Giải thích các bước giải:
a.Ta có:
$\dfrac{x^2}{9}+\dfrac{16}{x^2}=\dfrac{10}{3}(\dfrac{x}{3}-\dfrac4x)$
$\to \dfrac{x^2}{9}+2\cdot \dfrac{x}3\cdot \dfrac4x+\dfrac{16}{x^2}-\dfrac83=\dfrac{10}{3}(\dfrac{x}{3}-\dfrac4x)$
$\to (\dfrac{x}3-\dfrac4x)^2-\dfrac83=\dfrac{10}{3}(\dfrac{x}{3}-\dfrac4x)$
$\to (\dfrac{x}3-\dfrac4x)^2-\dfrac{10}{3}(\dfrac{x}{3}-\dfrac4x)-\dfrac83=0$
$\to (\dfrac{x}{3}-\dfrac4x+\dfrac23)(\dfrac{x}{3}-\dfrac4x-4)=0$
$\to \dfrac{x}{3}-\dfrac4x+\dfrac23=0$
$\to x^2-12+2x=0$
$\to x=-1\pm\sqrt{13}$
Hoặc $\dfrac{x}{3}-\dfrac4x-4=0$
$\to x^2-12-12x=0$
$\to x=6\pm4\sqrt3$
b.ĐKXĐ: $x\ne 0,-1,-2$
Ta có:
$\dfrac{1}{x(x+2)}-\dfrac{1}{(x+1)^2}=\dfrac{1}{12}$
$\to \dfrac{1}{x^2+2x}-\dfrac{1}{x^2+2x+1}=\dfrac{1}{12}$
Đặt $x^2+2x+1=a\to a=(x+1)^2\ge 0$
$\to \dfrac1{a-1}-\dfrac1a=\dfrac1{12}$
$\to 12a-12\left(a-1\right)=a\left(a-1\right)$
$\to 12=a^2-a$
$\to a^2-a-12=0$
$\to (a-4)(a+3)=0$
$\to a-4=0$ vì $a\ge 0\to a+3>0$
$\to a=4$
$\to (x+1)^2=4$
$\to x+1=2\to x=1$ hoặc $x+1=-2\to x=-3$
c.Ta có:
$2x^3-3x^2-11x+6=0$
$\to \left(x+2\right)\left(2x-1\right)\left(x-3\right)=0$
$\to x\in\{-2,\dfrac12,3\}$
