Question

solve it the attachment

Answer 1

[3-2√2]² - 1/[3-2√2]² 
=[3² -2×3×2√2+{2√2}²]- 1/[3²-2×3×2√2×{2√2}²]
=[9-12√2+8] -1/[9-12√2+8]
=17-12√2 -1/17-12√2
=17-12√2-1/17-12√2×17+12√2
=17²-{12√2}²-1/17²-{12√2}²
=289-288-1/289-288
=1-1/1
=1-1
=0

Answer 2

x = 3 - 2√2
so 1 / x = 1 / 3 - 2√2
By rationalisation of 1 / x
1 / ( 3 - 2√2 ) * ( 3 + 2√2 ) / ( 3+ 2√2 )
1 * ( 3+ 2√2 ) / ( 3 - 2√2 ) * ( 3 + 2√2 )
Using ( a + b ) ( a - b ) = a² - b²
so 3 + 2√2 / ( 3² - 2√2² )
= 3 + 2√2 / 9 - 8
= 3 + 2√2
So x = 3 - 2√2 ,
1 / x = 3 + 2√2
so x ² = ( 3 - 2√2 )²
= 3² + 2√2² - 2(3)(2√2)
= 9 + 8 - 12√2
= 17 - 12√2
and 1 / x² = ( 3 + 2√2 )²
= 3² + 2√2² + 2(3)(2√2)
= 9 + 8 + 12√2
= 17 + 12√2
so x² - 1 / x² = ( 17 - 12√2 ) - ( 17 + 12√2 )
= 17 - 12√2 - 17 - 12√2
= - 24√2