Simplification Questions and answers

\( \Large x = \frac{\sqrt{3}+1}{\sqrt{3}-1} \)

\( \Large y = \frac{\sqrt{3}-1}{\sqrt{3}+1} \)

\( \Large x + y = \frac{ \left(\sqrt{3}+1\right)^{2} + \left(\sqrt{3}-1\right)^{2} }{ \left(\sqrt{3}-1\right) \left(\sqrt{3}-1\right) } \)

= \( \Large \frac{3+1+2\sqrt{3}+3+1-2\sqrt{3}}{3 - 1} = \frac{8}{2} \)

= x + y = 4

and xy = \( \Large \frac{\sqrt{3}+1}{\sqrt{3}-1} \times \frac{\sqrt{3}-1}{\sqrt{3}+1} = 1 \)

Therefore, \( \Large \frac{x^{2}}{y}+\frac{y^{2}}{x} = \frac{x^{3} + y^{3}}{xy} \)

= \( \Large \frac{ \left(x+y\right)^{3} - 3xy \left(x+y\right) }{xy} \)

= \( \Large \frac{ \left(4\right)^{3} - 3 \times 1 \times 4 }{1} \)

= \( \Large \frac{64 - 12}{1} = 52 \)

\( \Large \frac{p}{q} + \frac{q}{p} = \frac{p^{2} + q^{2}}{pq} \)

\( \Large = \frac{ \left(p + q\right)^{2} - 2pq }{pq} = \frac{ \left(10\right)^{2} - 2 \times 5 }{5} \)

\( \Large = \frac{100 - 10}{5} = \frac{90}{5} = 18 \)

Given expression

= \( \Large \left(2-\frac{1}{3}\right) \left(2-\frac{3}{5}\right) \left(2-\frac{5}{7}\right).......\left(2-\frac{997}{999}\right) \)

= \( \Large \frac{5}{3} \times \frac{7}{5} \times \frac{9}{7} \times .... \times \frac{1001}{999}

= \frac{1001}{3} \)

Given expression

= \( \Large \left(1+\frac{1}{2}\right) \left(1+\frac{1}{3}\right) \left(1+\frac{1}{4}\right).... \left(1+\frac{1}{150}\right) \)

= \( \Large \frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} \times .... \times \frac{151}{150}

= \frac{151}{2} = 75.5 \)

Given expression

\( \Large \frac{1 + x}{1+\sqrt{1+x}} + \frac{1-x}{1-\sqrt{1-x}} \)

On putting \( \Large x = \frac{\sqrt{3}}{2} \), we get

\( \Large \frac{1 + \frac{\sqrt{3}}{2}}{1+\sqrt{1 + \frac{\sqrt{3}}2}} + \frac{1 - \frac{\sqrt{3}}{2}}{1 - \sqrt{1-\frac{\sqrt{3}}2{}}} \)

= \( \Large \frac{2 + \sqrt{3}}{\sqrt{2} \left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right) } + \frac{2 - \sqrt{3}}{\sqrt{2} \left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right) } \)

= \( \Large \frac{2 + \sqrt{3}}{2 + \sqrt{4 + 2\sqrt{3}}} + \frac{2 - \sqrt{3}}{2 - \sqrt{4-2\sqrt{3}}} \)

=> \( \Large \frac{2 + \sqrt{3}}{2 + \sqrt{ \left(1\right)^{2} + \left(\sqrt{3}\right)^{2} } + 2\sqrt{3}} + \frac{2 - \sqrt{3}}{2 - \sqrt{ \left(1\right)^{2} + \left(\sqrt{3}\right)^{2} - 2\sqrt{3} }} \)

=> \( \Large \frac{2 + \sqrt{3}}{2 + \sqrt{ \left(1 + \sqrt{3}\right) ^{2}}} + \frac{2 - \sqrt{3}}{2 - \sqrt{ \left(\sqrt{3} - 1\right)^{2} }} \)

= \( \Large \frac{2 + \sqrt{3}}{2 + \left(1 + \sqrt{3}\right) } + \frac{2 - \sqrt{3}}{2 - \left(\sqrt{3} - 1\right) } \)

= \( \Large \frac{2+\sqrt{3}}{3+\sqrt{3}} + \frac{2-\sqrt{3}}{3-\sqrt{3}} \)

= \( \Large \frac{ \left(2+\sqrt{3}\right) \left(3 - \sqrt{3}\right) + \left(2 - \sqrt{3}\right) \left(3 + \sqrt{3}\right) }{ \left(3 + \sqrt{3}\right) \left(3 - \sqrt{3}\right) } \)

=> \( \Large \frac{6+3\sqrt{3}-2\sqrt{3}-3+6-3\sqrt{3}+2\sqrt{3}-3}{ \left(3 + \sqrt{3}\right) \left(3 - \sqrt{3}\right) } \)

=> \( \Large \frac{6}{9 + 3\sqrt{3} - 3\sqrt{3} - 3} = \frac{6}{6} = 1 \)

Given expression = \( \Large \frac{2x^{2}+2}{3x^{2}+3+5x} \)

On dividing numerator and denominator by x, we get

\( \Large \frac{2 \left(x+\frac{1}{x}\right) }{3 \left(x+\frac{1}{x}\right)+5 } = \frac{2 \times 2}{3 \times 2 + 5} = \frac{4}{11} \) [given \( \Large x + \frac{1}{x} \) =2]

\( \Large a + \frac{1}{b} = 1 \)

=> \( \Large a = \left(1 - \frac{1}{b}\right) = \frac{b - 1}{b} \)

=> \( \Large \frac{1}{a} = \frac{b}{ \left(b - 1\right) } \) [reciprocal]

and \( \Large b + \frac{1}{c} = 1 \)

=> \( \Large \frac{1}{c} = \left(1 - b\right) \)

=> \( \Large c = \frac{1}{ \left(1 - b\right) } \)

Therefore, \( \Large c + \frac{1}{a} = \frac{1}{ \left(1 - b\right)} + \frac{b}{ \left(b - 1\right)  } \)

= \( \Large \frac{1}{ \left(1 - b\right)} - \frac{b}{ \left(1 - b\right)  } \)

= \( \Large \frac{ \left(1 - b\right) }{ \left(1 - b\right) } = 1 \)

\( \Large x + \frac{a}{x} = b => \frac{x^{2}+a}{x} = b \)

=> \( \Large x^{2} + a = bx \) ...(i)

Now, \( \Large \frac{x^{2}+bx+a}{bx^{2}-x^{3}} => \frac{ \left(x^{2}+a\right) + bx} {bx^{2}-x^{3}} \) [using Eq. (i)]

= \( \Large \frac{2bx}{bx^{2} - x^{3}} = \frac{2b}{bx - x^{2}} = \frac{2b}{a} \)

\( \Large Because, x^{2} + bx \)

\( \Large bx - x^{2} = a \)