Question

[WTFシSnow]

giải phương trình : \(x^2+\sqrt{2x+1}+\sqrt{x-3}=5x\)

Answer 1

[Nguyễn Minh Đăng]

đk: \(x\ge3\)

Ta có: \(x^2+\sqrt{2x+1}+\sqrt{x-3}=5x\)

\(\Leftrightarrow\left(x^2-16\right)+\left(\sqrt{2x+1}-3\right)+\left(\sqrt{x-3}-1\right)-\left(5x-20\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+4\right)+\frac{2x-8}{\sqrt{2x+1}+3}+\frac{x-4}{\sqrt{x-3}+1}-5\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+4+\frac{2}{\sqrt{2x+1}+3}+\frac{1}{\sqrt{x-3}+1}-5\right)=0\)

Vì \(\hept{\begin{cases}x+4\ge7\\\frac{2}{\sqrt{2x+1}+3}>0\\\frac{1}{\sqrt{x-3}+1}>0\end{cases}}\left(\forall x\ge3\right)\) nên từ đó:

\(\Rightarrow x+4+\frac{2}{\sqrt{2x+1}+3}+\frac{1}{\sqrt{x-3}+1}-5>0\left(\forall x\ge3\right)\)

\(\Rightarrow x-4=0\Rightarrow x=4\)

Vậy x = 4