Đáp án:
Giải thích các bước giải:
Bài 1
a, ĐKXĐ x $\neq$ 0
b, x $\neq$ ±1
c, x $\neq$ $\frac{1}{5}$
x $\neq$ $\frac{-1}{5}$
d, x $\neq$ -1
x $\neq$ - 2
Bài 2
a, $\frac{10.(x-2)}{(x+2)(x-2)}$ = $\frac{x^2 -4}{(x+2)(x-2)}$ + $\frac{x+2}{(x+2)(x-2)}$
==> 10x - 20 = $x^2$ - 4 + x + 2
<=>10x - x - $x^2$ -4 + 2 + 20 = 0
$x^2$ + 9x + 18 = 0
(x+3)(x+6) = 0
==> x= -3 ; x= -6
b, $\frac{x+2}{(x+2)^2}$ + $\frac{x-2}{(x-2)^2}$ = $\frac{32}{x(x^4 + 4x^2+16)}$
<=> $\frac{1}{x+2}$ + $\frac{1}{x-2}$ = $\frac{32}{x(x^4 + 4x^2+16)}$
<=> $\frac{x-2}{(x+2)(x-2)}$ + $\frac{x+2}{(x+2)(x-2)}$ = $\frac{32}{x(x^4 + 4x^2+16)}$
<=> $\frac{2x}{(x+2)(x-2)}$ = $\frac{32}{x(x^4 + 4x^2+16)}$
<=> 2x.x$(x^4 + 4x^2+16)$= ($x^{2}-4$ ) . 32
<=> $2x^{6}$ + $8x^{4}$ + $32x^{2}$ = $32x^{2}$ -128
<=> $2x^{6}$ + $8x^{4}$ = -128
<=>$2x^{6}$ + $8x^{4}$ +128 = 0
<=> 2( $x^{6}$ + $4x^{4}$ + 64)= 0
Bài 3
a, $\frac{2x-1}{x+2}$ = $\frac{x+2}{x+2}$ - $\frac{x}{x+2}$
==> 2x-1 = x+2 - x
<=> 2x - x +x = 2 + 1
<=> 2x = 3
<=> x = $\frac{3}{2}$
b, $\frac{x^2}{x+2}$ = $\frac{x(x+2)}{x+2}$ - $\frac{3-x}{x+2}$
==> $x^{2}$ = $x^{2}$ + 2x - 3 + x
<=> -3x = -3
<=> x =1
Bài 4
==> $x^2 + 2x + 3 $ = 0
<=> $x^2 + 2x + 1 +2 $ = 0
<=> $(x+1)^2$ = -2 (vô lý)
Vậy pt vô nghiệm
b, $\frac{x}{x+2}$ + $\frac{4}{x-2}$ = $\frac{4}{(x-2)(x+2)}$
<=>$\frac{x(x-2)}{(x+2)(x-2)}$ + $\frac{4(x+2)}{(x-2)(x+2)}$ = $\frac{4}{(x-2)(x+2)}$
==> $x^2 - 2x$ + $4x + 8$ = 4
<=> $x^2 + 2x + 1 + 3$ = 0
<=>$(x+1)^2$ = -3 (vô lý)
Vậy pt vô nghiệm
