Đáp án:
\(\begin{array}{l}
1)x = 1\\
2)\left[ \begin{array}{l}
x = 0\\
x = 7
\end{array} \right.\\
3)\left[ \begin{array}{l}
x = 4\\
x = - 3
\end{array} \right.\\
4)\left[ \begin{array}{l}
x = 7\\
x = - 1
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)DK:x \ge \dfrac{2}{3}\\
\sqrt {3x - 2} = \sqrt {25 - 2.5.\sqrt 3 + 3} - \sqrt {16 - 2.4.\sqrt 3 + 3} \\
\to \sqrt {3x - 2} = \sqrt {{{\left( {5 - \sqrt 3 } \right)}^2}} - \sqrt {{{\left( {4 - \sqrt 3 } \right)}^2}} \\
\to \sqrt {3x - 2} = 5 - \sqrt 3 - \left( {4 - \sqrt 3 } \right)\\
\to \sqrt {3x - 2} = 1\\
\to 3x - 2 = 1\\
\to x = 1\\
2)\sqrt {{x^2} - 7x + 16} = \sqrt {9 - 2.3.\sqrt 2 + 2} + \sqrt {2 + 2\sqrt 2 .1 + 1} \\
\to \sqrt {{x^2} - 7x + 16} = \sqrt {{{\left( {3 - \sqrt 2 } \right)}^2}} + \sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} \\
\to \sqrt {{x^2} - 7x + 16} = 3 - \sqrt 2 + \sqrt 2 + 1\\
\to \sqrt {{x^2} - 7x + 16} = 4\\
\to {x^2} - 7x + 16 = 16\\
\to {x^2} - 7x = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = 7
\end{array} \right.\\
3)DK:{x^2} - x - 3 \ge 0 \to \left[ \begin{array}{l}
x \ge \dfrac{{1 + \sqrt {13} }}{2}\\
x \le \dfrac{{1 - \sqrt {13} }}{2}
\end{array} \right.\\
\sqrt {{x^2} - x - 3} = \sqrt {5 + 2.\sqrt 5 .1 + 1} - \left| {\sqrt 5 - 2} \right|\\
\to \sqrt {{x^2} - x - 3} = \sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} - \sqrt 5 + 2\\
\to \sqrt {{x^2} - x - 3} = \sqrt 5 + 1 - \sqrt 5 + 2\\
\to \sqrt {{x^2} - x - 3} = 3\\
\to {x^2} - x - 3 = 9\\
\to {x^2} - x - 12 = 0\\
\to \left( {x - 4} \right)\left( {x + 3} \right) = 0\\
\to \left[ \begin{array}{l}
x = 4\\
x = - 3
\end{array} \right.\\
4)\sqrt {{x^2} - 6x + 9} = \sqrt {9 - 2.3.\sqrt 7 + 7} + \sqrt {7 + 2\sqrt 7 .1 + 1} \\
\to \sqrt {{x^2} - 6x + 9} = \sqrt {{{\left( {3 - \sqrt 7 } \right)}^2}} + \sqrt {{{\left( {\sqrt 7 + 1} \right)}^2}} \\
\to \sqrt {{x^2} - 6x + 9} = 3 - \sqrt 7 + \sqrt 7 + 1\\
\to \sqrt {{x^2} - 6x + 9} = 4\\
\to {x^2} - 6x + 9 = 16\\
\to {x^2} - 6x - 7 = 0\\
\to \left( {x - 7} \right)\left( {x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 7\\
x = - 1
\end{array} \right.
\end{array}\)
