Bài 103800

Question

Rút gọn các biểu thức :
$ a)$ $\frac{{{a^{ - 1}} + {{(b + c)}^{ - 1}}}}{{{a^{ - 1}} - {{(b + c)}^{ - 1}}}}\left( {1 + \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}} \right){(a + b + c)^2}$
$b)$ $\left( {\frac{{{a^{\frac{1}{2}}} + 2}}{{a + 2{a^{\frac{1}{2}}} + 1}} - \frac{{{a^{\frac{1}{2}}} - 2}}{{a - 1}}} \right)\left( {\frac{{{a^{\frac{1}{2}}} + 1}}{{{a^{\frac{1}{2}}}}}} \right)$

score 0 · Answer 1

$a)$ Ta có $\frac{{{a^{ - 1}} + {{(b + c)}^{ - 1}}}}{{{a^{ - 1}} - {{(b + c)}^{ - 1}}}} =
\frac{{\frac{1}{a} + \frac{1}{{b + c}}}}{{\frac{1}{a} - \frac{1}{{b + c}}}} = \frac{{a + b +
c}}{{b + c - a}}$
Và: $1 + \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}} = \frac{{(2bc + {b^2} + {c^2}) -
{a^2}}}{{2bc}}$
$ = \frac{{{{(b + c)}^2} - {a^2}}}{{2bc}} = \frac{{(b + c +
a)(b + c - a)}}{{2bc}}$
$ \Rightarrow \frac{{{a^{ - 1}} + {{(b + c)}^{ - 1}}}}{{{a^{ - 1}} - {{(b + c)}^{ - 1}}}}\left( {1
+ \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}} \right){(a + b + c)^2}$
$ = \frac{{a + b + c}}{{b + c - a}}.\frac{{(b + c + a)(b + c - a)}}{{2bc}}.\frac{1}{{{{(a + b +

c)}^2}}} = \frac{1}{{2bc}}$
$b)$ $\left( {\frac{{{a^{\frac{1}{2}}} + 2}}{{a + 2{{\rm{a}}^{\frac{1}{2}}} + 1}} -
\frac{{{a^{\frac{1}{2}}} - 2}}{{a - 1}}} \right)\left( {\frac{{{a^{\frac{1}{2}}} +
1}}{{{a^{\frac{1}{2}}}}}} \right)$
$ = \left[ {\frac{{\sqrt a + 2}}{{{{\left( {\sqrt a + 1} \right)}^2}}} - \frac{{\sqrt a - 2}}{{\left(
{\sqrt a + 1} \right)\left( {\sqrt a - 1} \right)}}} \right]\left( {\frac{{\sqrt a + 1}}{{\sqrt a }}}
\right) = \left[ {\frac{{\sqrt a + 2}}{{\sqrt a + 1}} - \frac{{\sqrt a - 2}}{{\sqrt a - 1}}}
\right]\frac{1}{{\sqrt a }}$
$ = \frac{{a + \sqrt a - 2 - \left( {a - \sqrt a - 2} \right)}}{{a - 1}}.\frac{1}{{\sqrt a }} =
\frac{2}{{a - 1}}$

Bài 103800

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Bài 103800

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