Chưng minh dăng thưc:

a) \(\frac{{\sqrt {{a^2} + {x^2}}  + \sqrt {{a^2} - {x^2}} }}{{\sqrt {{a^2} + {x^2}}  - \sqrt {{a^2} - {x^2}} }} - \sqrt {\frac{{{a^4}}}{{{x^4}}} - 1}  = \frac{{{a^2}}}{{{x^2}}};\) với \(|a| > |x|\)

b) \({\left( {\frac{{5 + 2\sqrt 6 }}{{\sqrt 3  + \sqrt 2 }}} \right)^2} - {\left( {\frac{{5 - 2\sqrt 6 }}{{\sqrt 3  - \sqrt 2 }}} \right)^2} = 4\sqrt 6 \)

c) \(\left( {\frac{{x\sqrt x  - y\sqrt y }}{{\sqrt x  - \sqrt y }} + \sqrt {xy} } \right):{(\sqrt x  + \sqrt y )^2} = 1\quad (x > 0,y > 0,x \ge y)\)

a) Ta có \(\sqrt {50}  - \sqrt {32}  + 3\sqrt 8  = \sqrt {25.2}  - \sqrt {16.2}  + 3\sqrt {4.2}  = 5\sqrt 2  - 4\sqrt 2  + 3.2.\sqrt 2  = 7\sqrt 2 \).

b) \(\sqrt {25a}  + 2\sqrt {160a}  - 3\sqrt {10a}  = \sqrt {25.10a}  + 2.\sqrt {16.10a}  - 3\sqrt {10a} \)

\( = 5\sqrt {10a}  + 2.4.\sqrt {10a}  - 3\sqrt {10a}  = 10\sqrt {10a} \).

c) \(\left( {2\sqrt 7  + \sqrt 3 } \right)\sqrt 7  - \sqrt {84}  = 2\sqrt 7 .\sqrt 7  + \sqrt 3 .\sqrt 7  - \sqrt {4.21} \)

\( = 2.7 + \sqrt {21}  - 2\sqrt {21}  = 14 - \sqrt {21} \).

d) \(\left( {\sqrt {63}  - \sqrt 8  - \sqrt 7 } \right)\sqrt 7  + 2\sqrt {14}  = \sqrt {63} .\sqrt 7  - \sqrt 8 .\sqrt 7  - \sqrt 7 .\sqrt 7  + 2\sqrt {14} \)

\( = \sqrt {9.7.7}  - 2\sqrt 2 .\sqrt 7  - 7 + 2\sqrt {14}  = 3.7 - 2\sqrt {14}  - 7 + 2\sqrt {14}  = 14\).

a). Ta có: \(\frac{{\sqrt 2  - 1}}{{\sqrt 2  + 2}} - \frac{2}{{2 + \sqrt 2 }} + \frac{{\sqrt 2  + 1}}{{\sqrt 2 }} = \frac{{\sqrt 2  - 1}}{{\sqrt 2 (1 + \sqrt 2 )}} - \frac{2}{{\sqrt 2 (\sqrt 2  + 1)}} + \frac{{{{\left( {\sqrt 2  + 1} \right)}^2}}}{{\sqrt 2 (\sqrt 2  + 1)}}\)

\( = \frac{{\sqrt 2  - 1 - 2 + 2 + 2\sqrt 2  + 1}}{{\sqrt 2 (\sqrt 2  + 1)}} = \frac{{3\sqrt 2 }}{{\sqrt 2 (\sqrt 2  + 1)}} = \frac{3}{{\sqrt 2  + 1}} = \frac{{3\left( {\sqrt 2  - 1} \right)}}{{2 - 1}} = 3\left( {\sqrt 2  - 1} \right)\)

b). Ta có: \(\frac{{2 + \sqrt 5 }}{{\sqrt 2  + \sqrt {3 + \sqrt 5 } }} + \frac{{2 - \sqrt 5 }}{{\sqrt 2  - \sqrt {3 - \sqrt 5 } }} = \frac{{2\sqrt 2  + \sqrt {10} }}{{2 + \sqrt {6 + 2\sqrt 5 } }} + \frac{{2\sqrt 2  - \sqrt {10} }}{{2 - \sqrt {6 - 2\sqrt 5 } }}\)

\( = \frac{{2\sqrt 2  + \sqrt {10} }}{{2 + \sqrt {{{\left( {\sqrt 5  + 1} \right)}^2}} }} + \frac{{2\sqrt 2  - \sqrt {10} }}{{2 - \sqrt {{{\left( {\sqrt 5  - 1} \right)}^2}} }}\) \( = \frac{{2\sqrt 2  + \sqrt {10} }}{{2 + \sqrt 5  + 1}} + \frac{{2\sqrt 2  - \sqrt {10} }}{{2 - \sqrt 5  + 1}} = \frac{{2\sqrt 2  + \sqrt {10} }}{{3 + \sqrt 5 }} + \frac{{2\sqrt 2  - \sqrt {10} }}{{3 - \sqrt 5 }}\)

\( = \frac{{\left( {2\sqrt 2  + \sqrt {10} } \right)\left( {3 - \sqrt 5 } \right) + \left( {2\sqrt 2  - \sqrt {10} } \right)\left( {3 + \sqrt 5 } \right)}}{{\left( {3 + \sqrt 5 } \right)\left( {3 - \sqrt 5 } \right)}}\)

\( = \frac{{6\sqrt 2  - 2\sqrt {10}  + 3\sqrt {10}  - \sqrt {50}  + 6\sqrt 2  + 2\sqrt {10}  - 3\sqrt {10}  - \sqrt {50} }}{{9 - 5}}\)

\( = \frac{{12\sqrt 2  - 2\sqrt {50} }}{4} = \frac{{12\sqrt 2  - 10\sqrt 2 }}{4} = \frac{{2\sqrt 2 }}{4} = \frac{{\sqrt 2 }}{2}.\)