Đáp án:
\(\begin{array}{l}
1,\\
a,\,\,\,6\sqrt 2 \\
b,\,\,\,1\\
2,\\
a,\\
x = 1\\
b,\\
x = 24
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
a,\\
6\sqrt {\dfrac{8}{9}} - 5\sqrt {\dfrac{{32}}{{25}}} + 14\sqrt {\dfrac{{18}}{{49}}} \\
= 6\sqrt {\dfrac{4}{9}.2} - 5\sqrt {\dfrac{{16}}{{25}}.2} + 14.\sqrt {\dfrac{9}{{49}}.2} \\
= 6.\sqrt {{{\left( {\dfrac{2}{3}} \right)}^2}.2} - 5\sqrt {{{\left( {\dfrac{4}{5}} \right)}^2}.2} + 14.\sqrt {{{\left( {\dfrac{3}{7}} \right)}^2}.2} \\
= 6.\dfrac{2}{3}.\sqrt 2 - 5.\dfrac{4}{5}\sqrt 2 + 14.\dfrac{3}{7}.\sqrt 2 \\
= 4\sqrt 2 - 4\sqrt 2 + 6\sqrt 2 \\
= 6\sqrt 2 \\
b,\\
\dfrac{{\sqrt 2 + 1}}{{\sqrt 2 - 1}} - \dfrac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}} - \dfrac{{8 - \sqrt 2 }}{{\sqrt 2 }}\\
= \dfrac{{{{\left( {\sqrt 2 + 1} \right)}^2} - {{\left( {\sqrt 2 - 1} \right)}^2}}}{{\left( {\sqrt 2 - 1} \right)\left( {\sqrt 2 + 1} \right)}} - \dfrac{{4.{{\sqrt 2 }^2} - \sqrt 2 }}{{\sqrt 2 }}\\
= \dfrac{{\left( {{{\sqrt 2 }^2} + 2.\sqrt 2 .1 + {1^2}} \right) - \left( {{{\sqrt 2 }^2} - 2.\sqrt 2 .1 + {1^2}} \right)}}{{{{\sqrt 2 }^2} - {1^2}}} - \dfrac{{\sqrt 2 \left( {4\sqrt 2 - 1} \right)}}{{\sqrt 2 }}\\
= \dfrac{{\left( {2 + 2\sqrt 2 + 1} \right) - \left( {2 - 2\sqrt 2 + 1} \right)}}{{2 - 1}} - \left( {4\sqrt 2 - 1} \right)\\
= \dfrac{{3 + 2\sqrt 2 - 3 + 2\sqrt 2 }}{1} - \left( {4\sqrt 2 - 1} \right)\\
= 4\sqrt 2 - \left( {4\sqrt 2 - 1} \right)\\
= 1\\
2,\\
a,\\
DKXD:\,\,\,x \ge - \dfrac{3}{2}\\
\sqrt {{x^2} + 4} = \sqrt {2x + 3} \\
\Leftrightarrow {\sqrt {{x^2} + 4} ^2} = {\sqrt {2x + 3} ^2}\\
\Leftrightarrow {x^2} + 4 = 2x + 3\\
\Leftrightarrow {x^2} + 4 - 2x - 3 = 0\\
\Leftrightarrow {x^2} - 2x + 1 = 0\\
\Leftrightarrow {x^2} - 2.x.1 + {1^2} = 0\\
\Leftrightarrow {\left( {x - 1} \right)^2} = 0\\
\Leftrightarrow x - 1 = 0\\
\Leftrightarrow x = 1\\
b,\\
DKXD:\,\,\,x \ge - 1\\
3\sqrt {4x + 4} - \sqrt {9x + 9} - 8\sqrt {\dfrac{{x + 1}}{{16}}} = 5\\
\Leftrightarrow 3\sqrt {4\left( {x + 1} \right)} - \sqrt {9\left( {x + 1} \right)} - 8\sqrt {\dfrac{1}{{16}}.\left( {x + 1} \right)} = 5\\
\Leftrightarrow 3\sqrt {{2^2}.\left( {x + 1} \right)} - \sqrt {{3^2}.\left( {x + 1} \right)} - 8\sqrt {{{\left( {\dfrac{1}{4}} \right)}^2}.\left( {x + 1} \right)} = 5\\
\Leftrightarrow 3.2\sqrt {x + 1} - 3\sqrt {x + 1} - 8.\dfrac{1}{4}.\sqrt {x + 1} = 5\\
\Leftrightarrow 6\sqrt {x + 1} - 3\sqrt {x + 1} - 2\sqrt {x + 1} = 5\\
\Leftrightarrow \sqrt {x + 1} = 5\\
\Leftrightarrow x + 1 = {5^2}\\
\Leftrightarrow x = 24
\end{array}\)
