ĐK: \(x\ge-2\)
Đặt: \(\sqrt{x+2}=t\ge0\) => \(x=t^2-2\)
pt <=> \(\left(3-t^2+2\right).t=2\left(t^2-2\right)-2\)
<=> \(5t-t^3=2t^2-6\)
<=> \(\left(t+1\right)\left(t+3\right)\left(t-2\right)=0\)
=> \(t=2\) \(\left(t\ge0\right)\)
=> \(\sqrt{x+2}=2\)
<=> \(x=2\)
Thử lại:.... (đúng)
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