Question

if x/y=(-1/3)^-3÷(2/3)^-4 find the value of (x/y+y/x)^-1​

Answer 1

Answer:

$(\frac{x}{y} + \frac{y}{x} )^{-1}$ $= -\frac{48}{265}$

Step-by-step explanation:

Given that, $\frac{x}{y} = (-\frac{1}{3} )^{-3}$ ÷ $(\frac{2}{3} )^{-4}$

$=> \frac{x}{y} = \frac{1}{(-\frac{1}{3} )^{3}}$ ÷ $\frac{1}{(\frac{2}{3} )^{4}}$  [∵ $a^{-n} = \frac{1}{a^n}$]

$=> \frac{x}{y} = (-27) . (\frac{2}{3} )^{4} = -\frac{2^4}{3} = - \frac{16}{3}$  . . . . . . . . . . . .  (i)

Now $(\frac{x}{y} + \frac{y}{x} )^{-1}$

$= (-\frac{16}{3} - \frac{3}{16} )^{-1}$

$= (-\frac{16^2 + 3^2}{(3)(16)})^{-1}$

$= -\frac{48}{16^2 + 3^2}$  [∵ $a^{-n} = \frac{1}{a^n}$]

$= -\frac{48}{256 + 9}$

$= -\frac{48}{265}$