Đáp án:
\(\begin{array}{l}
2,\\
a,\\
8\sqrt 2 \\
b,\\
- 1\\
c,\\
\dfrac{{147}}{{10}}\\
3,\\
a,\\
- \dfrac{{292}}{9}\\
b,\\
\dfrac{1}{{b - a}}\\
4,\\
a,\\
\left[ \begin{array}{l}
x = \dfrac{8}{3}\\
x = 0
\end{array} \right.\\
b,\\
Phương\,\,\,trình\,\,\,vô\,\,\,nghiệm
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2,\\
a,\\
2\sqrt 2 - 3\sqrt {18} + 5\sqrt {32} - \sqrt {50} \\
= 2\sqrt 2 - 3.\sqrt {9.2} + 5.\sqrt {16.2} - \sqrt {25.2} \\
= 2\sqrt 2 - 3\sqrt {{3^2}.2} + 5\sqrt {{4^2}.2} - \sqrt {{5^2}.2} \\
= 2\sqrt 2 - 3.3\sqrt 2 + 5.4\sqrt 2 - 5\sqrt 2 \\
= 2\sqrt 2 - 9\sqrt 2 + 20\sqrt 2 - 5\sqrt 2 \\
= 8\sqrt 2 \\
b,\\
\dfrac{{\sqrt {0,36} - \sqrt {0,49} }}{{\sqrt {0,04} - \sqrt {0,01} }} = \dfrac{{\sqrt {{{0,6}^2}} - \sqrt {{{0,7}^2}} }}{{\sqrt {{{0,2}^2}} - \sqrt {{{0,1}^2}} }} = \dfrac{{0,6 - 0,7}}{{0,2 - 0,1}} = \dfrac{{ - 0,1}}{{0,1}} = - 1\\
c,\\
\sqrt {\dfrac{{81}}{{25}}.3\dfrac{1}{{16}}.\dfrac{{196}}{9}} = \sqrt {\dfrac{{81}}{{25}}.\dfrac{{49}}{{16}}.\dfrac{{196}}{9}} = \sqrt {\dfrac{{{9^2}}}{{{5^2}}}.\dfrac{{{7^2}}}{{{4^2}}}.\dfrac{{{{14}^2}}}{{{3^2}}}} \\
= \dfrac{{9.7.14}}{{5.4.3}} = \dfrac{{147}}{{10}}\\
3,\\
a,\\
\left( {\sqrt {\dfrac{1}{3}} .\sqrt {\dfrac{1}{3}} - \dfrac{3}{4}.\sqrt {36} + \dfrac{1}{6}.\sqrt {\dfrac{{36}}{{81}}} } \right):\sqrt {{{\left( { - \dfrac{1}{8}} \right)}^2}} \\
= \left( {{{\sqrt {\dfrac{1}{3}} }^2} - \dfrac{3}{4}.\sqrt {{6^2}} + \dfrac{1}{6}.\sqrt {\dfrac{{{6^2}}}{{{9^2}}}} } \right):\left| { - \dfrac{1}{8}} \right|\\
= \left( {\dfrac{1}{3} - \dfrac{3}{4}.6 + \dfrac{1}{6}.\dfrac{6}{9}} \right):\dfrac{1}{8}\\
= \left( {\dfrac{1}{3} - \dfrac{9}{2} + \dfrac{1}{9}} \right):\dfrac{1}{8}\\
= \left( { - \dfrac{{73}}{{18}}} \right).8\\
= - \dfrac{{292}}{9}\\
b,\\
a,b \ge 0 \Rightarrow a + b \ge 0 \Rightarrow \left| {a + b} \right| = a + b\\
\dfrac{{ - 2}}{{{a^2} - {b^2}}}.\sqrt {\dfrac{{{{\left( {a + b} \right)}^2}}}{4}} \\
= \dfrac{{ - 2}}{{\left( {a - b} \right)\left( {a + b} \right)}}.\sqrt {\dfrac{{{{\left( {a + b} \right)}^2}}}{{{2^2}}}} \\
= \dfrac{{ - 2}}{{\left( {a - b} \right)\left( {a + b} \right)}}.\dfrac{{\left| {a + b} \right|}}{2}\\
= \dfrac{{ - 2}}{{\left( {a - b} \right)\left( {a + b} \right)}}.\dfrac{{a + b}}{2}\\
= \dfrac{{ - 1}}{{a - b}}\\
= \dfrac{1}{{b - a}}\\
4,\\
a,\\
\sqrt {{{\left( {3x - 4} \right)}^2}} = 4\\
\Leftrightarrow \left| {3x - 4} \right| = 4\\
\Leftrightarrow \left[ \begin{array}{l}
3x - 4 = 4\\
3x - 4 = - 4
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
3x = 8\\
3x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{8}{3}\\
x = 0
\end{array} \right.\\
b,\\
DKXD:\,\,\,x \ge 2\\
\sqrt {4x - 8} - 5\sqrt {x - 2} + \sqrt {9x - 18} = 20\\
\Leftrightarrow \sqrt {4\left( {x - 2} \right)} - 5\sqrt {x - 2} + \sqrt {9\left( {x - 2} \right)} = 20\\
\Leftrightarrow 2\sqrt {x - 2} - 5\sqrt {x - 2} + 3\sqrt {x - 2} = 20\\
\Leftrightarrow 0.\sqrt {x - 2} = 20\\
\Leftrightarrow 0 = 20\\
\Rightarrow Phương\,\,\,trình\,\,\,vô\,\,\,nghiệm
\end{array}\)
