Question

plz tell me this question frnds​

Answer 1

Hope it helps..........

Answer 2

${x}^{4} + \frac{1}{ {x}^{4} } = 119$

Now

We know

(a+b)²=a²+b²-2ab

Let a=x²

B=1/x²

Now

$( {x}^{2} + \frac{1}{ {x}^{2} } ) {}^{2} =( {x}^{2} ) {}^{2} +( \frac{1}{ {x}^{2} } ) {}^{2} + 2 \\ \\ ( {x}^{2} + \frac{1}{ {x}^{2} } ) {}^{2} = {x}^{4} + \frac{1}{ {x}^{4} } + 2 \\ \\ ( {x}^{2} + \frac{1}{ {x}^{2} } ) {}^{2} = 119 + 2 \\ \\ ( {x}^{2} + \frac{1}{ {x}^{2} } ) {}^{2} = 121 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } = \sqrt{121} \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } = 11$

Now we know

(a-b)²=a²+b²-2ab

Here

a=X

B=1/X

$(x - \frac{1}{x} ) {}^{2} = {x}^{2} + \frac{1}{ {x}^{2} } - 2 \\ \\ (x - \frac{1}{x} ) {}^{2} = 11 - 2 \\ \\ x - \frac{1}{x} = \sqrt{9} \\ \\ x - \frac{1}{ x } = 3$

$\boxed{\mathbf{\huge{X-\frac{1}{X}</p><p>=3}}}$