Toán 2/x^2-4 – x-1/x(x-2) + x-4/x(x+2) giúp pt này 17/07/2021 By Madeline 2/x^2-4 – x-1/x(x-2) + x-4/x(x+2) giúp pt này
$\dfrac{2}{x^2-4}-\dfrac{x-1}{x(x-2)}+\dfrac{x-4}{x(x+2)}=0$ ĐKXĐ: `x ne 0 `; `x ne ±2` `<=>`$\dfrac{2x}{x(x-2)(x+2)}-\dfrac{(x-1)(x+2)}{x(x-2)(x+2)}+\dfrac{(x-4)(x-2)}{x(x+2)(x-2)}=0$ `⇒2x-(x-1)(x+2)+(x-4)(x-2)=0` `⇔2x-(x^2+2x-x-2)+x^2-2x-4x+8=0` `⇔2x-x^2-2x+x+2+x^2-2x-4x+8=0` `⇔ 5x =10` `⇔x=2(loại)` Vậy `S=∅` Trả lời
Đáp án: `S=∅` Giải thích các bước giải: `2/(x^2-4) – (x-1)/(x(x-2)) + (x-4)/(x(x+2)) (\text{ĐK}: x`$\neq0; x $$\neq±2)$ $\Leftrightarrow\dfrac{2x\left(x-1\right)}{x\left(x-2\right)\left(x+2\right)}-\dfrac{\left(x-2\right)\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}-\dfrac{\left(4+x\right)\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}=0$ `<=>` $\dfrac{2}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x\left(x-1\right)}-\dfrac{4+x}{x\left(x+2\right)}=0$ $\Rightarrow2x^2-2x-x^2+4-x^2+2x+8=0$ $\Leftrightarrow12=0$ (vô lý) `=> S=∅` Trả lời
$\dfrac{2}{x^2-4}-\dfrac{x-1}{x(x-2)}+\dfrac{x-4}{x(x+2)}=0$
ĐKXĐ: `x ne 0 `; `x ne ±2`
`<=>`$\dfrac{2x}{x(x-2)(x+2)}-\dfrac{(x-1)(x+2)}{x(x-2)(x+2)}+\dfrac{(x-4)(x-2)}{x(x+2)(x-2)}=0$
`⇒2x-(x-1)(x+2)+(x-4)(x-2)=0`
`⇔2x-(x^2+2x-x-2)+x^2-2x-4x+8=0`
`⇔2x-x^2-2x+x+2+x^2-2x-4x+8=0`
`⇔ 5x =10`
`⇔x=2(loại)`
Vậy `S=∅`
Đáp án:
`S=∅`
Giải thích các bước giải:
`2/(x^2-4) – (x-1)/(x(x-2)) + (x-4)/(x(x+2)) (\text{ĐK}: x`$\neq0; x $$\neq±2)$
$\Leftrightarrow\dfrac{2x\left(x-1\right)}{x\left(x-2\right)\left(x+2\right)}-\dfrac{\left(x-2\right)\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}-\dfrac{\left(4+x\right)\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}=0$
`<=>` $\dfrac{2}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x\left(x-1\right)}-\dfrac{4+x}{x\left(x+2\right)}=0$
$\Rightarrow2x^2-2x-x^2+4-x^2+2x+8=0$
$\Leftrightarrow12=0$ (vô lý)
`=> S=∅`