Math, asked by 19presa0145, 3 months ago

please give for problems solutions 2 and 3​

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Answers

Answered by nickkaushiknick
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}

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