Advertisements
Advertisements
प्रश्न
Prove that:
`sqrt(3xx5^-3)divroot3(3^-1)sqrt5xxroot6(3xx5^6)=3/5`
Advertisements
उत्तर
we have to prove that `sqrt(3xx5^-3)/(root3(3^-1)sqrt5)xxroot6(3xx5^6)=3/5`
By using rational exponents `a^-n=1/a^n` we get,
`sqrt(3xx5^-3)/(root3(3^-1)sqrt5)xxroot6(3xx5^6)=sqrt(3xx1/5^3)/(root3(1/3)sqrt5)xxroot6(3xx5^6)`
`=(3^(1/2)xx1/5^(3xx1/2))/(1/3^(1/3)xx5^(1/2))xx3^(1/6)xx5^(6xx1/6)`
`=(3^(1/2)/5^(3/2))/(5^(1/2)/3^(1/3))xx3^(1/6)xx5^1`
`=3^(1/2)/5^(3/2)xx3^(1/3)/5^(1/2)xx3^(1/6)xx5^1`
`=3^(1/2)xx3^(1/3)xx5^(-3/2)xx5^(-1/2)xx3^(1/6)xx5^1`
`=3^(1/2+1/3+1/6)xx5^(-3/2-1/2+1)`
`=3^((1xx3)/(2xx3)+(1xx2)/(3xx2)+1/6)xx5^(-3/2-1/2+(1xx2)/(1xx2))`
`=3^((3+2+1)/6)xx5^((-3-1+2)/2)`
`=3^1xx5^-1`
`=3xx1/5`
`=3/5`
Hence `sqrt(3xx5^-3)/(root3(3^-1)sqrt5)xxroot6(3xx5^6)=3/5`
APPEARS IN
संबंधित प्रश्न
Simplify the following
`((x^2y^2)/(a^2b^3))^n`
If a = 3 and b = -2, find the values of :
ab + ba
Prove that:
`1/(1 + x^(b - a) + x^(c - a)) + 1/(1 + x^(a - b) + x^(c - b)) + 1/(1 + x^(b - c) + x^(a - c)) = 1`
If 49392 = a4b2c3, find the values of a, b and c, where a, b and c are different positive primes.
Simplify:
`(16^(-1/5))^(5/2)`
Show that:
`((a+1/b)^mxx(a-1/b)^n)/((b+1/a)^mxx(b-1/a)^n)=(a/b)^(m+n)`
If \[8^{x + 1}\] = 64 , what is the value of \[3^{2x + 1}\] ?
If x-2 = 64, then x1/3+x0 =
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 =
The value of \[\frac{\sqrt{48} + \sqrt{32}}{\sqrt{27} + \sqrt{18}}\] is
