Register Now

Login

Lost Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: Tìm x biết : $\frac{2x+1}{x^{2}-1}$ + $\frac{3}{1-x}$ – $\frac{-x}{x+1}$ = 1

Toán Lớp 8: Tìm x biết :
$\frac{2x+1}{x^{2}-1}$ + $\frac{3}{1-x}$ – $\frac{-x}{x+1}$ = 1

Comments ( 2 )

  1. [ 2x + 1 ] / [ x² – 1 ] + 3 / [ 1 – x ] – [ -x ] / [ x + 1 ] = 1
    [ 2x + 1 ] / [ ( x – 1 ) ( x + 1 ) ] – 3 / [ x – 1 ] + x / [ x + 1 ] = 1
    [ 2x + 1 ] / [ ( x – 1 ) ( x + 1 ) ] – [ 3 ( x + 1 ) ] / [ ( x – 1 ) ( x + 1 ) ] + [ x ( x – 1 ) ] / [ ( x – 1 ) ( x + 1 ) ] = 1
    [ 2x + 1 – 3 ( x + 1 ) + x ( x – 1 ) ] / [ ( x – 1 ) ( x + 1 ) ] = 1
    [ 2x + 1 – 3x – 3 + x^2 – x ] / [ ( x – 1 ) ( x + 1 ) ] = 1
    [ x² – 2x + 1 ] / [ ( x – 1 ) ( x + 1 ) ] = 1
    [ ( x – 1 )² ] / [ ( x – 1 ) ( x + 1 ) ] = 1
    [ x – 1 ] / [ x + 1 ] = 1
    x – 1 = x + 1
    -1 = 1 ( vô lý )
     Vậy phương trình vô nghiệm

  2. $\dfrac{2x + 1}{x^2 – 1}$ + $\dfrac{3}{1 – x}$ – $\dfrac{- x}{x + 1}$ = 1
    <=> $\dfrac{2x + 1}{(x – 1)(x + 1)}$ – $\dfrac{3}{x – 1}$ + $\dfrac{x}{x + 1}$ = 1
    <=> $\dfrac{2x + 1}{(x – 1)(x + 1)}$ – $\dfrac{3(x + 1)}{(x – 1)(x + 1)}$ + $\dfrac{x(x – 1)}{(x – 1)(x + 1)}$ = 1
    <=> $\dfrac{2x + 1 – 3(x + 1) + x(x – 1)}{(x – 1)(x + 1)}$ = 1
    <=> $\dfrac{2x + 1 – 3x – 3 + x^2 – x  }{(x – 1)(x + 1)}$ = 1
    <=> $\dfrac{x^2 – 2x + 1}{(x – 1)(x + 1)}$ = 1
    <=>  $\dfrac{(x – 1)^2}{(x – 1)(x + 1)}$ = 1
    <=> $\dfrac{x – 1}{x + 1}$ = 1
    <=> x – 1 = x + 1
    <=> x – x = 1 + 1
    <=> $\dfrac{x – 1}{x + 1}$ = 1
    <=> x – 1 = x + 1
    <=> x – x = 1 + 1
    <=> 0 = 2 (vô lí)
    KL: Không tìm được x với điều kiện x $\geq$  0, x $\neq$ 1 
    $#Ling$

Leave a reply

222-9+11+12:2*14+14 = ? ( )