ĐKXĐ: \(\begin{cases}x-3\ge0\\ 2x+1\ge0\end{cases}\)
=>x>=3
Ta có: \(\sqrt{2x+1}=2+\sqrt{x-3}\)
=>\(\sqrt{2x+1}-\sqrt{x-3}=2\)
=>\(\sqrt{2x+1}-3+1-\sqrt{x-3}=2-3+1=3-3=0\)
=>\(\frac{2x+1-9}{\sqrt{2x+1}+3}+\frac{1-\left(x-3\right)}{1+\sqrt{x-3}}=0\)
=>\(\frac{2x-8}{\sqrt{2x+1}+3}-\frac{x-4}{\sqrt{x-3}+1}=0\)
=>\(\left(x-4\right)\left(\frac{2}{\sqrt{2x+1}+3}-\frac{1}{\sqrt{x-3}+1}\right)=0\)
=>x-4=0
=>x=4(nhận)
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