Đáp án:
\(\begin{array}{l}
4.1\\
\left[ \begin{array}{l}
x = \sqrt 5 + 2\\
x = - \sqrt 5 - 2
\end{array} \right.\\
4.2\\
\left[ \begin{array}{l}
x = \sqrt 3 \\
x = 2 - \sqrt 3
\end{array} \right.\\
4.3\\
\left[ \begin{array}{l}
x = 2\sqrt 2 + 5\\
x = - 2\sqrt 2 + 5
\end{array} \right.\\
4.4\\
\left[ \begin{array}{l}
x = 10\\
x = - 4
\end{array} \right.\\
4.5\\
\left[ \begin{array}{l}
x = - 1\\
x = 1
\end{array} \right.\\
4.6\\
x \le 4\\
4.7\\
\left[ \begin{array}{l}
x = - 4\\
x = 5
\end{array} \right.\\
4.8\\
\left[ \begin{array}{l}
x = 4\\
x = 9
\end{array} \right.\\
4.9\\
x = \dfrac{{25}}{{144}}\\
4.10\\
x = 2\\
4.11\\
x = - 1\\
4.12\\
x = 4
\end{array}\)
\(\begin{array}{l}
4.13\\
x = 5\\
4.14\\
x = 3\\
4.15\\
x = 14\\
4.16\\
x = 4
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
4.1\\
{x^2} = 9 + 4\sqrt 5 \\
\Leftrightarrow {x^2} = 5 + 4\sqrt 5 + 4\\
\Leftrightarrow {x^2} = {\sqrt 5 ^2} + 2.\sqrt 5 .2 + {2^2}\\
\Leftrightarrow {x^2} = {\left( {\sqrt 5 + 2} \right)^2}\\
\Leftrightarrow \left[ \begin{array}{l}
x = \sqrt 5 + 2\\
x = - \left( {\sqrt 5 + 2} \right)
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \sqrt 5 + 2\\
x = - \sqrt 5 - 2
\end{array} \right.\\
4.2\\
{x^2} - 2x + 1 = 4 - 2\sqrt 3 \\
\Leftrightarrow {x^2} - 2.x.1 + {1^2} = 3 - 2\sqrt 3 + 1\\
\Leftrightarrow {\left( {x - 1} \right)^2} = {\sqrt 3 ^2} - 2.\sqrt 3 .1 + {1^2}\\
\Leftrightarrow {\left( {x - 1} \right)^2} = {\left( {\sqrt 3 - 1} \right)^2}\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = \sqrt 3 - 1\\
x - 1 = - \left( {\sqrt 3 - 1} \right)
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x - 1 = \sqrt 3 - 1\\
x - 1 = - \sqrt 3 + 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \sqrt 3 - 1 + 1\\
x = - \sqrt 3 + 1 + 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \sqrt 3 \\
x = 2 - \sqrt 3
\end{array} \right.\\
4.3\\
{x^2} - 10x + 25 = 8\\
\Leftrightarrow {x^2} - 2.x.5 + {5^2} = {2^2}.2\\
\Leftrightarrow {\left( {x - 5} \right)^2} = {\left( {2\sqrt 2 } \right)^2}\\
\Leftrightarrow \left[ \begin{array}{l}
x - 5 = 2\sqrt 2 \\
x - 5 = - 2\sqrt 2
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\sqrt 2 + 5\\
x = - 2\sqrt 2 + 5
\end{array} \right.\\
4.4\\
DKXD:\,\,\,{x^2} - 6x + 9 \ge 0\\
\sqrt {{x^2} - 6x + 9} = 7\\
\Leftrightarrow \sqrt {{x^2} - 2.x.3 + {3^2}} = 7\\
\Leftrightarrow \sqrt {{{\left( {x - 3} \right)}^2}} = 7\\
\Leftrightarrow \left| {x - 3} \right| = 7\\
\Leftrightarrow \left[ \begin{array}{l}
x - 3 = 7\\
x - 3 = - 7
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 7 + 3\\
x = - 7 + 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 10\\
x = - 4
\end{array} \right.\\
4.5\\
DKXD:\,\,\,9{x^2} - 6x + 1 \ge 0\\
\sqrt {9{x^2} - 6x + 1} = \left| {x - 3} \right|\\
\Leftrightarrow \sqrt {{{\left( {3x} \right)}^2} - 2.3x.1 + {1^2}} = \left| {x - 3} \right|\\
\Leftrightarrow \sqrt {{{\left( {3x - 1} \right)}^2}} = \left| {x - 3} \right|\\
\Leftrightarrow \left| {3x - 1} \right| = \left| {x - 3} \right|\\
\Leftrightarrow \left[ \begin{array}{l}
3x - 1 = x - 3\\
3x - 1 = 3 - x
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
3x - 1 - x + 3 = 0\\
3x - 1 - 3 + x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x + 2 = 0\\
4x - 4 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = - 1\\
x = 1
\end{array} \right.\\
4.6\\
DKXD:\,\,\,{x^2} - 8x + 16 \ge 0\\
\sqrt {{x^2} - 8x + 16} = 4 - x\\
\Leftrightarrow \sqrt {{x^2} - 2.x.4 + {4^2}} = 4 - x\\
\Leftrightarrow \sqrt {{{\left( {x - 4} \right)}^2}} = 4 - x\\
\Leftrightarrow \left| {x - 4} \right| = 4 - x\\
\Leftrightarrow \left\{ \begin{array}{l}
4 - x \ge 0\\
\left[ \begin{array}{l}
x - 4 = 4 - x\\
x - 4 = - \left( {4 - x} \right)
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \le 4\\
\left[ \begin{array}{l}
x - 4 - 4 + x = 0\\
x - 4 = - 4 + x
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \le 4\\
\left[ \begin{array}{l}
2x - 8 = 0\\
x - 4 = x - 4
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \le 4\\
\left[ \begin{array}{l}
x = 4\\
\forall x
\end{array} \right.
\end{array} \right. \Leftrightarrow x \le 4\\
4.7\\
DKXD:\,\,\,\left\{ \begin{array}{l}
{x^2} - 16 \ge 0\\
x + 4 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{x^2} \ge 16\\
x \ge - 4
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 4\\
x \le - 4
\end{array} \right.\\
x \ge - 4
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x \ge 4\\
x = - 4
\end{array} \right.\\
\sqrt {{x^2} - 16} - \sqrt {x + 4} = 0\\
\Leftrightarrow \sqrt {\left( {x - 4} \right)\left( {x + 4} \right)} - \sqrt {x + 4} = 0\\
\Leftrightarrow \sqrt {x + 4} .\left( {\sqrt {x - 4} - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {x + 4} = 0\\
\sqrt {x - 4} - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sqrt {x + 4} = 0\\
\sqrt {x - 4} = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x + 4 = 0\\
x - 4 = 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = - 4\\
x = 5
\end{array} \right.\\
4.8\\
DKXD:\,\,\,x \ge 0\\
x - 5\sqrt x + 6 = 0\\
\Leftrightarrow \left( {x - 2\sqrt x } \right) + \left( { - 3\sqrt x + 6} \right) = 0\\
\Leftrightarrow \sqrt x .\left( {\sqrt x - 2} \right) - 3.\left( {\sqrt x - 2} \right) = 0\\
\Leftrightarrow \left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x - 2 = 0\\
\sqrt x - 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sqrt x = 2\\
\sqrt x = 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 4\\
x = 9
\end{array} \right.\\
4.9\\
DKXD:\,\,\,x \ge 0\\
- 5x + 7\sqrt x + 12 = 0\\
\Leftrightarrow \left( { - 5x - 5\sqrt x } \right) + \left( {12\sqrt x + 12} \right) = 0\\
\Leftrightarrow - 5\sqrt x \left( {\sqrt x + 1} \right) + 12.\left( {\sqrt x + 1} \right) = 0\\
\Leftrightarrow \left( {\sqrt x + 1} \right)\left( { - 5\sqrt x + 12} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x + 1 = 0\\
- 5\sqrt x + 12 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sqrt x = - 1\,\,\,\,\,\left( L \right)\\
\sqrt x = \dfrac{5}{{12}}
\end{array} \right.\\
\Rightarrow \sqrt x = \dfrac{5}{{12}} \Leftrightarrow x = {\left( {\dfrac{5}{{12}}} \right)^2} \Leftrightarrow x = \dfrac{{25}}{{144}}\\
4.10\\
DKXD:\,\,{x^2} - 2x \ge 0 \Leftrightarrow x\left( {x - 2} \right) \ge 0\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge 0\\
x - 2 \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le 0\\
x - 2 \le 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge 0\\
x \ge 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le 0\\
x \le 2
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x \ge 2\\
x \le 0
\end{array} \right.\\
\sqrt {{x^2} - 2x} = 2 - x\\
\Leftrightarrow \left\{ \begin{array}{l}
2 - x \ge 0\\
{\sqrt {{x^2} - 2x} ^2} = {\left( {2 - x} \right)^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \le 2\\
{x^2} - 2x = {\left( {x - 2} \right)^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \le 2\\
x\left( {x - 2} \right) = {\left( {x - 2} \right)^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \le 2\\
\left[ \begin{array}{l}
x - 2 = 0\\
x = x - 2
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \le 2\\
\left[ \begin{array}{l}
x = 2\\
0 = - 2\,\,\,\,\left( L \right)
\end{array} \right.
\end{array} \right. \Leftrightarrow x = 2\\
4.11\\
DKXD:\,\,\,x \ge - \dfrac{{27}}{2}\\
\sqrt {2x + 27} - x = 6\\
\Leftrightarrow \sqrt {2x + 27} = x + 6\\
\Leftrightarrow \left\{ \begin{array}{l}
x + 6 \ge 0\\
2x + 27 = {\left( {x + 6} \right)^2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge - 6\\
2x + 27 = {x^2} + 2.x.6 + {6^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge - 6\\
2x + 27 = {x^2} + 12x + 36
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge - 6\\
{x^2} + 10x + 9 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge - 6\\
\left( {{x^2} + x} \right) + \left( {9x + 9} \right) = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge - 6\\
x\left( {x + 1} \right) + 9\left( {x + 1} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge - 6\\
\left( {x + 1} \right)\left( {x + 9} \right) = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge - 6\\
\left[ \begin{array}{l}
x + 1 = 0\\
x + 9 = 0
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge - 6\\
\left[ \begin{array}{l}
x = - 1\\
x = - 9
\end{array} \right.
\end{array} \right. \Leftrightarrow x = - 1\\
4.12\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x + 3 \ge 0\\
2x - 1 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge - 3\\
x \ge \dfrac{1}{2}
\end{array} \right. \Leftrightarrow x \ge \dfrac{1}{2}\\
\sqrt {x + 3} = \sqrt {2x - 1} \\
\Leftrightarrow x + 3 = 2x - 1\\
\Leftrightarrow 2x - 1 - x - 3 = 0\\
\Leftrightarrow x - 4 = 0\\
\Leftrightarrow x = 4
\end{array}\)
\(\begin{array}{l}
4.13\\
DKXD:\,\,\,x \ge 1\\
\sqrt {x - 1} + \dfrac{3}{2}\sqrt {4x - 4} - \dfrac{2}{5}\sqrt {25x - 25} - 4 = 0\\
\Leftrightarrow \sqrt {x - 1} + \dfrac{3}{2}\sqrt {4.\left( {x - 1} \right)} - \dfrac{2}{5}\sqrt {25.\left( {x - 1} \right)} - 4 = 0\\
\Leftrightarrow \sqrt {x - 1} + \dfrac{3}{2}\sqrt {{2^2}.\left( {x - 1} \right)} - \dfrac{2}{5}\sqrt {{5^2}.\left( {x - 1} \right)} - 4 = 0\\
\Leftrightarrow \sqrt {x - 1} + \dfrac{3}{2}.2\sqrt {x - 1} - \dfrac{2}{5}.5\sqrt {x - 1} - 4 = 0\\
\Leftrightarrow \sqrt {x - 1} + 3\sqrt {x - 1} - 2\sqrt {x - 1} - 4 = 0\\
\Leftrightarrow 2\sqrt {x - 1} - 4 = 0\\
\Leftrightarrow \sqrt {x - 1} = 2\\
\Leftrightarrow x - 1 = {2^2}\\
\Leftrightarrow x = 5\\
4.14\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x + 1 \ge 0\\
x + 6 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge - 1\\
x \ge - 6
\end{array} \right. \Leftrightarrow x \ge - 1\\
\sqrt {x + 1} + \sqrt {x + 6} = 5\\
\Leftrightarrow {\left( {\sqrt {x + 1} + \sqrt {x + 6} } \right)^2} = {5^2}\\
\Leftrightarrow {\sqrt {x + 1} ^2} + 2.\sqrt {x + 1} .\sqrt {x + 6} + {\sqrt {x + 6} ^2} = 25\\
\Leftrightarrow x + 1 + 2.\sqrt {\left( {x + 1} \right)\left( {x + 6} \right)} + x + 6 = 25\\
\Leftrightarrow 2x + 7 + 2.\sqrt {{x^2} + 6x + x + 6} = 25\\
\Leftrightarrow 2.\sqrt {{x^2} + 7x + 6} = 18 - 2x\\
\Leftrightarrow \sqrt {{x^2} + 7x + 6} = 9 - x\\
\Leftrightarrow \left\{ \begin{array}{l}
9 - x \ge 0\\
{x^2} + 7x + 6 = {\left( {9 - x} \right)^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \le 9\\
{x^2} + 7x + 6 = 81 - 18x + {x^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \le 9\\
7x + 6 - 81 + 18x = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \le 9\\
25x - 75 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \le 9\\
x = 3
\end{array} \right. \Leftrightarrow x = 3\\
4.15\\
DKXD:\,\,\,x \ge 5\\
\sqrt {4x - 20} + 3\sqrt {\dfrac{{x - 5}}{9}} - \dfrac{1}{3}\sqrt {9x - 45} = 6\\
\Leftrightarrow \sqrt {4.\left( {x - 5} \right)} + 3\sqrt {\dfrac{1}{9}.\left( {x - 5} \right)} - \dfrac{1}{3}\sqrt {9\left( {x - 5} \right)} = 6\\
\Leftrightarrow \sqrt {{2^2}.\left( {x - 5} \right)} + 3\sqrt {{{\left( {\dfrac{1}{3}} \right)}^2}.\left( {x - 5} \right)} - \dfrac{1}{3}\sqrt {{3^2}.\left( {x - 5} \right)} = 6\\
\Leftrightarrow 2\sqrt {x - 5} + 3.\dfrac{1}{3}\sqrt {x - 5} - \dfrac{1}{3}.3\sqrt {x - 5} = 6\\
\Leftrightarrow 2\sqrt {x - 5} + \sqrt {x - 5} - \sqrt {x - 5} = 6\\
\Leftrightarrow 2\sqrt {x - 5} = 6\\
\Leftrightarrow \sqrt {x - 5} = 3\\
\Leftrightarrow x - 5 = {3^2}\\
\Leftrightarrow x = 14\\
4.16\\
DKXD:\,\,\,x \ge 0\\
3\sqrt x - 3 + 2.\left( {2\sqrt x + 1} \right) = 13\\
\Leftrightarrow 3\sqrt x - 3 + 4\sqrt x + 2 = 13\\
\Leftrightarrow 7\sqrt x - 1 = 13\\
\Leftrightarrow 7\sqrt x = 14\\
\Leftrightarrow \sqrt x = 2\\
\Leftrightarrow x = {2^2}\\
\Leftrightarrow x = 4
\end{array}\)
