Toán Lớp 9: $\frac{ \sqrt{6} – \sqrt{3}}{ \sqrt{2} – 1 }$ + $\sqrt{\frac{2}{4 + \sqrt[n]{15}} }$
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Toán Lớp 9: $\frac{ \sqrt{6} – \sqrt{3}}{ \sqrt{2} – 1 }$ + $\sqrt{\frac{2}{4 + \sqrt[n]{15}} }$
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\quad \dfrac{\sqrt6 – \sqrt3}{\sqrt2 – 1}+ \sqrt{\dfrac{2}{4 + \sqrt{15}}}\\
= \dfrac{\sqrt3\left(\sqrt2 – 1\right)}{\sqrt2 – 1} + \sqrt{\dfrac{4}{8 + 2\sqrt{15}}}\\
= \sqrt3 + \dfrac{2}{\sqrt{5 + 2.\sqrt5.\sqrt3 + 3}}\\
= \sqrt3 + \dfrac{2}{\sqrt{\left(\sqrt5 + \sqrt3\right)^2}}\\
= \sqrt3 + \dfrac{2}{\sqrt5 + \sqrt3}\\
= \sqrt3 + \dfrac{2\left(\sqrt5 – \sqrt3\right)}{\left(\sqrt5 – \sqrt3\right)\left(\sqrt5 +\sqrt3\right)}\\
= \sqrt3 + \dfrac{2\left(\sqrt5 – \sqrt3\right)}{5 – 3}\\
= \sqrt3 + \left(\sqrt5 – \sqrt3\right)\\
= \sqrt5
\end{array}\)