Simplify by rationalising the denominator. 2underrot6-5underrot …
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Sia ? 6 years, 5 months ago
We have,
{tex}\frac{2 \sqrt{6}-\sqrt{5}}{3 \sqrt{5}-2 \sqrt{6}}{/tex} = {tex}\frac{2 \sqrt{6}-\sqrt{5}}{3 \sqrt{5}-2 \sqrt{6}} \times \frac{3 \sqrt{5}+2 \sqrt{6}}{3 \sqrt{5}+2 \sqrt{6}}{/tex}
= {tex}\frac{(2 \sqrt{6}-\sqrt{5})(3 \sqrt{5}+2 \sqrt{6})}{(3 \sqrt{5}-2 \sqrt{6})(3 \sqrt{5}+2 \sqrt{6})}{/tex}
= {tex}\frac{2 \sqrt{6}(3 \sqrt{5}+2 \sqrt{6})-\sqrt{5}(3 \sqrt{5}+2 \sqrt{6})}{(3 \sqrt{5})^{2}-(2 \sqrt{6})^{2}}{/tex}
= {tex}\frac{6 \sqrt{30}+24-15-2 \sqrt{30}}{45-24}{/tex}
= {tex}\frac{4 \sqrt{30}+9}{21}{/tex}
{tex}\therefore{/tex} {tex}\frac{2 \sqrt{6}-\sqrt{5}}{3 \sqrt{5}-2 \sqrt{6}}{/tex} = {tex}\frac{4 \sqrt{30}+9}{21}{/tex}
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