$\frac{x+1}{x}$ -$\frac{2}{x-1}$ =$\frac{2x-5}{x.(x+1)}$

Question

$\frac{x+1}{x}$ -$\frac{2}{x-1}$ =$\frac{2x-5}{x.(x+1)}$

in progress 0
Brielle 1 tháng 2021-11-04T05:20:32+00:00 2 Answers 4 views 0

Answers ( )

    0
    2021-11-04T05:21:34+00:00

    $\dfrac{x+1}{x}-\dfrac{2}{x-1}=\dfrac{2x-5}{x(x+1)}$

    $ĐK:x \neq 0; x \neq 1; x \neq -1$

    $PT⇔ (x-1)(x+1)^2-2x(x+1)=(2x-5)(x-1)$

        $⇔ (x-1)(x^2+2x+1)-2x^2-2x=2x^2-6x+5$

        $⇔ x^3-x^2-3x-1=2x^2-6x+5$

        $⇔ x^3-x^2-3x-1-2x^2+6x-5=0$

        $⇔ x^3-3x^2+3x-6=0$

    $hmm$

    $sorry$ $bn$

    $đến$ $đây$ $mk$ $pí$ $mất$

    0
    2021-11-04T05:21:57+00:00

    Đáp án:

     xin câu trả lời hay nhất nha

    Giải thích các bước giải:

     $\dfrac{x+1}{x} – \dfrac{2}{x – 1} = \dfrac{2x – 5}{x.(x+1)}$ (đkxđ : $x\neq0;1;-1$ )

    ⇒ $(x+1)(x-1) – 2x(x+1) = (2x-5)(x-1)$

    ⇔ $x² – 1 – 2x² – 2x =2x² – 7x + 5 $

    ⇔ $x² – 2x² -2x² + 5x – 6 = 0$

    ⇔$-3x² + 5x – 5 = 0$

    ko có x

Leave an answer

Browse

35:5x4+1-9:3 = ? ( )