Advertisements
Advertisements
प्रश्न
The value of \[\frac{\sqrt{48} + \sqrt{32}}{\sqrt{27} + \sqrt{18}}\] is
विकल्प
\[\frac{4}{3}\]
4
3
`3/4`
Advertisements
उत्तर
Given that `(sqrt48+sqrt32)/(sqrt27 +sqrt18)`
We know that rationalization factor for `sqrt27 +sqrt18` is`sqrt27 - sqrt18` We will multiply numerator and denominator of the given expression `(sqrt48+sqrt32)/(sqrt27 +sqrt18)` by`sqrt27 - sqrt18`, to get
`(sqrt48+sqrt32)/(sqrt27 +sqrt18) xx (sqrt27-sqrt18)/(sqrt27 -sqrt18) = (sqrt48 xx sqrt27 - sqrt48 xx sqrt18 + sqrt32 xx sqrt27 - sqrt32 xx sqrt18)/ ((sqrt27)^2 - (sqrt18)^2)`
We can factor irrational terms as
` (sqrt(3) xx sqrt16 xx sqrt9 xx sqrt3 - sqrt3 xx sqrt16 xx sqrt9 xx sqrt2 +sqrt2 xx sqrt16 xx sqrt3 xx sqrt9 - sqrt2 xx sqrt16 xx sqrt9 xx sqrt2)/((sqrt27)^2 - (sqrt18)^2)`
`= ((sqrt3)^2 xx 4 xx 3 - sqrt(3xx2)xx 4 xx 3 + sqrt(2 xx3) xx 4 xx3 - (sqrt2)^2 xx 4 xx 3)/(27-18) `
`= (3xx12-12xxsqrt6+12 xxsqrt6 -2 xx12)/(27-18) `
`= (36-12sqrt6+12sqrt6-24)/(27-18)`
`= 12/9`
` = 4/3`
APPEARS IN
संबंधित प्रश्न
Simplify the following:
`(2x^-2y^3)^3`
Solve the following equation:
`3^(x+1)=27xx3^4`
If a and b are different positive primes such that
`((a^-1b^2)/(a^2b^-4))^7div((a^3b^-5)/(a^-2b^3))=a^xb^y,` find x and y.
If 3x-1 = 9 and 4y+2 = 64, what is the value of \[\frac{x}{y}\] ?
If a, b, c are positive real numbers, then \[\sqrt[5]{3125 a^{10} b^5 c^{10}}\] is equal to
The simplest rationalising factor of \[\sqrt[3]{500}\] is
If x= \[\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}\] and y = \[\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}\] , then x2 + y +y2 =
If \[\frac{5 - \sqrt{3}}{2 + \sqrt{3}} = x + y\sqrt{3}\] , then
Simplify:-
`(1/3^3)^7`
Simplify:
`(9^(1/3) xx 27^(-1/2))/(3^(1/6) xx 3^(- 2/3))`
