# Selina Solutions Class 9 Mathematics Solutions for Exercise 1(D) in Chapter 1 - Chapter 1- Rational and Irrational Numbers

Question 9 Exercise 1(D)

Show that:

\frac{4-\sqrt{5}}{4+\sqrt{5}}+\frac{2}{5+\sqrt{3}}+\frac{4+\sqrt{5}}{4-\sqrt{5}}+\frac{2}{5-\sqrt{3}}=\frac{52}{11}

\begin{aligned} &\frac{4-\sqrt{5}}{4+\sqrt{5}}+\frac{2}{5+\sqrt{3}}+\frac{4+\sqrt{5}}{4-\sqrt{5}}+\frac{2}{5-\sqrt{3}}=\frac{52}{11}\\ &\text { Here, }\\ &\text { Considering } \frac{4-\sqrt{5}}{4+\sqrt{5}}\\ &\Rightarrow \frac{4-\sqrt{5}}{4+\sqrt{5}} \times \frac{4-\sqrt{5}}{4-\sqrt{5}}=\frac{(4-\sqrt{5})^{2}}{16-5}=\frac{(4-\sqrt{5})^{2}}{11}\\ &\text {Now, Considering } \frac{2}{5+\sqrt{3}}\\ &\Rightarrow \frac{2}{5+\sqrt{3}} \times \frac{5-\sqrt{3}}{5-\sqrt{3}}=\frac{10-2 \sqrt{3}}{25-3}=\frac{10-2 \sqrt{3}}{22}\\ &\text {Now, Considering} \frac{4+\sqrt{5}}{4-\sqrt{5}}\\ &\Rightarrow \frac{4+\sqrt{5}}{4-\sqrt{5}} \times \frac{4+\sqrt{5}}{4+\sqrt{5}}=\frac{(4+\sqrt{5})^{2}}{16-5}=\frac{(4+\sqrt{5})^{2}}{11}\\ &\text {Now, Considering } \frac{2}{5-\sqrt{3}}\\ &\Rightarrow \frac{2}{5-\sqrt{3}} \times \frac{5+\sqrt{3}}{5+\sqrt{3}}-\frac{10+2 \sqrt{3}}{25-3}-\frac{10+2 \sqrt{3}}{22}\\ &\therefore \frac{4-\sqrt{5}}{4+\sqrt{5}}+\frac{2}{5+\sqrt{3}}+\frac{4+\sqrt{5}}{4-\sqrt{5}}+\frac{2}{5-\sqrt{3}}\\ &=\frac{(4-\sqrt{5})^{2}}{11}+\frac{10-2 \sqrt{3}}{22}+\frac{(4+\sqrt{5})^{2}}{11}+\frac{10+2 \sqrt{3}}{22}\\ &=\frac{(4-\sqrt{5})^{2}}{11}+\frac{5-\sqrt{3}}{11}+\frac{(4+\sqrt{5})^{2}}{11}+\frac{5+\sqrt{3}}{11}\\ &=\frac{16-8 \sqrt{5}+5+5-\sqrt{3}+16+8 \sqrt{5}+5+5+\sqrt{3}}{11}\\ &\begin{array}{l} =\frac{52}{11} \\ \text { Hence proved } \end{array} \end{aligned}

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