āϝāĻĻāĻŋ \(x=2+√3\) āĻāĻŦāĻ‚ \(y=2-√3\) āĻšā§Ÿ, āϤāĻŦ⧇ \(y^2+\cfrac{1}{y^2}\) āĻāϰ āĻŽāĻžāύ āύāĻŋāĻ°ā§āϪ⧟ āĻ•āϰāĻŋ āĨ¤
Loading content...

\(\cfrac{1}{y}=\cfrac{1}{(2-√3)} \)
\(=\cfrac{(2+√3)}{(2-√3)(2+√3)}\)
\(=\cfrac{(2+√3)}{(2)^2-(√3)^2} \)
\(=\cfrac{2+√3}{4-3} \)
\(=2+√3\)

\(∴ y+\cfrac{1}{y}=(2-√3)+(2+√3)\)
\(=2-√3+2+√3\)
\(=4\)

\(\therefore y^2+\cfrac{1}{y^2}=\left(y+\cfrac{1}{y}\right)^2-2\cdot \cancel{y} \cdot \cfrac{1}{\cancel{y}} \)
\(=4^2-2=16-2=14\) [Answer]

đŸšĢ Don't Click. Ad Inside 😈

Similar Questions