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

1 Đáp án

Lý thuyết liên quan

Bài 103856

1 Đáp án

Liên quan

Lý thuyết liên quan