Question

please give for problems solutions 2 and 3​

Answer 1

Answer:

(ii) $6-\sqrt{35}$ (iii) $\frac{3\sqrt2+2\sqrt3}{6}$

Step-by-step explanation:

(ii) $\frac{\sqrt7-\sqrt5}{\sqrt7+\sqrt5}$

To rationalise the denominator we will multiply Numerator and Denominator by the  conjugate of the denominator

$\frac{\sqrt7-\sqrt5}{\sqrt7+\sqrt5} \times \frac{\sqrt7-\sqrt5}{\sqrt7-\sqrt5}$

$\frac{(\sqrt7-\sqrt5)^2}{(\sqrt7)^2-(\sqrt5)^2}\ \ [\because\ (a+b)(a-b)=a^2-b^2]$

$\frac{(\sqrt7)^2+(\sqrt5)^2-2\times \sqrt7\times \sqrt5}{7-5}\ \ [\because\ (a-b)^2=a^2+b^2-2ab]$

$\frac{7+5-2\sqrt{35}}{2}$

$\frac{12-2\sqrt{35}}{2}$

$\frac{2(6-\sqrt{35})}{2}$

$6-\sqrt{35}$

(iii) $\frac{1}{3\sqrt2-2\sqrt3}$

To rationalise the denominator we will multiply Numerator and Denominator by the  conjugate of the denominator

$\frac{1}{3\sqrt2-2\sqrt3}\times \frac{3\sqrt2+2\sqrt3}{3\sqrt2+2\sqrt3}$

$\frac{3\sqrt2+2\sqrt3}{(3\sqrt2)^2-(2\sqrt3)^2}\ \ [\because\ (a+b)(a-b)=a^2-b^2]$

$\frac{3\sqrt2+2\sqrt3}{18-12}$

$\frac{3\sqrt2+2\sqrt3}{6}$