+1)}`
`={(x-1)^2}/[2(x-1)(x+1)}`
`=[x-1]/[2(x+1)]`
Đáp án+Giải thích các bước giải:
$\left \{ {{2x-2=2(x-1)} \atop {x^2-1=(x-1)(x+1)}} \right.$
$⇔MTC=2(x-1)(x+1)$
$*)\dfrac{x+1}{2x-2}+ \dfrac{-2x}{x^2-1}$
$=\dfrac{x+1}{2(x-1)}+\dfrac{-2x}{(x-1)(x+1)}$
$=\dfrac{(x+1)(x+1)}{2(x-1)(x+1)}+\dfrac{2(-2x)}{2(x-1)(x+1)}$
$=\dfrac{(x+1)^2-4x}{2(x-1)(x+1)}$
$=\dfrac{x^2-2x+1}{2(x-1)(x+1)}$
$=\dfrac{(x-1)^2}{2(x-1)(x+1)}$
$=\dfrac{x-1}{2(x+1)}$