Question

Giải pt sau: (x+1)² =(x+5)(x+2-\(\sqrt{2x+3}\))

Answer 1

ĐK \(x\ge-\frac{3}{2}\)

Nhân liên hợp ta có

\(\left(x+1\right)^2\left(x+2+\sqrt{2x+3}\right)=\left(x+5\right)\left[\left(x+2\right)^2-2x-3\right]\)

<=> \(\left(x+1\right)^2\left(x+2+\sqrt{2x+3}\right)=\left(x+5\right)\left(x+1\right)^2\)

<=> \(\left[{}\begin{matrix}x=-1\\x+2+\sqrt{2x+3}=x+5\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=-1\\\sqrt{2x+3}=3\end{matrix}\right.\)=> \(\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)(tm ĐK)

vậy \(S=\left\{-1;3\right\}\)

Answer 2

Mọi người giúp mình với ạ