If x=1x−5x = \dfrac{1}{x - 5}x=x−51 and x≠ 5, find : x2−1x2x^2 - \dfrac{1}{x^2}x2−x21.
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Given,
⇒x=1x−5⇒x(x−5)=1⇒x2−5x=1⇒x2−1=5x⇒x2−1x=5xx⇒x−1x=5.\Rightarrow x = \dfrac{1}{x - 5} \\[1em] \Rightarrow x(x - 5) = 1 \\[1em] \Rightarrow x^2 - 5x = 1 \\[1em] \Rightarrow x^2 - 1 = 5x \\[1em] \Rightarrow \dfrac{x^2 - 1}{x} = \dfrac{5x}{x} \\[1em] \Rightarrow x - \dfrac{1}{x} = 5.⇒x=x−51⇒x(x−5)=1⇒x2−5x=1⇒x2−1=5x⇒xx2−1=x5x⇒x−x1=5.
By formula,
⇒(x+1x)2−(x−1x)2=4⇒(x+1x)2−52=4⇒(x+1x)2−25=4⇒(x+1x)2=25+4⇒(x+1x)2=29⇒x+1x=±29.\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 - \Big(x - \dfrac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 - 5^2 = 4 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 - 25 = 4 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 25 + 4 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 29 \\[1em] \Rightarrow x + \dfrac{1}{x} = \pm \sqrt{29}.⇒(x+x1)2−(x−x1)2=4⇒(x+x1)2−52=4⇒(x+x1)2−25=4⇒(x+x1)2=25+4⇒(x+x1)2=29⇒x+x1=±29.
⇒x2−1x2=(x−1x)(x+1x)=5×±29=±529.\Rightarrow x^2 - \dfrac{1}{x^2} = \Big(x - \dfrac{1}{x}\Big)\Big(x + \dfrac{1}{x}\Big)\\[1em] = 5 \times \pm \sqrt{29} \\[1em] = \pm 5\sqrt{29}.⇒x2−x21=(x−x1)(x+x1)=5×±29=±529.
Hence, x2−1x2=±529x^2 - \dfrac{1}{x^2} = \pm 5\sqrt{29}x2−x21=±529.
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