Question

Giai  phương trình

  \(\sqrt{2x+3+\sqrt{x+2}}+\sqrt{2x+2-\sqrt{x+2}}=1+2\sqrt{x+2}\)

 

Answer 1

Đk:  \(\hept{\begin{cases}x\ge2\\2x+3+\sqrt{x+2}\ge0\\2x+2-\sqrt{x+2}\ge0\end{cases}}\)

Đặt \(\sqrt{x+2}=t\left(t\ge0\right)\Rightarrow x=t^2-2\)

\(pt\Leftrightarrow\sqrt{2t^2-1+t}+\sqrt{2t^2-2-t}=1+2t\)

\(\Leftrightarrow4t^2-3+2\sqrt{\left(2t^2+t-1\right)\left(2t^2-t-2\right)}=4t^2+4t+1\)

\(\Leftrightarrow\sqrt{\left(2t^2+t-1\right)\left(2t^2-t-2\right)}=2t+2\)

\(\Leftrightarrow4t^4-11t^2-9t-2=0\)

\(\Leftrightarrow\left(2t+1\right)^2\left(t-2\right)\left(t+1\right)=0\)

Do \(t\ge0\) nên t = 2. Vậy \(\sqrt{x+2}=2\Rightarrow x=2\left(tm\right)\)

Vậy pt có nghiệm x = 2.

Chúc em học tốt!