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.
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Given,
x=1−2∴1x=11−2×1+21+2⇒1+21−(2)2⇒1+21−2⇒−(1+2)∴x−1x=1−2+1+2⇒x−1x=2∴(x−1x)4=24=16x = 1 - \sqrt{2} \\[1.5em] \therefore \dfrac{1}{x} = \dfrac{1}{1 - \sqrt{2}} × \dfrac{1 + \sqrt{2}}{1 + \sqrt{2}} \\[1.5em] \Rightarrow \dfrac{1 + \sqrt{2}}{1 - (\sqrt{2})^2} \\[1.5em] \Rightarrow \dfrac{1 + \sqrt{2}}{1 - 2} \\[1.5em] \Rightarrow - (1 +\sqrt{2}) \\[1.5em] \therefore x - \dfrac{1}{x} = 1 - \sqrt{2} + 1 + \sqrt{2} \\[1.5em] \Rightarrow x - \dfrac{1}{x} = 2 \\[1.5em] \therefore \bold{{\Big(x-\dfrac{1}{x}\Big)}^4 = 2 ^4 = 16}x=1−2∴x1=1−21×1+21+2⇒1−(2)21+2⇒1−21+2⇒−(1+2)∴x−x1=1−2+1+2⇒x−x1=2∴(x−x1)4=24=16
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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 a = 2 + 3\sqrt{3}3 , then find the value a−1aa - \dfrac{1}{a}a−a1.
If x = 5−265 - 2\sqrt{6}5−26, find the value of x2+1x2x^2 + \dfrac{1}{x^2}x2+x21.
If p = 2−52+5\dfrac{2-\sqrt{5}}{2+\sqrt{5}}2+52−5 and q = 2+52−5\dfrac{2+\sqrt{5}}{2-\sqrt{5}}2−52+5 find the values of :
(i) p + q
(ii) p - q
(iii) p2 + q2
(iv) p2 - q2
