Đáp án:
\(\begin{array}{l}
1,\,\,\,\,x - 1\\
2,\,\,\,7a\\
3,\,\,\,2\\
4,\,\,\,1\\
5,\,\,\,5\\
6,\,\,\,5 - x
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
x \ge 1 \Rightarrow x - 1 \ge 0 \Rightarrow \left| {x - 1} \right| = x - 1\\
\sqrt {{{\left( {x - 1} \right)}^2}} = \left| {x - 1} \right| = x - 1\\
2,\\
a \ge 0 \Rightarrow \left| a \right| = a\\
\sqrt {144{a^2}} - 5a = \sqrt {{{12}^2}.{a^2}} - 5a = 12.\left| a \right| - 5a = 12a - 5a = 7a\\
3,\\
1 \le a \le 2 \Leftrightarrow 0 \le a - 1 \le 1 \Leftrightarrow 0 \le \sqrt {a - 1} \le 1\\
\Rightarrow \sqrt {a - 1} - 1 \le 0 \Rightarrow \left| {\sqrt {a - 1} - 1} \right| = 1 - \sqrt {a - 1} \\
\sqrt {a + 2\sqrt {a - 1} } + \sqrt {a - 2\sqrt {a - 1} } \\
= \sqrt {\left( {a - 1} \right) + 2\sqrt {a - 1} + 1} + \sqrt {\left( {a - 1} \right) - 2\sqrt {a - 1} + 1} \\
= \sqrt {{{\sqrt {a - 1} }^2} + 2.\sqrt {a - 1} .1 + {1^2}} + \sqrt {{{\sqrt {a - 1} }^2} - 2.\sqrt {a - 1} .1 + {1^2}} \\
= \sqrt {{{\left( {\sqrt {a - 1} + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt {a - 1} - 1} \right)}^2}} \\
= \left| {\sqrt {a - 1} + 1} \right| + \left| {\sqrt {a - 1} - 1} \right|\\
= \left( {\sqrt {a - 1} + 1} \right) + \left( {1 - \sqrt {a - 1} } \right)\\
= 2\\
4,\\
x > \dfrac{1}{2} \Leftrightarrow 2x > 1 \Leftrightarrow 2x - 1 > 0\\
\Rightarrow \left| {2x - 1} \right| = 2x - 1\\
2x - \sqrt {4{x^2} - 4x + 1} \\
= 2x - \sqrt {{{\left( {2x} \right)}^2} - 2.2x.1 + {1^2}} \\
= 2x - \sqrt {{{\left( {2x - 1} \right)}^2}} \\
= 2x - \left| {2x - 1} \right|\\
= 2x - \left( {2x - 1} \right)\\
= 1\\
5,\\
\sqrt {x + 2\sqrt {x - 1} } - \sqrt {x - 1} + 4\\
= \sqrt {\left( {x - 1} \right) + 2\sqrt {x - 1} + 1} - \sqrt {x - 1} + 4\\
= \sqrt {{{\sqrt {x - 1} }^2} + 2\sqrt {x - 1} .1 + {1^2}} - \sqrt {x - 1} + 4\\
= \sqrt {{{\left( {\sqrt {x - 1} + 1} \right)}^2}} - \sqrt {x - 1} + 4\\
= \left| {\sqrt {x - 1} + 1} \right| - \sqrt {x - 1} + 4\\
= \sqrt {x - 1} + 1 - \sqrt {x - 1} + 4\\
= 5\\
6,\\
x < 4 \Rightarrow x - 4 < 0 \Rightarrow \left| {x - 4} \right| = \left| {4 - x} \right| = 4 - x\\
\left| {4 - x} \right| + \dfrac{{4 - x}}{{\sqrt {{x^2} - 8x + 16} }}\\
= 4 - x + \dfrac{{4 - x}}{{\sqrt {{x^2} - 2.x.4 + {4^2}} }}\\
= 4 - x + \dfrac{{4 - x}}{{\sqrt {{{\left( {x - 4} \right)}^2}} }}\\
= 4 - x + \dfrac{{4 - x}}{{\left| {x - 4} \right|}}\\
= 4 - x + \dfrac{{4 - x}}{{4 - x}}\\
= 4 - x + 1\\
= 5 - x
\end{array}\)
