If x-110-10-3- Find x-1xx-1x x-1x2-

x= 1 10√( 10 ) 3 = 10√( 10 ) +3 1000 9 = 10√( 10 ) +3 991 1 x =10√( 10 ) 3 ∴ x+ 1 x = 10√( 10 ) +3 991 + 9910√( 10 ) 2973 991 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard VIII

Mathematics

Expansion of (x^2 - y^2)

If x=110√10...

Question

If $x = \frac{1}{10 \sqrt{10} - 3}$ . Find $(x + \frac{1}{x}) (x - \frac{1}{x})$ & ${(x + \frac{1}{x})}^{2}$ .

Open in App

Solution

$x = \frac{1}{10 \sqrt{10} - 3} = \frac{10 \sqrt{10} + 3}{1000 - 9} = \frac{10 \sqrt{10} + 3}{991}$
$\frac{1}{x} = 10 \sqrt{10} - 3$
$∴ x + \frac{1}{x} = \frac{10 \sqrt{10} + 3}{991} + \frac{9910 \sqrt{10} - 2973}{991}$
$= \frac{9920 \sqrt{10} - 2970}{991}$
$x - \frac{1}{x} = \frac{10 \sqrt{10} + 3 - 9910 \sqrt{10} + 2973}{991} = \frac{2976 - 9900 \sqrt{10}}{991}$
${(x + \frac{1}{x})}^{2} = \frac{984064000 + 8820900 - 58924800 \sqrt{10}}{982081}$
$= \frac{992884900 - 58924800 \sqrt{10}}{982081}$

Suggest Corrections

Similar questions

Q. The term independent of

x (x > 0, x

\neq

1)

in the expansion of

{[\frac{(x + 1)}{(x^{2 / 3} - x^{1 / 3} + 1)} - \frac{(x - 1)}{(x - \sqrt{x})}]}^{10}

The term independent of $x (x > 0, x \neq 1)$ in the expansion of ${[\frac{(x + 1)}{(x^{2 / 3} - x^{1 / 3} + 1)} - \frac{(x - 1)}{(x - \sqrt{x})}]}^{10} i s$

\frac{x}{x - 1} \div \frac{x^{2}}{x^{2} - 1}

= .

Q. If

(x + \frac{1}{x}) = \frac{10}{3}

, then

{(x - \frac{1}{x})}^{2}

If (x−1/x)=10, then find the value of x²+1/x².

Join BYJU'S Learning Program

If x=110√10−3. Find (x+1x)(x−1x) & (x+1x)2.

x=110√10−3=10√10+31000−9=10√10+39911x=10√10−3∴x+1x=10√10+3991+9910√10−2973991=9920√10−2970991x−1x=10√10+3−9910√10+2973991=2976−9900√10991(x+1x)2=984064000+8820900−58924800√10982081=992884900−58924800√10982081

If $x = \frac{1}{10 \sqrt{10} - 3}$ . Find $(x + \frac{1}{x}) (x - \frac{1}{x})$ & ${(x + \frac{1}{x})}^{2}$ .