Advertisements
Advertisements
Question
If x = `2sqrt3 + 2sqrt2`, find: `1/x`
Advertisements
Solution
`1/x = 1/[2sqrt3 + 2sqrt2] xx [2sqrt3 - 2sqrt2]/[2sqrt3 - 2sqrt2]`
= `[2sqrt3 - 2sqrt2]/[(2sqrt3)^2 - (2sqrt2)^2]`
= `[2sqrt3 - 2sqrt2]/(12 - 8)`
= `[cancel(2)^1(sqrt3 - sqrt2)]/cancel(4)_2`
= `(sqrt3 - sqrt2)/2`
APPEARS IN
RELATED QUESTIONS
Rationalize the denominator.
`1/sqrt14`
Rationalize the denominator.
`5/sqrt 7`
Rationalize the denominator.
`6/(9sqrt 3)`
Write the lowest rationalising factor of 5√2.
Write the lowest rationalising factor of : √5 - √2
Write the lowest rationalising factor of : 3√2 + 2√3
Find the values of 'a' and 'b' in each of the following:
`3/[ sqrt3 - sqrt2 ] = asqrt3 - bsqrt2`
If m = `1/[ 3 - 2sqrt2 ] and n = 1/[ 3 + 2sqrt2 ],` find m2
Show that :
`1/[ 3 - 2√2] - 1/[ 2√2 - √7 ] + 1/[ √7 - √6 ] - 1/[ √6 - √5 ] + 1/[√5 - 2] = 5`
Rationalise the denominator and simplify `(sqrt(48) + sqrt(32))/(sqrt(27) - sqrt(18))`
