Câu hỏi của Anh Mai

Question

A = $\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right).\left(\sqrt{a}+\frac{1}{\sqrt{a}}\right)$

a) rút gọn A

b) Tính A với a  = \(\left(4+\sqrt{15}\right)\left(\s...

Nguyễn Thành Trương · Accepted Answer

$\begin{array}{l}
a)A = \left( {\dfrac{{\sqrt a  + 1}}{{\sqrt a  - 1}} - \dfrac{{\sqrt a  - 1}}{{\sqrt a  + 1}} + 4\sqrt a } \right).\left( {\sqrt a  + \dfrac{1}{{\sqrt a }}} \right)\
 = \left[ {\dfrac{{{{\left( {\sqrt a  + 1} \right)}^2} - {{\left( {\sqrt a  - 1} \right)}^2}}}{{\left( {\sqrt a  - 1} \right)\left( {\sqrt a  + 1} \right)}} + 4\sqrt a } \right].\dfrac{{a + 1}}{{\sqrt a }}\
 = \left[ {\dfrac{{4\sqrt a }}{{\left( {\sqrt a  - 1} \right)\left( {\sqrt a  + 1} \right)}} + 4\sqrt a } \right].\dfrac{{a + 1}}{{\sqrt a }}\
 = \dfrac{{4\sqrt a  + 4\sqrt a \left( {\sqrt a  - 1} \right)\left( {\sqrt a  + 1} \right)}}{{\left( {\sqrt a  - 1} \right)\left( {\sqrt a  + 1} \right)}}.\dfrac{{a + 1}}{{\sqrt a }}\
 = \dfrac{{4a\sqrt a }}{{a - 1}}.\dfrac{{a + 1}}{{\sqrt a }} = \dfrac{{4a}}{{a - 1}}\left( {a + 1} \right) = \dfrac{{4{a^2} + 4a}}{{a - 1}}
\end{array}$

$b)$Thay $a=\left( 4+\sqrt{15} \right)\left( \sqrt{10}-\sqrt{6} \right)\left( \sqrt{4-\sqrt{15}} \right)$ vào ta được:

$A=\dfrac{4{{\left[ \left( 4+\sqrt{15} \right)\left( \sqrt{10}-\sqrt{6} \right)\left( \sqrt{4-\sqrt{15}} \right) \right]}^{2}}+4\left[ \left( 4+\sqrt{15} \right)\left( \sqrt{10}-\sqrt{6} \right)\left( \sqrt{4-\sqrt{15}} \right) \right]}{\left( 4+\sqrt{15} \right)\left( \sqrt{10}-\sqrt{6} \right)\left( \sqrt{4-\sqrt{15}} \right)-1}=12$Câu trả lời: $\begin{array}{l}
a)A = \left( {\dfrac{{\sqrt a  + 1}}{{\sqrt a  - 1}} - \dfrac{{\sqrt a  - 1}}{{\sqrt a  + 1}} + 4\sqrt a } \right).\left( {\sqrt a  + \dfrac{1}{{\sqrt a }}} \right)\
 = \left[ {\dfrac{{{{\left( {\sqrt a  + 1} \right)}^2} - {{\left( {\sqrt a  - 1} \right)}^2}}}{{\left( {\sqrt a  - 1} \right)\left( {\sqrt a  + 1} \right)}} + 4\sqrt a } \right].\dfrac{{a + 1}}{{\sqrt a }}\
 = \left[ {\dfrac{{4\sqrt a }}{{\left( {\sqrt a  - 1} \right)\left( {\sqrt a  + 1} \right)}} + 4\sqrt a } \right].\dfrac{{a + 1}}{{\sqrt a }}\
 = \dfrac{{4\sqrt a  + 4\sqrt a \left( {\sqrt a  - 1} \right)\left( {\sqrt a  + 1} \right)}}{{\left( {\sqrt a  - 1} \right)\left( {\sqrt a  + 1} \right)}}.\dfrac{{a + 1}}{{\sqrt a }}\
 = \dfrac{{4a\sqrt a }}{{a - 1}}.\dfrac{{a + 1}}{{\sqrt a }} = \dfrac{{4a}}{{a - 1}}\left( {a + 1} \right) = \dfrac{{4{a^2} + 4a}}{{a - 1}}
\end{array}$

$b)$Thay $a=\left( 4+\sqrt{15} \right)\left( \sqrt{10}-\sqrt{6} \right)\left( \sqrt{4-\sqrt{15}} \right)$ vào ta được:

$A=\dfrac{4{{\left[ \left( 4+\sqrt{15} \right)\left( \sqrt{10}-\sqrt{6} \right)\left( \sqrt{4-\sqrt{15}} \right) \right]}^{2}}+4\left[ \left( 4+\sqrt{15} \right)\left( \sqrt{10}-\sqrt{6} \right)\left( \sqrt{4-\sqrt{15}} \right) \right]}{\left( 4+\sqrt{15} \right)\left( \sqrt{10}-\sqrt{6} \right)\left( \sqrt{4-\sqrt{15}} \right)-1}=12$

Nguyễn Thành Trương · Answer

$\begin{align}

& a)A=\left( \dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a} ight).\left( \sqrt{a}+\dfrac{1}{\sqrt{a}} ight) \

& =\left[ \dfrac{{{\left( \sqrt{a}+1 ight)}^{2}}-{{\left( \sqrt{a}-1 ight)}^{2}}}{\left( \sqrt{a}-1 ight)\left( \sqrt{a}+1 ight)}+4\sqrt{a} ight].\dfrac{a+1}{\sqrt{a}} \

& =\left[ \dfrac{4\sqrt{a}}{\left( \sqrt{a}-1 ight)\left( \sqrt{a}+1 ight)}+4\sqrt{a} ight].\dfrac{a+1}{\sqrt{a}} \

& =\dfrac{4\sqrt{a}+4\sqrt{a}\left( \sqrt{a}-1 ight)\left( \sqrt{a}+1 ight)}{\left( \sqrt{a}-1 ight)\left( \sqrt{a}+1 ight)}.\dfrac{a+1}{\sqrt{a}} \

& =\dfrac{4a\sqrt{a}}{a-1}.\dfrac{a+1}{\sqrt{a}}=\dfrac{4a}{a-1}\left( a+1 ight)=\dfrac{4{{a}^{2}}+4a}{a-1} \

\end{align}$

$b)$Thay $a=\left( 4+\sqrt{15} ight)\left( \sqrt{10}-\sqrt{6} ight)\left( \sqrt{4-\sqrt{15}} ight)$ vào ta được:

$A=\dfrac{4{{\left[ \left( 4+\sqrt{15} ight)\left( \sqrt{10}-\sqrt{6} ight)\left( \sqrt{4-\sqrt{15}} ight) ight]}^{2}}+4\left[ \left( 4+\sqrt{15} ight)\left( \sqrt{10}-\sqrt{6} ight)\left( \sqrt{4-\sqrt{15}} ight) ight]}{\left( 4+\sqrt{15} ight)\left( \sqrt{10}-\sqrt{6} ight)\left( \sqrt{4-\sqrt{15}} ight)-1}=12$Câu trả lời: $\begin{align}