Giải thích các bước giải:
Ta có:
\(\begin{array}{l}d,\\
\sqrt {x - 7} + \sqrt {7 - x} = 7\\
{\rm{DK:}}\,\,\,\,\left\{ \begin{array}{l}
x - 7 \ge 0\\
7 - x \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 7\\
x \le 7
\end{array} \right. \Rightarrow x = 7\\
x = 7 \Rightarrow \sqrt {x - 7} + \sqrt {7 - x} = 0\\
\Rightarrow ptvn\\
e,\\
\dfrac{{\sqrt {x + 5} }}{{\sqrt {x + 4} }} = \dfrac{{\sqrt {x - 2} }}{{\sqrt {x + 3} }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge 2} \right)\\
\Leftrightarrow \sqrt {x + 5} .\sqrt {x + 3} = \sqrt {x - 2} .\sqrt {x + 4} \\
\Leftrightarrow \left( {x + 5} \right)\left( {x + 3} \right) = \left( {x - 2} \right)\left( {x + 4} \right)\\
\Leftrightarrow {x^2} + 8x + 15 = {x^2} + 2x - 8\\
\Leftrightarrow 6x = - 23\\
\Leftrightarrow x = - \dfrac{{23}}{6}\,\,\,\,\,\,\,\left( {L,\,\,\,x \ge 2} \right)\\
\Rightarrow ptvn\\
f,\\
\sqrt {4{x^2} - 4x + 1} - 2\sqrt {9 - 12x + 4{x^2}} - x + 2 = 0\\
\Leftrightarrow \sqrt {{{\left( {2x - 1} \right)}^2}} - 2.\sqrt {{{\left( {2x - 3} \right)}^2}} - x + 2 = 0\\
\Leftrightarrow \left| {2x - 1} \right| - 2.\left| {2x - 3} \right| - x + 2 = 0\\
TH1:\,\,\,x < \dfrac{1}{2} \Rightarrow \left\{ \begin{array}{l}
2x - 1 < 0\\
2x - 3 < 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\left| {2x - 1} \right| = 1 - 2x\\
\left| {2x - 3} \right| = 3 - 2x
\end{array} \right.\\
\Rightarrow \left( {1 - 2x} \right) - 2.\left( {3 - 2x} \right) - x + 2 = 0\\
\Leftrightarrow 1 - 2x - 6 + 4x - x + 2 = 0\\
\Leftrightarrow x = 3\,\,\,\,\,\,\,\,\left( {L,\,\,x < \dfrac{1}{2}} \right)\\
TH2:\,\,\,\,\dfrac{1}{2} \le x \le \dfrac{3}{2} \Rightarrow \left\{ \begin{array}{l}
2x - 1 \ge 0\\
2x - 3 \le 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\left| {2x - 1} \right| = 2x - 1\\
\left| {2x - 3} \right| = 3 - 2x
\end{array} \right.\\
\Rightarrow \left( {2x - 1} \right) - 2.\left( {3 - 2x} \right) - x + 2 = 0\\
\Leftrightarrow 2x - 1 - 6 + 4x - x + 2 = 0\\
\Leftrightarrow x = 1\,\,\,\,\,\,\,\,\,\,\left( {t/m} \right)\\
TH3:\,\,\,x > \dfrac{3}{2} \Rightarrow \left\{ \begin{array}{l}
2x - 1 > 0\\
2x - 3 > 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\left| {2x - 1} \right| = 2x - 1\\
\left| {2x - 3} \right| = 2x - 3
\end{array} \right.\\
\Rightarrow \left( {2x - 1} \right) - 2.\left( {2x - 3} \right) - x + 2 = 0\\
\Leftrightarrow 2x - 1 - 4x + 6 - x + 2 = 0\\
\Leftrightarrow x = \dfrac{7}{3}\,\,\,\,\,\,\,\,\,\,\left( {t/m} \right)\\
g,\\
\sqrt {x + 2\sqrt {x - 1} } + \sqrt {x - 2\sqrt {x - 1} } = 2\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge 1} \right)\\
\Leftrightarrow \sqrt {\left( {x - 1} \right) + 2\sqrt {x - 1} + 1} + \sqrt {\left( {x - 1} \right) - 2\sqrt {x - 1} + 1} = 2\\
\Leftrightarrow \sqrt {{{\left( {\sqrt {x - 1} + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt {x - 1} - 1} \right)}^2}} = 2\\
\Leftrightarrow \sqrt {x - 1} + 1 + \left| {\sqrt {x - 1} - 1} \right| = 2\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\\
TH1:\,\,\,\,1 \le x \le 2 \Rightarrow 0 \le \sqrt {x - 1} \le 1 \Rightarrow \sqrt {x - 1} - 1 \le 0\\
\left( 1 \right) \Leftrightarrow \sqrt {x - 1} + 1 + \left( {1 - \sqrt {x - 1} } \right) = 2\\
\Leftrightarrow 2 = 2,\,\,\,\,\forall x \in \left[ {1;2} \right]\\
TH2:\,\,\,\,x > 2 \Rightarrow \sqrt {x - 1} > 1 \Rightarrow \sqrt {x - 1} - 1 > 0\\
\left( 1 \right) \Leftrightarrow \sqrt {x - 1} + 1 + \sqrt {x - 1} - 1 = 2\\
\Leftrightarrow \sqrt {x - 1} = 1\\
\Leftrightarrow x = 2\\
h,\\
\sqrt {3{x^2} + 6x + 7} + \sqrt {5{x^2} + 10x + 21} = 5 - 2x - {x^2}\\
\Leftrightarrow \sqrt {3.\left( {{x^2} + 2x + 1} \right) + 4} + \sqrt {5\left( {{x^2} + 2x + 1} \right) + 16} + {x^2} + 2x + 1 = 6\\
\Leftrightarrow \sqrt {3.{{\left( {x + 1} \right)}^2} + 4} + \sqrt {5.{{\left( {x + 1} \right)}^2} + 16} + {\left( {x + 1} \right)^2} = 6\\
\sqrt {3.{{\left( {x + 1} \right)}^2} + 4} + \sqrt {5.{{\left( {x + 1} \right)}^2} + 16} + {\left( {x + 1} \right)^2} \ge \sqrt {3.0 + 4} + \sqrt {5.0 + 16} + 0 = 6\\
\Rightarrow {\left( {x + 1} \right)^2} = 0 \Leftrightarrow x = - 1
\end{array}\)
