If a = 2 + 3\sqrt{3}3 , then find the value a−1aa - \dfrac{1}{a}a−a1.
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Given,
a=2+3∴1a=12+3×2−32−3⇒2−322−(3)2⇒2−34−3⇒(2−3)∴a−1a=2+3−2+3⇒a−1a=23a = 2 + \sqrt{3} \\[1.5em] \therefore \dfrac{1}{a} = \dfrac{1}{2 + \sqrt{3}} ×\dfrac{2 - \sqrt{3}}{2 - \sqrt{3}} \\[1.5em] \Rightarrow \dfrac{2 - \sqrt{3}}{2^2 - (\sqrt{3})^2} \\[1.5em] \Rightarrow \dfrac{2 - \sqrt{3}}{4 - 3} \\[1.5em] \Rightarrow (2 - \sqrt{3}) \\[1.5em] \therefore a - \dfrac{1}{a} = 2 + \sqrt{3} - 2 + \sqrt{3} \\[1.5em] \Rightarrow\bold{ a - \dfrac{1}{a} = 2\sqrt{3}} \\[1.5em]a=2+3∴a1=2+31×2−32−3⇒22−(3)22−3⇒4−32−3⇒(2−3)∴a−a1=2+3−2+3⇒a−a1=23
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If 7+353+5−7−353−5=p+q5\dfrac{7 + 3\sqrt{5}}{3 + \sqrt{5}} - \dfrac{7 - 3\sqrt{5}}{3 - \sqrt{5}} = p + q\sqrt{5}3+57+35−3−57−35=p+q5 , find the values of p and q where p and q are rational numbers.
Rationalise the denominator of the following and hence evaluate by taking 2\sqrt{2}2 = 1.414 and 3\sqrt{3}3 = 1.732 , upto three places of decimal:
(i)22+2(ii)13+2\begin{matrix} \text{(i)} & \dfrac{\sqrt2}{2+\sqrt2} \\[1.5em] \text{(ii)} & \dfrac{1}{\sqrt3+\sqrt2} \\[1.5em] \end{matrix}(i)(ii)2+223+21
If x=1−2x = 1-\sqrt{2}x=1−2, find the value of (x−1x)4{\Big(x-\dfrac{1}{x}\Big)}^4(x−x1)4.
If x = 5−265 - 2\sqrt{6}5−26, find the value of x2+1x2x^2 + \dfrac{1}{x^2}x2+x21.
