Advertisements
Advertisements
प्रश्न
If a, b, c are positive real numbers, then \[\sqrt{a^{- 1} b} \times \sqrt{b^{- 1} c} \times \sqrt{c^{- 1} a}\] is equal to
विकल्प
1
abc
\[\sqrt{abc}\]
\[\frac{1}{abc}\]
उत्तर
We have to find the value of `sqrt(a^-1b)xx sqrt (b^-1c) xx sqrt(c^-1 a)` when a, b, c are positive real numbers.
So,
`sqrt(a^-1b)xx sqrt (b^-1c) xx sqrt(c^-1 a) =sqrt(1/a xxb)xx sqrt(1/b xx c) xx sqrt(1/c xx a)`
`sqrt(b/a) xx sqrt (c/b) xx sqrt(a/c)`
Taking square root as common we get
\[\sqrt{a^{- 1} b} \times \sqrt{b^{- 1} c} \times \sqrt{c^{- 1} a} = \sqrt{\frac{b}{a} \times \frac{c}{b} \times \frac{a}{c}}\]
\[\sqrt{a^{- 1} b} \times \sqrt{b^{- 1} c} \times \sqrt{c^{- 1} a} = 1\]
APPEARS IN
संबंधित प्रश्न
Find:-
`9^(3/2)`
Solve the following equations for x:
`3^(2x+4)+1=2.3^(x+2)`
Simplify:
`(0.001)^(1/3)`
Simplify:
`((25)^(3/2)xx(243)^(3/5))/((16)^(5/4)xx(8)^(4/3))`
Prove that:
`sqrt(3xx5^-3)divroot3(3^-1)sqrt5xxroot6(3xx5^6)=3/5`
Find the value of x in the following:
`(root3 4)^(2x+1/2)=1/32`
If (23)2 = 4x, then 3x =
The simplest rationalising factor of \[2\sqrt{5}-\]\[\sqrt{3}\] is
If x = \[\frac{2}{3 + \sqrt{7}}\],then (x−3)2 =
Which of the following is equal to x?