Question

[Conan edogawa]

\(4\sqrt{x+5}-\sqrt{x+1}=9+x\)

Giari phương trình trên

Answer 1

[Victorique de Blois]

\(đk:x\ge-1\)

\(4\sqrt{x+5}-\sqrt{x+1}=9+x\)

\(\Leftrightarrow4\sqrt{x+5}-\sqrt{x+1}-x-9=0\)

\(\Leftrightarrow4\frac{\left(\sqrt{x+5}-2\right)\left(\sqrt{x+5}+2\right)}{\sqrt{x+5}+2}-\sqrt{x+1}-x-1=0\)

\(\Leftrightarrow4\cdot\frac{x+5-4}{\sqrt{x+5}+2}-\sqrt{x+1}-\left(x+1\right)=0\)

\(\Leftrightarrow4\cdot\frac{x+1}{\sqrt{x+5}+2}-\sqrt{x+1}-\left(x+1\right)=0\)

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

th1: \(\frac{4}{\sqrt{x+5}+2}-1-\sqrt{x+1}=0\)

có : \(x\ge-1\Rightarrow\frac{4}{\sqrt{x+5}+2}\le1\Rightarrow\frac{4}{\sqrt{x+5}+2}-1\le0\) và \(-\sqrt{x+1}\le0\) nên

\(\hept{\begin{cases}\frac{4}{\sqrt{x+5}+2}-1=0\\\sqrt{x+1}=0\end{cases}}\Leftrightarrow x=-1\left(tm\right)\)

th2 : \(\sqrt{x+1}=0\Leftrightarrow x=-1\left(tm\right)\)

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