Question

find the inverse of (1 -1 +1,2-3-3,6-2-1)​

Answer 1

Answer:

Here is your solution

Given :-

x=2+√3

Now

\begin{gathered} \frac{1}{x} = \frac{1}{2 +\sqrt{3} } \times \frac{2 -\sqrt{3} }{2 - \sqrt{3} } \\ \frac{1}{x} = \frac{2 + \sqrt{3} }{2 {}^{2} - ( \sqrt{3}) {}^{2} } \\ \frac{1}{x} = \frac{2 + \sqrt{3} }{4 - 3} \\ \frac{1}{x} = 2 - \sqrt{3} \\ \\ \\ \end{gathered}x1=2+31×2−32−3x1=22−(3)22+3x1=4−32+3x1=2−3

\begin{gathered} x + \frac{1}{x} = 2 - \sqrt{3} + 2 + \sqrt{3} \\ x + \frac{1}{x} = 4 \\ Both \: sides \: squaring. \: \\ (x + \frac{1}{x} ) {}^{2} = 4 {}^{2} \\ x {}^{2} + \frac{1}{x {}^{2} } + 2 = 16 \\ x {}^{2} + \frac{1}{x {}^{2} } = 16 - 2 \\ x {}^{2} + \frac{1}{x {}^{2} } = 14\end{gathered}x+x1=2−3+2+3x+x1=4Bothsidessquaring.(x+x1)2=42x2+x21+2=16x2+x21=16−2x2+x21=14