Tìm kiếm
Nam
b) $\frac{1}{(x+1)^{2}} + \frac{1}{(x+2)^{2}} = \frac{5}{4}$
ĐK: $x\ne -1; x\ne -2$
Đặt $t=x+\dfrac{3}{2}$, $t\ne \pm \dfrac{1}{2}$
Pt trở thành:
$\dfrac{1}{\left ( t-\dfrac{1}{2} \right )^2}+\dfrac{1}{\left ( t+\dfrac{1}{2} \right )^2}=\dfrac{5}{4}$
$\Leftrightarrow \frac{2t^2+\dfrac{1}{2}}{\left ( t^2-\dfrac{1}{4} \right )^2}=\dfrac{5}{4}$
$\Leftrightarrow 2a+\dfrac{1}{2}=\dfrac{5}{4}\left ( a^2-\dfrac{1}{2}a+\dfrac{1}{16} \right )$ với $a=t^2, a\ge 0, a\ne \dfrac{1}{4}$
$\Leftrightarrow a=\dfrac{9}{4}$
$\Rightarrow t=\pm \dfrac{3}{2}$ (t/m)
$\boxed{x=0~ (v)~ x=-3}$
