Math, asked by abhishek58163, 6 hours ago

If x minus one upon x is equal to make 8 find the value of x square + 1 upon x square and X square 4 plus 1/x square 4

Answers

Answered by 12thpáìn
6

Given

  • \sf{x -  \dfrac{1}{x}  = 8}

To Find

  • \sf{ {x  +  \dfrac{1}{x}   }  }

___________Solution___________

~~~\implies~~~   \sf{x -  \dfrac{1}{x}  = 8}

  • On Squaring Both side

~~~\implies~~~   \sf{ {\bigg(x -  \dfrac{1}{x}  \bigg) }^{2} =  {8}^{2} }

~~~\implies~~~   \sf{ {x  +  \dfrac{1}{x}   } - 2 \times x  \times \dfrac{1}{x} = 64 }

~~~\implies~~~   \sf{ {x  +  \dfrac{1}{x}   }  = 64 + 2 }

~~~\implies~~~   \sf{ {x  +  \dfrac{1}{x}   }  = 66 }

Answered by rosoni28
34

\huge\fcolorbox{black}{magenta}{question☺᭄✍︎}

\tt\red{x - \dfrac{1}{x} = 8}

\huge\pink{\boxed{\green {\mathbb{\overbrace {\underbrace{\fcolorbox{red}{aqua}{\underline{\pink{†◖ \: ⚆to \: find⚆◗†}}}}}}}}}

  \tt \pink{ {x + \dfrac{1}{x} } }

\huge\fcolorbox{pink}{Red}{ ★A᭄ΠSWΣR✍︎}

~~\implies~~~ \tt \orange{x - \dfrac{1}{x} = 8}

On Squaring Both side

~~\implies~~~ \tt\purple{ {\bigg(x - \dfrac{1}{x} \bigg) }^{2}}

~~\implies~~~ \tt \green{ {x + \dfrac{1}{x} } - 2 \times x \times \dfrac{1}{x} = 64 }

~~\implies~~~ \tt\pink{ {x + \dfrac{1}{x} } = 64 + 2 }

~~\implies~~~ \tt\red{ {x + \dfrac{1}{x} } = 66 }

\huge\pink{\mid{\fbox{\tt{thanks}}\mid}}

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