Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
DKXD:\,\,\,x \ge - \dfrac{1}{3}\\
\sqrt {x + 1} + \sqrt {3x + 1} = 4\\
\Leftrightarrow {\left( {\sqrt {x + 1} + \sqrt {3x + 1} } \right)^2} = {4^2}\\
\Leftrightarrow x + 1 + 2.\sqrt {\left( {x + 1} \right)\left( {3x + 1} \right)} + 3x + 1 = 16\\
\Leftrightarrow 4x + 2 + 2.\sqrt {3{x^2} + 4x + 1} = 16\\
\Leftrightarrow 2\sqrt {3{x^2} + 4x + 1} = 14 - 4x\\
\Leftrightarrow \sqrt {3{x^2} + 4x + 1} = 7 - 2x\\
\Leftrightarrow \left\{ \begin{array}{l}
7 - 2x \ge 0\\
3{x^2} + 4x + 1 = {\left( {7 - 2x} \right)^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \le \dfrac{7}{2}\\
3{x^2} + 4x + 1 = 49 - 28x + 4{x^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \le \dfrac{7}{2}\\
{x^2} - 32x + 48 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \le \dfrac{7}{2}\\
\left[ \begin{array}{l}
x = 16 + 4\sqrt 3 \\
x = 16 - 4\sqrt 3
\end{array} \right.
\end{array} \right. \Leftrightarrow x = 16 - 4\sqrt 3 \\
b,\\
DKXD:\,\,\,\,x \ge - 3\\
\sqrt {{x^2} + 10x + 21} = 3.\sqrt {x + 3} + 2\sqrt {x + 7} - 6\\
\Leftrightarrow \sqrt {{x^2} + 10x + 21} - 3\sqrt {x + 3} - 2\sqrt {x + 7} + 6 = 0\\
\Leftrightarrow \sqrt {\left( {x + 3} \right)\left( {x + 7} \right)} - 3.\sqrt {x + 3} - 2.\sqrt {x + 7} + 6 = 0\\
\Leftrightarrow \left( {\sqrt {\left( {x + 3} \right)\left( {x + 7} \right)} - 3\sqrt {x + 3} } \right) - \left( {2\sqrt {x + 7} - 6} \right) = 0\\
\Leftrightarrow \sqrt {x + 3} \left( {\sqrt {x + 7} - 3} \right) - 2.\left( {\sqrt {x + 7} - 3} \right) = 0\\
\Leftrightarrow \left( {\sqrt {x + 3} - 2} \right).\left( {\sqrt {x + 7} - 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {x + 3} = 2\\
\sqrt {x + 7} = 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x + 3 = 2\\
x + 7 = 9
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = - 1\\
x = 2
\end{array} \right.
\end{array}\)
