a) $\Large\frac{x + 6}{x − 5}$ + $\Large\frac{x - 5}{x + 6}$ = $\Large\frac{2x^2 + 23x + 61}{x^2 + x − 30}$
<=> $\Large\frac{(x + 6)^2}{(x − 5)(x+6)}$ + $\Large\frac{(x - 5)^2}{(x − 5)(x+6)}$ = $\Large\frac{2x^2 + 23x + 61}{(x − 5)(x+6)}$
<=> $(x + 6)^2 + (x - 5)^2 = 2x^2 + 23x + 61$
<=> $x^2 + 12x + 36 + x^2 - 10x + 25= 2x^2 + 23x + 61$
<=> $2x^2 + 2x + 61 = 2x^2 + 23x + 61$
<=> $2x = 23x$
<=> $x = 0$
.
b) $\Large\frac{6}{x − 5}$ + $\Large\frac{x + 2}{x -8}$ = $\Large\frac{18}{(x-5)(8-x)}$ - $1$ (ĐK: x khác 5; -2; 8)
<=> $\Large\frac{6x-48}{(x-5)(x-8)}$ + $\Large\frac{x^2-3x-10}{(x-5)(x-8)}$ = $\Large\frac{-18}{(x-5)(x-8)}$ - $\Large\frac{(x-5)(x-8)}{(x-5)(x-8)}$
<=> $\Large\frac{6x-48+x^2-3x-10}{(x-5)(x-8)}$ = $\Large\frac{-18-(x-5)(x-8)}{(x-5)(x-8)}$
<=> $6x-48+x^2-3x-10 = -18-(x-5)(x-8)$
<=> $6x-48+x^2-3x-10 = -18-x^2+13x-40$
<=> $2x^2-10x = 0$
<=> $2x(x-5) = 0$
<=> x=0 hoặc x=5
x=5 không thỏa mãn ĐK
=> x=0
.
c) $\Large\frac{x − 4}{x − 1}$ + $\Large\frac{x + 4}{x + 1}$ = $2$
<=> $\Large\frac{(x − 4)(x+1)}{(x − 1)(x+1)}$ + $\Large\frac{(x + 4)(x-1)}{(x − 1)(x+1)}$ = $\Large\frac{2(x − 1)(x+1)}{(x − 1)(x+1)}$
<=> $\Large\frac{x^2-3x-4}{(x − 1)(x+1)}$ + $\Large\frac{x^2+3x-4}{(x − 1)(x+1)}$ = $\Large\frac{2x^2-2}{(x − 1)(x+1)}$
<=> $x^2-3x-4 + x^2+3x-4 = 2x^2-2$
<=> $-4 -4 = -2$
<=> $-8 = -2$ (vô lý)
=> PT vô nghiệm
.
d) $\Large\frac{3}{x+1}$ - $\Large\frac{1}{x - 2}$ = $\Large\frac{9}{(x+1)(2-x)}$ (ĐK: x khác 1; 2)
<=> $\Large\frac{3x-6}{(x+1)(x-2)}$ - $\Large\frac{x+1}{(x+1)(x-2)}$ = $\Large\frac{-9}{(x+1)(x-2)}$
<=> $3x-6-x-1 = -9$
<=> $2x-7 = -9$
<=> $x = -1$ (k thỏa mãn)
=> PT vô nghiệm
