Question

Tìm x, (ghi đk nếu có)

a) \(3\left(\dfrac{x+3}{x-2}\right)^2+168\left(\dfrac{x-3}{x+2}\right)^2-46\left(\dfrac{x^2-9}{x^2-4}\right)=0\)

b) \(\left(x+2\right)\left(x+3\right)\left(x+8\right)\left(x+12\right)=-2x^{^2}\)

Answer 1

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 $a+x=0\Leftrightarrow x^2+12x+24=0\Rightarrow x=-6\pm 2\sqrt{3}$

Answer 2

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 $a+x=0\Leftrightarrow x^2+12x+24=0\Rightarrow x=-6\pm 2\sqrt{3}$