ĐKXĐ: \(x\ge-\frac{1}{2}\)
Đặt \(\sqrt{2x+1}+\sqrt{3x+4}=a\ge0\)
\(\Rightarrow a^2=5x+5+2\sqrt{6x^2+11x+4}\)
\(\Rightarrow5x+2\sqrt{6x^2+11x+4}=a^2-5\)
Phương trình trở thành:
\(a^2-5=4a+16\)
\(\Leftrightarrow a^2-4a-21=0\)\(\Rightarrow\left[{}\begin{matrix}a=7\\a=-3< 0\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2x+1}+\sqrt{3x+4}=7\)
\(\Leftrightarrow\sqrt{2x+1}-3+\sqrt{3x+4}-4=0\)
\(\Leftrightarrow\frac{2\left(x-4\right)}{\sqrt{2x+1}+3}+\frac{3\left(x-4\right)}{\sqrt{3x+4}+4}=0\)
\(\Leftrightarrow\left(x-4\right)\left(\frac{2}{\sqrt{2x+1}+3}+\frac{3}{\sqrt{3x+4}+4}\right)=0\)
\(\Rightarrow x=4\)
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