Question

plz someone help to do this one​

Answer 1

Solution

$(x + y) {}^{ - 1} (x {}^{ - 1} + y {}^{ - 1} ) = x {}^{p} y {}^{q} \\ = > \frac{1}{x + y} \times ( \frac{1}{x} + \frac{1}{y} ) = x {}^{p} y {}^{q} \\ = > \frac{1}{x + y} \times \frac{x + y}{xy} = x {}^{p} y {}^{q} \\ = > \frac{1}{xy} =x {}^{p} y {}^{q} \\ = > x {}^{ - 1} y {}^{ - 1} = x {}^{p} y {}^{q} \\ comparing \: both \: sides \: we \: get...... \\ p = - 1 \\ q = - 1$

therefore....

p+q=(-1)+(-1)=-2

ANSWER= -2

Hope this helps you......khus rhe