Đáp án:
\(\begin{array}{l}
1,\\
x \ge 2\\
2,\\
\left\{ \begin{array}{l}
x = 12\\
x = - 6
\end{array} \right.\\
3,\\
\left[ \begin{array}{l}
x = \dfrac{5}{2}\\
x = - \dfrac{7}{2}
\end{array} \right.\\
4,\\
\left[ \begin{array}{l}
x = 8\\
x = - 2
\end{array} \right.\\
5,\\
\left[ \begin{array}{l}
x = 2\\
x = 6
\end{array} \right.\\
6,\\
x = \dfrac{5}{4}\\
7,\\
x = 19\\
8,\\
x = \sqrt {11} \\
9,\\
x = 15\\
10,\\
x = 25
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
DKXD:\,\,\,4 - 4x + {x^2} \ge 0,\,\,\,\forall x\\
\sqrt {4 - 4x + {x^2}} = x - 2\\
\Leftrightarrow \sqrt {{2^2} - 2.2.x + {x^2}} = x - 2\\
\Leftrightarrow \sqrt {{{\left( {2 - x} \right)}^2}} = x - 2\\
\Leftrightarrow \left| {2 - x} \right| = x - 2\\
\Leftrightarrow \left\{ \begin{array}{l}
x - 2 \ge 0\\
\left[ \begin{array}{l}
2 - x = x - 2\\
2 - x = - \left( {x - 2} \right)
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 2\\
\left[ \begin{array}{l}
2 + 2 = x + x\\
2 - x = 2 - x
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x \ge 2\\
\left[ \begin{array}{l}
x = 2\\
\forall x
\end{array} \right.
\end{array} \right. \Leftrightarrow x \ge 2\\
2,\\
DKXD:\,\,\,{\left( {x - 3} \right)^2} \ge 0,\,\,\,\forall x\\
\sqrt {{{\left( {x - 3} \right)}^2}} = 9\\
\Leftrightarrow \left| {x - 3} \right| = 9\\
\Leftrightarrow \left[ \begin{array}{l}
x - 3 = 9\\
x - 3 = - 9
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 9 + 3\\
x = - 9 + 3
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 12\\
x = - 6
\end{array} \right.\\
3,\\
DKXD:\,\,\,4{x^2} + 4x + 1 \ge 0,\,\,\,\forall x\\
\sqrt {4{x^2} + 4x + 1} = 6\\
\Leftrightarrow \sqrt {{{\left( {2x} \right)}^2} + 2.2x.1 + {1^2}} = 6\\
\Leftrightarrow \sqrt {{{\left( {2x + 1} \right)}^2}} = 6\\
\Leftrightarrow \left| {2x + 1} \right| = 6\\
\Leftrightarrow \left[ \begin{array}{l}
2x + 1 = 6\\
2x + 1 = - 6
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x = 5\\
2x = - 7
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{5}{2}\\
x = - \dfrac{7}{2}
\end{array} \right.\\
4,\\
DKXD:\,\,\,{x^2} - 6x + 9 \ge 0,\,\,\forall x\\
\sqrt {{x^2} - 6x + 9} = 5\\
\Leftrightarrow \sqrt {{x^2} - 2.x.3 + {3^2}} = 5\\
\Leftrightarrow \sqrt {{{\left( {x - 3} \right)}^2}} = 5\\
\Leftrightarrow \left| {x - 3} \right| = 5\\
\Leftrightarrow \left[ \begin{array}{l}
x - 3 = 5\\
x - 3 = - 5
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 5 + 3\\
x = - 5 + 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 8\\
x = - 2
\end{array} \right.\\
5,\\
DKXD:\,\,\,16 - 8x + {x^2} \ge 0,\,\,\,\forall x\\
\sqrt {16 - 8x + {x^2}} + 5 = 7\\
\Leftrightarrow \sqrt {16 - 8x + {x^2}} = 2\\
\Leftrightarrow \sqrt {{4^2} - 2.4.x + {x^2}} = 2\\
\Leftrightarrow \sqrt {{{\left( {4 - x} \right)}^2}} = 2\\
\Leftrightarrow \left| {4 - x} \right| = 2\\
\Leftrightarrow \left[ \begin{array}{l}
4 - x = 2\\
4 - x = - 2
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = 6
\end{array} \right.\\
6,\\
DKXD:\,\,\,x \ge 1\\
\sqrt {4x - 4} + \sqrt {9x - 9} = 3 - \sqrt {x - 1} \\
\Leftrightarrow \sqrt {4.\left( {x - 1} \right)} + \sqrt {9\left( {x - 1} \right)} = 3 - \sqrt {x - 1} \\
\Leftrightarrow \sqrt {{2^2}.\left( {x - 1} \right)} + \sqrt {{3^2}.\left( {x - 1} \right)} = 3 - \sqrt {x - 1} \\
\Leftrightarrow 2\sqrt {x - 1} + 3\sqrt {x - 1} = 3 - \sqrt {x - 1} \\
\Leftrightarrow 2\sqrt {x - 1} + 3\sqrt {x - 1} + \sqrt {x - 1} = 3\\
\Leftrightarrow 6\sqrt {x - 1} = 3\\
\Leftrightarrow \sqrt {x - 1} = \dfrac{1}{2}\\
\Leftrightarrow x - 1 = {\left( {\dfrac{1}{2}} \right)^2}\\
\Leftrightarrow x - 1 = \dfrac{1}{4}\\
\Leftrightarrow x = \dfrac{5}{4}\\
7,\\
DKXD:\,\,\,x \ge 3\\
\sqrt {x - 3} + \sqrt {4x - 12} = 12\\
\Leftrightarrow \sqrt {x - 3} + \sqrt {4.\left( {x - 3} \right)} = 12\\
\Leftrightarrow \sqrt {x - 3} + \sqrt {{2^2}.\left( {x - 3} \right)} = 12\\
\Leftrightarrow \sqrt {x - 3} + 2\sqrt {x - 3} = 12\\
\Leftrightarrow 3\sqrt {x - 3} = 12\\
\Leftrightarrow \sqrt {x - 3} = 4\\
\Leftrightarrow x - 3 = {4^2}\\
\Leftrightarrow x - 3 = 16\\
\Leftrightarrow x = 19\\
8,\\
{x^2} - 2\sqrt {11} x + 11 = 0\\
\Leftrightarrow {x^2} - 2.x.\sqrt {11} + {\sqrt {11} ^2} = 0\\
\Leftrightarrow {\left( {x - \sqrt {11} } \right)^2} = 0\\
\Leftrightarrow x - \sqrt {11} = 0\\
\Leftrightarrow x = \sqrt {11} \\
9,\\
DKXD:\,\,\,x \ge - 1\\
\sqrt {16x + 16} - \sqrt {9x + 9} + \sqrt {4x + 4} + \sqrt {x + 1} = 16\\
\Leftrightarrow \sqrt {16.\left( {x + 1} \right)} - \sqrt {9\left( {x + 1} \right)} + \sqrt {4.\left( {x + 1} \right)} + \sqrt {x + 1} = 16\\
\Leftrightarrow \sqrt {{4^2}.\left( {x + 1} \right)} - \sqrt {{3^2}.\left( {x + 1} \right)} + \sqrt {{2^2}.\left( {x + 1} \right)} + \sqrt {x + 1} = 16\\
\Leftrightarrow 4\sqrt {x + 1} - 3\sqrt {x + 1} + 2\sqrt {x + 1} + \sqrt {x + 1} = 16\\
\Leftrightarrow \sqrt {x + 1} .\left( {4 - 3 + 2 + 1} \right) = 16\\
\Leftrightarrow \sqrt {x + 1} .4 = 16\\
\Leftrightarrow \sqrt {x + 1} = 4\\
\Leftrightarrow x + 1 = {4^2}\\
\Leftrightarrow x + 1 = 16\\
\Leftrightarrow x = 15\\
10,\\
DKXD:\,\,\,x \ge 0\\
3\sqrt x - 2\sqrt {9x} + \sqrt {16x} = 5\\
\Leftrightarrow 3\sqrt x - 2\sqrt {{3^2}.x} + \sqrt {{4^2}.x} = 5\\
\Leftrightarrow 3\sqrt x - 2.3\sqrt x + 4\sqrt x = 5\\
\Leftrightarrow 3\sqrt x - 6\sqrt x + 4\sqrt x = 5\\
\Leftrightarrow \sqrt x = 5\\
\Leftrightarrow x = {5^2}\\
\Leftrightarrow x = 25
\end{array}\)
