$\frac{(123-2)^{1/2}(78+3)^{1/4}}{243^{(1/5)}}$

Challenger App

★

1M+ Downloads

A10

B11

C12

D13

Answer:

$\frac{(123-2)^{1/2}(78+3)^{1/4}}{243^{(1/5)}}$

$=\frac{(121)^{1/2}(81)^{1/4}}{243^{(1/5)}}$

$=\frac{11 \times (3^4)^{(1/4)}}{(3^5)^{(1/5)}}$

$=\frac{11 \times 3}{3}$

$=11$

Related Questions:

(-2)^7 ÷ [(-2)^4 x (-2) ^2] =

$(-1)^{2}+(-1)^{3} + (-1)^{4} + (-1)^{5}$ = ____

$(\frac{1}{2})^3\times(\frac{1}{2})^4\times(\frac{1}{2})^5= ?$

$2^{50}$ ന്ടെ പകുതി എത്രയാണ്?