Giải thích các bước giải:
\(2{x^2} - 11x + 23 = 4\sqrt {x + 1} \)
ĐK: \(x \ge - 1\)
Ta có:
\(\begin{array}{l}
2{x^2} - 11x + 23 = 4\sqrt {x + 1} \\
\Leftrightarrow 2{x^2} - 11x + 23 - x - 5 = 4\sqrt {x + 1} - x - 5\\
\Leftrightarrow \left( {2{x^2} - 12x + 18} \right) + \left( {x + 5} \right) - 4\sqrt {x + 1} = 0\\
\Leftrightarrow 2{\left( {x - 3} \right)^2} + \frac{{{{\left( {x + 5} \right)}^2} - 16\left( {x + 1} \right)}}{{\left( {x + 5} \right) + 4\sqrt {x + 1} }} = 0\\
\Leftrightarrow 2{\left( {x - 3} \right)^2} + \frac{{{x^2} - 6x + 9}}{{\left( {x + 5} \right) + 4\sqrt {x + 1} }} = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{\left( {x - 3} \right)^2} = 0\\
2 + \frac{1}{{x + 5 + 4\sqrt {x + 1} }} = 0
\end{array} \right.\\
x \ge - 1 \Rightarrow 2 + \frac{1}{{x + 5 + 4\sqrt {x + 1} }} > 0 \Rightarrow x = 3
\end{array}\)
