Math, asked by Harsh112233445566, 11 months ago

Genius please answer on a notebook ​

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Answered by Anonymous
3

 \sf{x = 3 - 2 \sqrt{2} }

 \sf \frac{1}{x}  =  \frac{1}{3 - 2 \sqrt{2} }

rationalise the denominator

 \sf \frac{1}{x}  =  \frac{1}{3 - 2 \sqrt{2} }  \times  \frac{3 + 2 \sqrt{2} }{3 + 2 \sqrt{2} }

 \sf \frac{1}{x}  =  \frac{3 + 2 \sqrt{2} }{{3}^{2}  - {(2 \sqrt{2} )}^{2}  }

 \sf \frac{1}{x}  =  \frac{3 + 2 \sqrt{2} }{9- 8}

 \sf \frac{1}{x}  =  {3 + 2 \sqrt{2} }

now,

 \sf{x -  \frac{1}{x}  = 3 - 2 \sqrt{2}  - (3 + 2  \sqrt{2}  ) }

 \sf{x -  \frac{1}{x}  = 3 - 2 \sqrt{2}  - 3 - 2  \sqrt{2}   }

\fbox{\sf{x -  \frac{1}{x}  =  - 4 \sqrt{2} }}

 \sf{x +  \frac{1}{x}  = 3 - 2 \sqrt{2}  + 3 + 2 \sqrt{2} }

\fbox{\star\:\sf{x +  \frac{1}{x}  = 6}}

 \sf{ {x}^{2} -  \frac{1}{ {x}^{2} }  = (x +  \frac{1}{x} )(x -  \frac{1}{x} ) }

 \sf{ {x}^{2} -  \frac{1}{ {x}^{2} }  = (6)( - 4 \sqrt{2}  ) }

 \fbox{ \sf{ {x}^{2} -  \frac{1}{ {x}^{2} }  = - 24 \sqrt{2}} }

Answered by Anonymous
2

\mathfrak{\large{\underline{\underline{Answer :}}}}

\boxed{{x}^{2} - \frac{1}{{x}^{2}} = - 24\sqrt{2}}

\mathfrak{\large{\underline{\underline{Explanation :}}}}

Refer to the attachment.

\textbf{\large{\underline{\underline{Algebraic Identities used :}}}}

  • (x + y)² = x² + 2xy + y²

  • (x + y)(x - y) = x² - y²

\textsf{\large{\underline{\underline{Additional Information :}}}}

  • An Algebraic identity is an equation, which is true for all values of the variables in the equation.

1) (x + y)² = x² + 2xy + y²

2) (x - y)² = x² - 2xy - y²

3) (x + y)(x - y) = x² - y²

4) (x + a)(x + b) = x² + (a + b)x + ab

5) (x + y)³ = x³ + 3x²y + 3xy² + y³

6) (x - y)³ = x³ - 3x²y + 3xy² - y³

7) x³ + y³ = (x + y)(x² - xy + y²)

8) x³ - y³ = (x - y)(x² + xy + y²)

9) (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2xz

10) x³ + y³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - xz

  • If the product of two two irrational numbers is rational number then each of the two is the Rationalising factor of the other.
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