Lời giải:
ĐK: $\xneq \pm 2$
Đặt $\frac{x+3}{x-2}=a; \frac{x-3}{x+2}=b$ thì PT trở thành:
$3a^2+168b^2-46ab=0$
$\Leftrightarrow 3a^2-18ab+168b^2-28ab=0$
$\Leftrightarrow 3a(a-6b)-28b(a-6b)=0$
$\Leftrightarrow (3a-28b)(a-6b)=0$
$\Rightarrow 3a=28b$ hoặc $a=6b$
Nếu $3a=28b\Leftrightarow \frac{3(x+3)}{x-2}=\frac{28(x-3)}{x+2}$
$\Rightarrow x=\frac{6}{5}$ hoặc $x=5$ (thỏa mãn)
Nếu $a=6b\Leftrightarrow \frac{x+3}{x-2}=\frac{6(x-3)}{x+2}$
$\Rightarrow x=1$ hoặc $x=6$ (thỏa mãn)
Vậy..........
b)
PT $\Leftrightarrow [(x+2)(x+12)][(x+3)(x+8)]=-2x^2$
$\Leftrightarrow (x^2+14x+24)(x^2+11x+24)=-2x^2$
Đặt $x^2+11x+24=a$ thì:
$(a+3x)a=-2x^2\Leftrightarrow a^2+3ax+2x^2=0$
$\Leftrightarrow a(a+x)+2x(a+x)=0\Leftrightarrow (a+2x)(a+x)=0$
Nếu $a+2x=0\Leftrightarrow x^2+13x+24=0\Rightarrow x=\frac{-13\pm \sqrt{73}}{2}$
Nếu $a+x=0\Leftrightarrow x^2+12x+24=0\Rightarrow x=-6\pm 2\sqrt{3}$
Lời giải:
ĐK: $\xneq \pm 2$
Đặt $\frac{x+3}{x-2}=a; \frac{x-3}{x+2}=b$ thì PT trở thành:
$3a^2+168b^2-46ab=0$
$\Leftrightarrow 3a^2-18ab+168b^2-28ab=0$
$\Leftrightarrow 3a(a-6b)-28b(a-6b)=0$
$\Leftrightarrow (3a-28b)(a-6b)=0$
$\Rightarrow 3a=28b$ hoặc $a=6b$
Nếu $3a=28b\Leftrightarow \frac{3(x+3)}{x-2}=\frac{28(x-3)}{x+2}$
$\Rightarrow x=\frac{6}{5}$ hoặc $x=5$ (thỏa mãn)
Nếu $a=6b\Leftrightarrow \frac{x+3}{x-2}=\frac{6(x-3)}{x+2}$
$\Rightarrow x=1$ hoặc $x=6$ (thỏa mãn)
Vậy..........
b)
PT $\Leftrightarrow [(x+2)(x+12)][(x+3)(x+8)]=-2x^2$
$\Leftrightarrow (x^2+14x+24)(x^2+11x+24)=-2x^2$
Đặt $x^2+11x+24=a$ thì:
$(a+3x)a=-2x^2\Leftrightarrow a^2+3ax+2x^2=0$
$\Leftrightarrow a(a+x)+2x(a+x)=0\Leftrightarrow (a+2x)(a+x)=0$
Nếu $a+2x=0\Leftrightarrow x^2+13x+24=0\Rightarrow x=\frac{-13\pm \sqrt{73}}{2}$
Nếu $a+x=0\Leftrightarrow x^2+12x+24=0\Rightarrow x=-6\pm 2\sqrt{3}$