Question

பகு‌தியை ‌வி‌கித‌ப்படு‌த்‌தி சுரு‌க்குக
(i) (√48+√32)/(√27-√18) .
ii) (5√3+√2)/(√3+√2)
(iii) (2√6-√5)/(3√5-2√6)
(iv) √5/(√6+2)-√5/(√6-2)

Answer 1

$i)\frac{4}{3}[5+2 \sqrt{6}]$   $ii)13-4 \sqrt{6}$  $iii)\frac{4 \sqrt{30}+9}{21}$ $iv)-2 \sqrt{5}$

விளக்கம்:

i) $\frac{\sqrt{48}+\sqrt{32}}{\sqrt{27}-\sqrt{18}}$

$\Rightarrow \frac{\sqrt{48}+\sqrt{32}}{\sqrt{27}-\sqrt{18}} \times \frac{\sqrt{27}+\sqrt{18}}{\sqrt{27}+\sqrt{18}}$

$\Rightarrow \frac{4(\sqrt{3}+4 \sqrt{2})(3 \sqrt{3}+3 \sqrt{2})}{27-18}$

$=\frac{36+12 \sqrt{6}+12 \sqrt{6}+24}{9}$

$=\frac{60+24 \sqrt{6}}{9}$

$=\frac{12[5+2 \sqrt{6}]}{9}$

$=\frac{4}{3}[5+2 \sqrt{6}]$

ii)$\frac{5 \sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

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

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

$=\frac{(5 \times 3)-5 \sqrt{6}+\sqrt{6}-2}{3-2}$

$=\frac{(15)-5 \sqrt{6}+\sqrt{6}-2}{1}$

$=(15-2)-\sqrt{6}(5-1)$

$=13-4 \sqrt{6}$

$\text { (iii) } \frac{2 \sqrt{6}-\sqrt{5}}{3 \sqrt{5}-2 \sqrt{6}}$

$=\frac{2 \sqrt{6}-\sqrt{5}}{3 \sqrt{5}-2 \sqrt{6}} \times \frac{3 \sqrt{5}+2 \sqrt{6}}{3 \sqrt{5}+2 \sqrt{6}}$

$=\frac{6 \sqrt{30}+4(6)-(5)(3)-2(\sqrt{30})}{45-24}$

$=\frac{4 \sqrt{30}+9}{21}$

$\text { (iv) } \frac{\sqrt{5}}{\sqrt{6}+2}-\frac{\sqrt{5}}{\sqrt{6}-2}$

$\Rightarrow \frac{\sqrt{5}(\sqrt{6}-2)-\sqrt{5}(\sqrt{6}+2)}{(\sqrt{6}+2)(\sqrt{6}-2)}$

$=\frac{\sqrt{30}-2 \sqrt{5}-\sqrt{30}-2 \sqrt{5}}{6-4}$

$=\frac{(-4 \sqrt{5})}{2}=\frac{2(-2 \sqrt{5})}{2}$

$=-2 \sqrt{5}$

Answer 2

Step-by-step explanation:

(\frac{\sqrt{48}+\sqrt{32}}{\sqrt{27}-\sqrt{18}}\) (ii) \(\frac{5\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\) (iii) \(\frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}\) (iv) \(\frac{\sqrt{5}}{\sqrt{6}+2}-\frac{\sqrt{5}}{\sqrt{6}-