KeerthanaPosted on यदि x2−x68+1=0x^{2}-x sqrt{68}+1=0x2−x68+1=0, तो (x−1x)?left(x-frac{1}{x} ight) ?(x−x1)? का मूल्य क्या है? a8 b62sqrt{62}62 c6 d66sqrt{66}66 Answer : Option AExplanation : x2−x68+1=0x^{2}-x sqrt{68}+1=0x2−x68+1=0 ⇒x2+1=x68Rightarrow x^{2}+1=x sqrt{68}⇒x2+1=x68 ⇒x+1x=68Rightarrow x+frac{1}{x}=sqrt{68}⇒x+x1=68 ∴(x−1x)2=(x+1x)2−4=68−4=64 hereforeleft(x-frac{1}{x} ight)^{2}=left(x+frac{1}{x} ight)^{2}-4=68-4=64∴(x−x1)2=(x+x1)2−4=68−4=64 ∴(x−1x)=64=8 hereforeleft(x-frac{1}{x} ight)=sqrt{64}=8∴(x−x1)=64=8 Rate This:NaN / 5 - 1 votesAdd comment