Question

Rationalize 2√3-√5/2√2+3√3

Answer 1

Answer:

$\frac{2\sqrt{3}-\sqrt{5}}{2\sqrt{2}+3\sqrt{3}}=\frac{18-4\sqrt{6}+2\sqrt{10}-3\sqrt{15}}{19}$

Step-by-step explanation:

Consider,

$\frac{2\sqrt{3}-\sqrt{5}}{2\sqrt{2}+3\sqrt{3}}$

We multiply and divide by 2√2 - 3√3

we get,

$\implies\frac{2\sqrt{3}-\sqrt{5}}{2\sqrt{2}+3\sqrt{3}}\times\frac{2\sqrt{2}-3\sqrt{3}}{2\sqrt{2}-3\sqrt{3}}$

$\implies\frac{(2\sqrt{3}-\sqrt{5})(2\sqrt{2}-3\sqrt{3})}{(2\sqrt{2}+3\sqrt{3})(2\sqrt{2}-3\sqrt{3})}$

$\implies\frac{4\sqrt{6}-6\times3-2\sqrt{10}+3\sqrt{15}}{(2\sqrt{2})^2-(3\sqrt{3})^2}$

$\implies\frac{4\sqrt{6}-18-2\sqrt{10}+3\sqrt{15}}{8-27}$

$\implies\frac{4\sqrt{6}-18-2\sqrt{10}+3\sqrt{15}}{-19}$

$\implies\frac{18-4\sqrt{6}+2\sqrt{10}-3\sqrt{15}}{19}$

Answer 2

Answer:

$\frac{2\sqrt{3}-\sqrt{5}}{2\sqrt{2}+3\sqrt{3}}=\frac{18-4\sqrt{6}+2\sqrt{10}-3\sqrt{15}}{19}$

Step-by-step explanation:

Consider,

$\frac{2\sqrt{3}-\sqrt{5}}{2\sqrt{2}+3\sqrt{3}}$

We multiply and divide by $2\sqrt{2}-3\sqrt{3}$

we get,

⇒$\frac{2\sqrt{3}-\sqrt{5}}{2\sqrt{2}+3\sqrt{3}}\times \frac{2\sqrt{2}-3\sqrt{3}}{2\sqrt{2}-3\sqrt{3}}$

⇒$\frac{(2\sqrt{3}-\sqrt{5})(2\sqrt{2}-3\sqrt{3})}{(2\sqrt{2}+3\sqrt{3})(2\sqrt{2}-3\sqrt{3})}$

⇒$\frac{4\sqrt{6}-6\times3-2\sqrt{10}+3\sqrt{15}}{(2\sqrt{2})^2-(3\sqrt{3})^2}$

⇒$\frac{4\sqrt{6}-18-2\sqrt{10}+3\sqrt{15}}{8-27}$

⇒$\frac{4\sqrt{6}-18-2\sqrt{10}+3\sqrt{15}}{19}$

⇒$\frac{18-4\sqrt{6}+2\sqrt{10}-3\sqrt{15}}{19}$

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