Advertisements
Advertisements
प्रश्न
Rationalise the denominators of : `1/(sqrt3 - sqrt2 )`
उत्तर
`1/(sqrt3 - sqrt2 ) xx ((sqrt3 + sqrt2)/(sqrt3 + sqrt2)) = sqrt( 3 + sqrt2)/[(sqrt3)^2- (sqrt2)^2] = [sqrt3 + sqrt2]/[ 3 - 2 ]`
= `sqrt3 + sqrt2`
APPEARS IN
संबंधित प्रश्न
Rationalize the denominator.
`4/(7+ 4 sqrt3)`
Rationalise the denominators of : `(2sqrt3)/sqrt5`
Simplify:
`sqrt2/[sqrt6 - sqrt2] - sqrt3/[sqrt6 + sqrt2]`
Rationalise the denominator of `1/[ √3 - √2 + 1]`
Simplify by rationalising the denominator in the following.
`(4 + sqrt(8))/(4 - sqrt(8)`
Simplify by rationalising the denominator in the following.
`(2sqrt(6) - sqrt(5))/(3sqrt(5) - 2sqrt(6)`
Simplify the following :
`(3sqrt(2))/(sqrt(6) - sqrt(3)) - (4sqrt(3))/(sqrt(6) - sqrt(2)) + (2sqrt(3))/(sqrt(6) + 2)`
If x = `(4 - sqrt(15))`, find the values of
`x + (1)/x`
If x = `((sqrt(3) + 1))/((sqrt(3) - 1)` and y = `((sqrt(3) - 1))/((sqrt(3) + 1)`, find the values of
x2 + y2
Using the following figure, show that BD = `sqrtx`.