\(\sqrt{x\left(x-2\right)}+\sqrt{x\left(x-5\right)}=\sqrt{x\left(x+3\right)}\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x-2}+\sqrt{x-5}-\sqrt{x+3}\right)=0\)
Trường hợp 1:
\(\sqrt{x}=0\)
\(\Leftrightarrow x=0\)
Trường hợp 2:
\(\sqrt{x-2}+\sqrt{x-5}-\sqrt{x+3}=0\)
\(\Leftrightarrow\left(\sqrt{x-2}-2\right)+\left(\sqrt{x-5}-1\right)-\left(\sqrt{x+3}-3\right)=0\)
\(\Leftrightarrow\dfrac{x-2-4}{\sqrt{x-2}+2}+\dfrac{x-5-1}{\sqrt{x-5}+1}-\dfrac{x+3-9}{\sqrt{x+3}+3}=0\)
\(\Leftrightarrow\left(\dfrac{1}{\sqrt{x-2}+2}+\dfrac{1}{\sqrt{x-5}+1}-\dfrac{1}{\sqrt{x+3}+3}\right)\left(x-6\right)=0\)
\(\Rightarrow x=6\)
Giải thích:
\(x-5< x+3\)
\(\Rightarrow\sqrt{x-5}+1< \sqrt{x+3}+3\)
\(\Rightarrow\dfrac{1}{\sqrt{x-5}+1}>\dfrac{1}{\sqrt{x+3}+3}\)
\(\Rightarrow\dfrac{1}{\sqrt{x-2}+2}+\dfrac{1}{\sqrt{x-5}+1}-\dfrac{1}{\sqrt{x+3}+3}>0\)
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