Question

rationalize the denominator of 1/root 6 -root 5​

Answer 1

Given :

• 1 / √6 - √5

To Find :

• Rationalize the denominator .

Solution :

$\longmapsto\tt\bf{\dfrac{1}{\sqrt{6}-\sqrt{5}}}$

$\longmapsto\tt{\dfrac{1}{\sqrt{6}-\sqrt{5}}\times\dfrac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}}$

Using Identity : (a-b) (a+b) = a² - b² :

$\longmapsto\tt{\dfrac{\sqrt{6}+\sqrt{5}}{\sqrt{(6)}^{2}-\sqrt{(5)}^{2}}}$

$\longmapsto\tt{\dfrac{\sqrt{6}+\sqrt{5}}{6-5}}$

$\longmapsto\tt\bf{\dfrac{\sqrt{6}+\sqrt{5}}{1}}$

So , The Final Answer is √6 + √5 /1 ...

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Some More Identities :

$\begin{gathered}\boxed{\begin{minipage}{7 cm}\boxed{\bigstar\:\:\textbf{\textsf{Algebric\:Identity}}\:\bigstar}\\\\1)\bf\:(A+B)^{2} = A^{2} + 2AB + B^{2}\\\\2)\sf\: (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\3)\bf\: A^{2} - B^{2} = (A+B)(A-B)\\\\4)\sf\: (A+B)^{2} = (A-B)^{2} + 4AB\\\\5)\bf\: (A-B)^{2} = (A+B)^{2} - 4AB\\\\6)\sf\: (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\7)\bf\:(A-B)^{3} = A^{3} - 3AB(A-B) + B^{3}\\\\8)\sf\: A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})\\\\\end{minipage}}\end{gathered}$

Answer 2

Step-by-step explanation:

Given:-

${\LARGE {{\frac {1}{{\sqrt {6}}-{\sqrt{5}}}}}}$

To do:-

Rationalize the denominator

Solution:-

Formula:-

${:}\longrightarrow$${\boxed{(a+b)(a-b)={a}^{2}-{b}^{2}}}$

Let's solve:-

${\Huge {{\frac {1}{{\sqrt {6}}-{\sqrt{5}}}}}}$

${:}\longrightarrow$${\Huge {{\frac {1 ({\sqrt {6}}+{\sqrt {5}})}{({\sqrt {6}}-{\sqrt {5}})({\sqrt {6}}+{\sqrt{5}})}}}}$

${:}\longrightarrow$${\Huge {{\frac {{\sqrt {6}}+{\sqrt {5}}}{{({\sqrt {6}})}^{2}-{({\sqrt {5}})}^{2}}}}}$

${:}\longrightarrow$${\Huge {{\frac {{\sqrt{6}}+{\sqrt {5}}}{6-5}}}}$

${:}\longrightarrow$${\Huge {{\frac {{\sqrt {6}}-{\sqrt{5}}}{1}}}}$

${:}\longrightarrow$${\Huge {{\underline{\boxed {\bf {{\sqrt {6}}-{\sqrt {5}}}}}}}}$