Question

14) Simplify: 2 root 6/root 2+ root 3 + 6 root 2/ root 6 + root 3 -- 8 root 3/root 6+ root 2

Answer 1

Answer:

$\green {0}$

Step-by-step explanation:

$\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}}+\frac{6\sqrt{2}}{\sqrt{6}+\sqrt{3}}-\frac{8\sqrt{3}}{\sqrt{6}+\sqrt{2}}$

$i) \frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}}\\=\frac{2\sqrt{6}}{\sqrt{3}+\sqrt{2}}\\=\frac{2\sqrt{6}(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}\\=\frac{2\sqrt{6}(\sqrt{3}-\sqrt{2})}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}\\=\frac{2\sqrt{6}(\sqrt{3}-\sqrt{2})}{3-2)}\\=2\sqrt{6}(\sqrt{3}-\sqrt{2})\\=2\sqrt{6}\times \sqrt{3} - 2\sqrt{6}\times \sqrt{2}\\=6\sqrt{2} - 4\sqrt{3} \:---(1)$

$ii) \frac{6\sqrt{2}}{\sqrt{6}+\sqrt{3}}\\=\frac{6\sqrt{2}(\sqrt{6}-\sqrt{3})}{(\sqrt{6}+\sqrt{3})(\sqrt{6}-\sqrt{3})}\\=\frac{6\sqrt{2}(\sqrt{6}-\sqrt{3})}{(\sqrt{6})^{2}-(\sqrt{3})^{2}}\\=\frac{6\sqrt{2}(\sqrt{6}-\sqrt{3})}{6-3)}\\=\frac{6\sqrt{2}(\sqrt{6}-\sqrt{3})}{6-3}\\=\frac{6\sqrt{2}(\sqrt{6}-\sqrt{3})}{3}\\=2\sqrt{2}(\sqrt{6}-\sqrt{3})\\=4\sqrt{3}-2\sqrt{6}\:---(2)$

$iii)\frac{8\sqrt{3}}{\sqrt{6}+\sqrt{2}}\\=\frac{8\sqrt{2}(\sqrt{6}-\sqrt{2})}{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})}\\=\frac{8\sqrt{2}(\sqrt{6}-\sqrt{2})}{(\sqrt{6})^{2}-(\sqrt{2}))^{2}}\\=\frac{8\sqrt{2}(\sqrt{6}-\sqrt{2})}{6-2}\\=\frac{8\sqrt{2}(\sqrt{6}-\sqrt{2})}{4}\\=2\sqrt{3}(\sqrt{6}-\sqrt{2})\\=6\sqrt{2}-4\:---(3)$

/* According to the problem given

$\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}}+\frac{6\sqrt{2}}{\sqrt{6}+\sqrt{3}}-\frac{8\sqrt{3}}{\sqrt{6}+\sqrt{2}}$

$= 6\sqrt{2} - 4\sqrt{3}</p><p>+4\sqrt{3}-2\sqrt{6}-(6\sqrt{2}-2\sqrt{6})$

$= 6\sqrt{2} - 4\sqrt{3}</p><p>+4\sqrt{3}-2\sqrt{6}-6\sqrt{2}+2\sqrt{6}$

$= 0$

Therefore.,

$\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}}+\frac{6\sqrt{2}}{\sqrt{6}+\sqrt{3}}-\frac{8\sqrt{3}}{\sqrt{6}+\sqrt{2}}$

$\green {= 0}$

•••♪

Answer 2

Answer:

Step-by-step explanation:

Please visit the attachment below

Sorry for bad writing

Mark as brilliant answer.