ĐKXĐ: \(x\ge\frac13\)
Ta có: \(\sqrt{2x+5}+\sqrt{3x-1}=5\)
=>\(2x+5+3x-1+2\sqrt{\left(2x+5\right)\left(3x-1\right)}=25\)
=>\(2\sqrt{\left(2x+5\right)\left(3x-1\right)}=25-5x-4=-5x+21\)
=>\(\sqrt{4\left(2x+5\right)\left(3x-1\right)}=-5x+21\)
=>\(\begin{cases}4\left(2x+5\right)\left(3x-1\right)=\left(-5x+21\right)^2\\ -5x+21\ge0\end{cases}\Rightarrow\begin{cases}25x^2-210x+441=4\left(6x^2-2x+15x-5\right)\\ -5x\ge-21\end{cases}\)
=>\(\) \(\begin{cases}25x^2-210x+441=24x^2+52x-20\\ \frac13\le x\le\frac{21}{5}\end{cases}\Rightarrow\begin{cases}x^2-262x+461=0\left(1\right)\\ \frac13\le x\le\frac{21}{5}\left(2\right)\end{cases}\)
\(\Delta=\left(262\right)^2-4\cdot1\cdot461=66800>0\)
Do đó: (1) có hai nghiệm phân biệt là:
\(\left[\begin{array}{l}x=\frac{262-20\sqrt{167}}{2\cdot1}=131-10\sqrt{167}\left(nhận\right)\\ x=\frac{262+20\sqrt{167}}{2\cdot1}=131+10\sqrt{167}\left(loại\right)\end{array}\right.\)
Vậy: \(x=131-10\sqrt{167}\)
