Rút gọn a,1/3 - 2√2 + 2√3/2+√5 b,1/√5+√7 +1/1-√7 c,7/3-√2 -2/√2-1
1 câu trả lời
Đáp án:
$\begin{array}{l}
a)\dfrac{1}{{3 - 2\sqrt 2 }} + \dfrac{{2\sqrt 3 }}{{2 + \sqrt 5 }}\\
= \dfrac{{3 + 2\sqrt 2 }}{{{3^2} - {{\left( {2\sqrt 2 } \right)}^2}}} + \dfrac{{2\sqrt 3 \left( {\sqrt 5 - 2} \right)}}{{5 - {2^2}}}\\
= \dfrac{{3 + 2\sqrt 3 }}{{9 - 8}} + \dfrac{{2\sqrt 3 \left( {\sqrt 5 - 2} \right)}}{{5 - 4}}\\
= 3 + 2\sqrt 3 + 2\sqrt {15} - 4\sqrt 3 \\
= 3 + 2\sqrt {15} - 2\sqrt 3 \\
b)\dfrac{1}{{\sqrt 5 + \sqrt 7 }} + \dfrac{1}{{1 - \sqrt 7 }}\\
= \dfrac{{\sqrt 7 - \sqrt 5 }}{{7 - 5}} + \dfrac{{1 + \sqrt 7 }}{{1 - 7}}\\
= \dfrac{{\sqrt 7 - \sqrt 5 }}{2} - \dfrac{{1 + \sqrt 7 }}{6}\\
= \dfrac{{3\sqrt 7 - 3\sqrt 5 - 1 - \sqrt 7 }}{6}\\
= \dfrac{{2\sqrt 7 - 3\sqrt 5 - 1}}{6}\\
c)\dfrac{7}{{3 - \sqrt 2 }} - \dfrac{2}{{\sqrt 2 - 1}}\\
= \dfrac{{7\left( {3 + \sqrt 2 } \right)}}{{{3^2} - 2}} - \dfrac{{2\left( {\sqrt 2 + 1} \right)}}{{2 - 1}}\\
= \dfrac{{7\left( {3 + \sqrt 2 } \right)}}{{9 - 2}} - 2\left( {\sqrt 2 + 1} \right)\\
= 3 + \sqrt 2 - 2\sqrt 2 - 2\\
= 1 - \sqrt 2
\end{array}$