giúp em với ạ em đang cần gấp
b1 rút gọn
a , 1/ √7 – √24 + 1 – 1/ √7 – √24 – 1
b, √3 / √3+1 – 1 – √3 / √3+1 + 1
c, √3+ √5 / 3- √5 + √3 – √5 / 3 + √5
giúp em với ạ em đang cần gấp
b1 rút gọn
a , 1/ √7 – √24 + 1 – 1/ √7 – √24 – 1
b, √3 / √3+1 – 1 – √3 / √3+1 + 1
c, √3+ √5 / 3- √5 + √3 – √5 / 3 + √5
Đáp án:
$\begin{array}{l}
a)\dfrac{1}{{\sqrt 7 – \sqrt {24} + 1}} – \dfrac{1}{{\sqrt 7 – \sqrt {24} – 1}}\\
= \dfrac{{\sqrt 7 – \sqrt {24} – 1 – \left( {\sqrt 7 – \sqrt {24} + 1} \right)}}{{\left( {\sqrt 7 – \sqrt {24} + 1} \right).\left( {\sqrt 7 – \sqrt {24} – 1} \right)}}\\
= \dfrac{{ – 2}}{{{{\left( {\sqrt 7 – \sqrt {24} } \right)}^2} – 1}}\\
= \dfrac{{ – 2}}{{7 – 2.\sqrt 7 .\sqrt {24} + 24 – 1}}\\
= \dfrac{{ – 2}}{{30 – 2.\sqrt 7 .2\sqrt 6 }}\\
= \dfrac{{ – 2}}{{30 – 4\sqrt {42} }}\\
= \dfrac{{ – 2\left( {30 + 4\sqrt {42} } \right)}}{{{{30}^2} – {{\left( {4\sqrt {42} } \right)}^2}}}\\
= \dfrac{{ – 15 – 2\sqrt {42} }}{{57}}\\
b)\dfrac{{\sqrt 3 }}{{\sqrt 3 + 1}} – \dfrac{{1 – \sqrt 3 }}{{\sqrt 3 + 1}} + 1\\
= \dfrac{{\sqrt 3 – 1 + \sqrt 3 }}{{\sqrt 3 + 1}} + 1\\
= \dfrac{{2\sqrt 3 – 1 – \sqrt 3 – 1}}{{\sqrt 3 + 1}}\\
= \dfrac{{\sqrt 3 – 2}}{{\sqrt 3 + 1}}\\
= \dfrac{{\left( {\sqrt 3 – 2} \right)\left( {\sqrt 3 – 1} \right)}}{{3 – 1}}\\
= \dfrac{{3 – \sqrt 3 – 2\sqrt 3 + 2}}{2}\\
= \dfrac{{5 – 3\sqrt 3 }}{2}\\
c)\dfrac{{\sqrt 3 + \sqrt 5 }}{{3 – \sqrt 5 }} + \dfrac{{\sqrt 3 – \sqrt 5 }}{{3 + \sqrt 5 }}\\
= \dfrac{{\left( {\sqrt 3 + \sqrt 5 } \right)\left( {3 + \sqrt 5 } \right) + \left( {\sqrt 3 – \sqrt 5 } \right)\left( {3 – \sqrt 5 } \right)}}{{\left( {3 – \sqrt 5 } \right)\left( {3 + \sqrt 5 } \right)}}\\
= \dfrac{{3\sqrt 3 + \sqrt {15} + 3\sqrt 5 + 5 + 3\sqrt 3 – \sqrt {15} – 3\sqrt 5 + 5}}{{9 – 5}}\\
= \dfrac{{6\sqrt 3 + 10}}{4}\\
= \dfrac{{3\sqrt 3 + 5}}{2}
\end{array}$