Toán Lớp 9: 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
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Toán Lớp 9: 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
Comments ( 1 )
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}$