Question

please answer
this question refer to the attachment

Answer 1

$\sf \dfrac{1}{\sqrt{7}-\sqrt{6}}+\dfrac{1}{\sqrt{7}-2}$

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Let's rationalize the 1st term ( denominator ) ,

$\to \sf \dfrac{1}{\sqrt{7}-\sqrt{6}}\times \dfrac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}$

• ( a + b ) ( a - b ) = a² - b²

$\to \sf \dfrac{\sqrt{7}+\sqrt{6}}{(\sqrt{7})^2-(\sqrt{6})^2}$

$\to \sf \dfrac{\sqrt{7}+\sqrt{6}}{7-6}\\\\\to \sf \dfrac{\sqrt{7}+\sqrt{6}}{1}\\\\\to \sf \sqrt{7}+\sqrt{6}\ \; \bigstar$

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Let's rationalize the 2nd term ( denominator ) ,

$\to \sf \dfrac{1}{\sqrt{7}-2}\times \dfrac{\sqrt{7}+2}{\sqrt{7}+2}$

• ( a - b ) ( a + b ) = a² - b²

$\to \sf \dfrac{\sqrt{7}+2}{(\sqrt{7})^2-(2)^2}\\\\\to \sf \dfrac{\sqrt{7}+2}{7-4}\\\\\to \sf \dfrac{\sqrt{7}+2}{3}\ \; \bigstar$

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Let's add both to get our answer ,

$\to \sf \sqrt{7}+\sqrt{6}+\dfrac{\sqrt{7}+2}{3}\\\\\to \sf \dfrac{3\sqrt{7}+3\sqrt{6}+\sqrt{7}+2}{3}\\\\\leadsto \sf \pink{\dfrac{4\sqrt{7}+3\sqrt{6}+2}{3}}\ \; \bigstar$

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

$\boxed{\bf{Rationalise\:\:the\:\:denominator\:\:of\:\::\dfrac{1}{\sqrt{7}-\sqrt{6}}+\dfrac{1}{\sqrt{7}-2}}}$

♣ ᴀɴꜱᴡᴇʀ :

$\boxed{\bf{\dfrac{4\sqrt{7}+3\sqrt{6}+2}{3}}}$

♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

→ Rationalise the first term $\sf{\dfrac{1}{\sqrt{7}-\sqrt{6}}}$

$\mathrm{Multiply\:by\:the\:conjugate}\:\dfrac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}$

$=\dfrac{1\cdot \left(\sqrt{7}+\sqrt{6}\right)}{\left(\sqrt{7}-\sqrt{6}\right)\left(\sqrt{7}+\sqrt{6}\right)}$

$1\cdot \left(\sqrt{7}+\sqrt{6}\right)=\sqrt{7}+\sqrt{6}$

$\sf{\left(\sqrt{7}-\sqrt{6}\right)\left(\sqrt{7}+\sqrt{6}\right)=1}$

$=\dfrac{\sqrt{7}+\sqrt{6}}{1}$

$\bf{=\sqrt{7}+\sqrt{6}}$

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→ Rationalise the second term $\sf{\dfrac{1}{\sqrt{7}-2}}$

$\mathrm{Multiply\:by\:the\:conjugate}\:\dfrac{\sqrt{7}+2}{\sqrt{7}+2}$

$=\dfrac{1\cdot \left(\sqrt{7}+2\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}$

$1\cdot \left(\sqrt{7}+2\right)=\sqrt{7}+2$

$\sf{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right) = 3}$

$\bf{=\dfrac{\sqrt{7}+2}{3}}$

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→ It's time to add

$\sf{\sqrt{7}+\sqrt{6}+\dfrac{\sqrt{7}+2}{3}}$

$\mathrm{Convert\:element\:to\:fraction}:\quad \sqrt{7}=\dfrac{\sqrt{7}\cdot \:3}{3},\:\sqrt{6}=\dfrac{\sqrt{6}\cdot \:3}{3}$

$=\dfrac{\sqrt{7}\cdot \:3}{3}+\dfrac{\sqrt{6}\cdot \:3}{3}+\dfrac{\sqrt{7}+2}{3}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \dfrac{a}{c}\pm \dfrac{b}{c}=\dfrac{a\pm \:b}{c}$

$=\dfrac{\sqrt{7}\cdot \:3+\sqrt{6}\cdot \:3+\sqrt{7}+2}{3}$

$\boxed{\bf{=\dfrac{4\sqrt{7}+3\sqrt{6}+2}{3}}}$