Question

rationalize the denominator 1/root2+root3+root5​

Answer 1

Step-by-step explanation:

step1:multiply 1 by root2+root3+root5 with root2 -root3-root5 by root 2-root3- root5

step2:observe the dinominator it is in the form (a-b-c) whole square =asq+bsq+csq-2ab+2bc-2ca

step 3: expand it and solve it .you will get the answer...

Answer 2

QUESTION-:

Rationalize  the denominator

$\rightarrow \bf\; \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$

EXPLANATION-:

The rationalizing factor of denominator i.e. √2+√3+√5 is (√2+√3)-(√5).Now,multiplying it in both numerator and denominator we get-:

$\rightarrow \bf\; \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\times \frac{(\sqrt{2}+\sqrt{3})-(\sqrt{5})}{(\sqrt{2}+\sqrt{3})-(\sqrt{5})}\\\\\rightarrow \bf\; \frac{(\sqrt{2}+\sqrt{3})-(\sqrt{5})}{(\sqrt{2}+\sqrt{3})^2-(\sqrt{5})^2}\\\\\rightarrow \bf\; \frac{(\sqrt{2}+\sqrt{3})-(\sqrt{5})}{5+2\sqrt{6}-5}\\\\\rightarrow \bf\; \frac{(\sqrt{2}+\sqrt{3})-(\sqrt{5})}{2\sqrt{6}}$

Similarily again rationalizing the denominator we have-:

$\rightarrow \bf\; \frac{(\sqrt{2}+\sqrt{3})-(\sqrt{5})}{2\sqrt{6}}\times \frac{2\sqrt{6}}{2\sqrt{6}}\\\\\rightarrow \bf\; \frac{2\sqrt{4 \times 3}+2\sqrt{9 \times 2}-2\sqrt{30}}{24}\rightarrow \bf\; \frac{4\sqrt{3}+6\sqrt{3}+2\sqrt{30}}{24}\\\\\rightarrow \bf\; \frac{2(2\sqrt{3}+3\sqrt{3}+\sqrt{30}}{24}\\\\\rightarrow \bf\; \frac{2\sqrt{3}+3\sqrt{3}+\sqrt{30}}{12}}$

So the denominator is rationalized !!

RELATED QUESTIONS-:

https://brainly.in/question/44375682