ĐKXĐ: \(x\ge-\dfrac{5}{3}\)
\(\sqrt{x^2+1}+\sqrt{3x+5}=\sqrt{x^2+6x+11}\\ \Rightarrow x^2+3x+6+2\sqrt{\left(x^2+1\right)\left(3x+5\right)}=x^2+6x+11\)
\(\Rightarrow2\sqrt{\left(x^2+1\right)\left(3x+5\right)}=3x+5\\ \Rightarrow4\left(x^2+1\right)\left(3x+5\right)=9x^2+30x+25\\ \Rightarrow4\left(3x^3+5x^2+3x+5\right)=9x^2+30x+25\\ \Rightarrow12x^3+20x^2+12x+20=9x^2+30x+25\)
\(\Rightarrow12x^3+11x^2-18x-5=0\\ \Rightarrow\left(12x^3-12x^2\right)+\left(23x^2-23x\right)+\left(5x-5\right)=0\\ \Rightarrow\left(x-1\right)\left(12x^2+23x+5\right)=0\\ \Rightarrow\left(x-1\right)\left[\left(12x^2+3x\right)+\left(20x+5\right)\right]=0\\ \Rightarrow\left(x-1\right)\left(3x+5\right)\left(4x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-\dfrac{5}{3}\left(tm\right)\\x=-\dfrac{1}{4}\left(tm\right)\end{matrix}\right.\)
\(ĐK:x\ge-\dfrac{5}{3}\\ PT\Leftrightarrow x^2+1+3x+5+2\sqrt{\left(x^2+1\right)\left(3x+5\right)}=x^2+6x+11\\ \Leftrightarrow2\sqrt{3x^3+5x^2+3x+5}=3x+5\\ \Leftrightarrow4\left(3x^3+5x^2+3x+5\right)=\left(3x+5\right)^2\\ \Leftrightarrow12x^3+20x^2+12x+20=9x^2+30x+25\\ \Leftrightarrow12x^3+11x^2-18x-5=0\\ \Leftrightarrow12x^3-12x^2+23x^2-23x+5x-5=0\\ \Leftrightarrow\left(x-1\right)\left(12x^2+23x+5\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x+5\right)\left(4x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-\dfrac{5}{3}\left(tm\right)\\x=-\dfrac{1}{4}\left(tm\right)\end{matrix}\right.\)