Advertisements
Advertisements
प्रश्न
The value of \[\frac{\sqrt{48} + \sqrt{32}}{\sqrt{27} + \sqrt{18}}\] is
पर्याय
\[\frac{4}{3}\]
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
Find:-
`9^(3/2)`
Simplify the following:
`(6(8)^(n+1)+16(2)^(3n-2))/(10(2)^(3n+1)-7(8)^n)`
If `a=xy^(p-1), b=xy^(q-1)` and `c=xy^(r-1),` prove that `a^(q-r)b^(r-p)c^(p-q)=1`
Show that:
`[{x^(a(a-b))/x^(a(a+b))}div{x^(b(b-a))/x^(b(b+a))}]^(a+b)=1`
Show that:
`{(x^(a-a^-1))^(1/(a-1))}^(a/(a+1))=x`
Show that:
`(a^(x+1)/a^(y+1))^(x+y)(a^(y+2)/a^(z+2))^(y+z)(a^(z+3)/a^(x+3))^(z+x)=1`
Which of the following is (are) not equal to \[\left\{ \left( \frac{5}{6} \right)^{1/5} \right\}^{- 1/6}\] ?
When simplified \[\left( - \frac{1}{27} \right)^{- 2/3}\] is
If \[\frac{3^{5x} \times {81}^2 \times 6561}{3^{2x}} = 3^7\] then x =
If \[x = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}\] and \[y = \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}\] then x + y +xy=
