Đáp án: $x=-1$
Giải thích các bước giải:
Đặt $(x+1)^2+1=t, t\ge 1$
$\to x^2+2t+2=t$
Ta có:
$\dfrac{1}{(x^2+2x+2)^2}+\dfrac{1}{(x^2+2x+3)^2}=\dfrac54$
$\to \dfrac{1}{(x^2+2x+2)^2}+\dfrac{1}{(x^2+2x+2+1)^2}=\dfrac54$
$\to \dfrac{1}{t^2}+\dfrac{1}{(t+1)^2}=\dfrac54$
$\to 4\left(t+1\right)^2+4t^2=5t^2\left(t+1\right)^2$
$\to 8t^2+8t+4=5t^4+10t^3+5t^2$
$\to 5t^4+10t^3-3t^2-8t-4=0$
$\to \left(t-1\right)\left(t+2\right)\left(5t^2+5t+2\right)=0$
Mà $t\ge 1$
$\to t+2>0, 5t^2+5t+2>0$
$\to t-1=0$
$\to t=1$
$\to (x+1)^2+1=1$
$\to (x+1)^2=0$
$\to x=-1$
