If (x2+125x2)=925\Big(x^2 + \dfrac{1}{25x^2}\Big) = 9\dfrac{2}{5}(x2+25x21)=952, find the value of (x−15x)\Big(x - \dfrac{1}{5x}\Big)(x−5x1).
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⇒(x−15x)2=[x2+(15x)2−2×x×15x]⇒(x−15x)2=[x2+125x2−25]⇒(x−15x)2=925−25⇒(x−15x)2=475−25⇒(x−15x)2=47−25⇒(x−15x)2=455⇒(x−15x)2=9⇒(x−15x)=9⇒(x−15x)=±3\Rightarrow \Big(x - \dfrac{1}{5x}\Big)^2 = \Big[x^2 + \Big(\dfrac{1}{5x}\Big)^2 - 2 \times x \times \dfrac{1}{5x}\Big] \\[1em] \Rightarrow \Big(x - \dfrac{1}{5x}\Big)^2 = \Big[x^2 + \dfrac{1}{25x^2} - \dfrac{2}{5}\Big] \\[1em] \Rightarrow \Big(x - \dfrac{1}{5x}\Big)^2 = 9\dfrac{2}{5} - \dfrac{2}{5} \\[1em] \Rightarrow \Big(x - \dfrac{1}{5x}\Big)^2 = \dfrac{47}{5} - \dfrac{2}{5} \\[1em] \Rightarrow \Big(x - \dfrac{1}{5x}\Big)^2 = \dfrac{47 - 2}{5} \\[1em] \Rightarrow \Big(x - \dfrac{1}{5x}\Big)^2 = \dfrac{45}{5} \\[1em] \Rightarrow \Big(x - \dfrac{1}{5x}\Big)^2 = 9 \\[1em] \Rightarrow (x - \dfrac{1}{5x}\Big) = \sqrt{9} \\[1em] \Rightarrow \Big(x - \dfrac{1}{5x}\Big) = \pm 3 \\[1em]⇒(x−5x1)2=[x2+(5x1)2−2×x×5x1]⇒(x−5x1)2=[x2+25x21−52]⇒(x−5x1)2=952−52⇒(x−5x1)2=547−52⇒(x−5x1)2=547−2⇒(x−5x1)2=545⇒(x−5x1)2=9⇒(x−5x1)=9⇒(x−5x1)=±3
Hence, (x−15x)=±3\Big(x - \dfrac{1}{5x}\Big) = \pm 3(x−5x1)=±3.
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If (x−1x)=8\Big(x - \dfrac{1}{x}\Big) = 8(x−x1)=8, find the values of :
(i) (x+1x)\Big(x + \dfrac{1}{x}\Big)(x+x1)
(ii) (x2−1x2)\Big(x^2 - \dfrac{1}{x^2}\Big)(x2−x21)
If (x2+1x2)=7\Big(x^2 + \dfrac{1}{x^2}\Big) = 7(x2+x21)=7, find the values of :
(ii) (x−1x)\Big(x - \dfrac{1}{x}\Big)(x−x1)
(iii) (2x2−2x2)\Big(2x^2 - \dfrac{2}{x^2}\Big)(2x2−x22).
If a2−4a−1=0a^2 - 4a - 1 = 0a2−4a−1=0 and a≠0a \neq 0a=0, find the values of:
(i) (a−1a)\Big(a - \dfrac{1}{a}\Big)(a−a1)
(ii) (a+1a)\Big(a + \dfrac{1}{a}\Big)(a+a1)
(iii) (a2−1a2)\Big(a^2 - \dfrac{1}{a^2}\Big)(a2−a21)
(iv) (a2+1a2)\Big(a^2 + \dfrac{1}{a^2}\Big)(a2+a21)
If a=1a−5a = \dfrac{1}{a - 5}a=a−51, where a≠5 and a≠0a \neq 5 \text{ and }a \neq 0a=5 and a=0, find the values of:
