Bài 103856

Question

Rút gọn các biểu thức :
$a)$ ${\left( {\frac{{\sqrt[4]{{ax^3}} - \sqrt[4]{{{a^3}x}}}}{{\sqrt a - \sqrt x }} - \frac{{1 + \sqrt {ax } }}{{\sqrt[4]{{ax}}}}} \right)^{ - 2}}\sqrt {1 + 2\sqrt {\frac{a}{x} + \frac{a}{x}} } $
(với $a > 0, x > 0, a \neq x$)
$b$) $\left( {\frac{{\sqrt {1 + a} }}{{\sqrt {1 + a} - \sqrt {1 - a} }} + \frac{{1 - a}}{{\sqrt {1 - {a^2}} - 1 + a}}} \right)\left( {\sqrt {\frac{1}{{{a^2}}}-1} - \frac{1}{a}} \right)$
(với $0 < a < 1$)

score 0 · Answer 1

$a)$
Ta có :     $A = \frac{{\sqrt[4]{{{\rm{a}}{{\rm{x}}^3}}} - \sqrt[4]{{{a^3}x}}}}{{\sqrt a - \sqrt
x }} - \frac{{1 + \sqrt {{\rm{ax}}} }}{{\sqrt[4]{{{\rm{ax}}}}}}$
= $\frac{{ - \sqrt[4]{{{\rm{ax}}}}\left( {\sqrt[4]{{{a^2}}} - \sqrt[4]{{{x^2}}}} \right)}}{{\sqrt a
- \sqrt x }} + \frac{{1 + \sqrt {{\rm{ax}}} }}{{\sqrt[4]{{{\rm{ax}}}}}}$
$ = \frac{{ - \sqrt[4]{{{\rm{ax}}}}\left( {\sqrt a - \sqrt x } \right)}}{{\sqrt a - \sqrt x }} +
\frac{{1 + \sqrt {{\rm{ax}}} }}{{\sqrt[4]{{{\rm{ax}}}}}}$
$ = - \sqrt[4]{{{\rm{ax}}}} + \frac{{1 + \sqrt {{\rm{ax}}} }}{{\sqrt[4]{{{\rm{ax}}}}}} = \frac{{
- \sqrt {{\rm{ax}}} + 1 + \sqrt {{\rm{ax}}} }}{{\sqrt[4]{{{\rm{ax}}}}}} =
\frac{1}{{\sqrt[4]{{{\rm{ax}}}}}}$
${A^{ - 2}} = {\left( {\frac{1}{{\sqrt[4]{{{\rm{ax}}}}}}} \right)^{ - 2}} = \sqrt {{\rm{ax}}} $
$B = \sqrt {1 + 2\sqrt {\frac{a}{x} + \frac{a}{x}} } = \sqrt {1 + 2\sqrt {\frac{a}{x} + {{\left(
{\sqrt {\frac{a}{x}} } \right)}^2}} }$
$ = \sqrt {{{\left( {1 + \sqrt {\frac{a}{x}} } \right)}^2}} = 1
+ \sqrt {\frac{a}{x}} $
$ \Rightarrow {A^{ - 2}}.B = \sqrt {{\rm{ax}}} \left( {1 + \sqrt {\frac{a}{x}} } \right) = \sqrt
{{\rm{ax}}} + a$
$b)$ E $ = \frac{{\sqrt {1 + a} }}{{\sqrt {1 + a} - \sqrt {1 - a} }} + \frac{{1 - a}}{{\sqrt {1 -
{a^2}} - 1 + a}}$
    $ = \frac{{\sqrt {1 + a} }}{{\sqrt {1 + a} - \sqrt {1 - a} }} + \frac{{{{\left( {\sqrt {1 -
a} } \right)}^2}}}{{\sqrt {1 - a} \sqrt {1 + a} - (1 - a)}}$
    $ = \frac{{\sqrt {1 + a} }}{{\sqrt {1 + a} - \sqrt {1 - a} }} + \frac{{{{\left( {\sqrt {1 -
a} } \right)}^2}}}{{\sqrt {1 - a} \left( {\sqrt {1 + a} - \sqrt {1 - a} } \right)}}$
    $ = \frac{{\sqrt {1 + a} }}{{\sqrt {1 + a} - \sqrt {1 - a} }} + \frac{{\sqrt {1 - a}
}}{{\sqrt {1 + a} - \sqrt {1 - a} }}$
    $ = \frac{{\sqrt {1 + a} + \sqrt {1 - a} }}{{\sqrt {1 + a} - \sqrt {1 - a} }} =
\frac{{{{\left( {\sqrt {1 + a} + \sqrt {1 - a} } \right)}^2}}}{{2{\rm{a}}}}$
    $ = \frac{{1 + a + 2\sqrt {1 - {a^2} + 1 - a} }}{{2{\rm{a}}}} = \frac{{1 + \sqrt {1 -
{a^2}} }}{a}$
    $F = \sqrt {\frac{1}{{{a^2}}} - 1} - \frac{1}{a} = \sqrt {\frac{{1 - {a^2}}}{{{a^2}}}}
- \frac{1}{a} = \frac{{\sqrt {1 - {a^2}} }}{a} - \frac{1}{a} = \frac{{\sqrt {1 - {a^2}} - 1}}{a}$
    $ \Rightarrow E.F = \frac{{1 + \sqrt {1 - {a^2}} }}{a}.\frac{{\sqrt {1 - {a^2}} - 1}}{a}
= \frac{{1 - {a^2} - 1}}{{{a^2}}} = - \frac{{{a^2}}}{{{a^2}}} = - 1$

Bài 103856

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Bài 103856

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