Question

`if \: x =( √(3) )/(2) then \: find \: \ \\ the \: \\ value \: of \\ \\ \\ \ \: ( √(1 + x) )/(1 + √(1 + x) ) + ( √(1 - x) )/(1 - √(1 - x) )`

If x=underoot 3/2,then find the value of underoot 1+x by 1+ underoot 1+x +underoot 1-x by 1- underoot 1-x.​

Answer 1

`\bf \underline{Given-} \\`

`\rm{x = ( √(3) )/(2) } \\`

`\bf \underline{To \: find-} \\`

`\mathsf{ (1 + x)/(1 + √(1 + x) ) + (1 - x)/(1 - √(1 - x) ) = ?} \\`

`\bf \underline{Solution-} \\`

`\rm{ √(1 + x) } \\`

`= \sqrt{1 + ( √(3) )/(2) } \:\:\:\: \: \: \: \sf{ [∵\: \:x = √3/2\:(Given)]}\\`

`= \sqrt{ ( 2 + √(3) )/(2) } \\`

`= \sqrt{ (4 + 2 √(3) )/(4) } \\`

`= \sqrt{ (3 + 1 + 2 * √(3) * 1)/(4) } \\`

`= \sqrt{ \frac{( √(3) + 1 {)}^(2) }{4} } \\`

`= ( √(3) + 1 )/(2) \\`

`\textsf{Similarly,} \\`

`\rm{ √(1 - x) } \\`

`\rm{ = ( √(3) - 1)/(2) } \\`

`\rm{\therefore \: (1 + x)/(1 + √(1 + x) ) + (1 - x)/(1 - √(1 - x) ) } \\`

`\rm{= (1 + ( √(3) )/(2) )/(1 + ( √(3) + 1 )/(2) ) + ( 1 - ( √(3) )/(2) )/(1 - \bigg( ( √(3) - 1 )/(2) \bigg) ) } \\`

`\rm{= (2 + √(3) )/(2 + √(3) + 1 ) + (2 - √(3) )/(2 - ( √(3) - 1)) } \\`

`\rm{= (2 + √(3) )/(3 + √(3) ) + (2 - √(3) )/(3 - √(3) ) } \\`

`\rm{= (2 + √(3) )/(3 + √(3) ) * (3 - √(3) )/(3 - √(3) ) + (2 - √(3) )/(3 - √(3) ) * (3 + √(3) )/(3 + √(3) ) } \\`

`\rm{= \frac{(2 + √(3) )(3 - √(3) ) + (2 - √(3))(3 + √(3) )}{(3 {)}^(2) - ( √(3) {)}^(2) } } \\`

`\rm{= (6 - 2 √(3) + 3 √(3) - 3 + 6 + 2 √(3) - 3 √(3) - 3)/(9 - 3) } \\`

`\rm{= (12 - 6)/(6) } \\`

`\rm{= (6)/(6) } \\`

`\rm{= 1} \\`

`\bf \underline{Answer-} \\`

`\mathsf{\underline{Hence, the \: value \: of \: (1 + x)/(1 + √(1 + x) ) + (1 - x)/(1 - √(1 - x) ) \: is \: 1}} \\`