Question

Find the value of $\frac{{\left(\frac{1}{x}+y\right)}^{\left(a+b\right)}{\left(\frac{1}{y}-x\right)}^{-\left(p+q\right)}}{{\left(\frac{1}{x}-y\right)}^{-\left(p+q\right)}{\left(x+\frac{1}{y}\right)}^{\left(a+b\right)}}$

• A. ${\left(\frac{x}{y}\right)}^{\left(a+b\right)+\left(p+q\right)}$
• B. ${\left(\frac{y}{x}\right)}^{\left(a+b\right)+\left(p+q\right)}$
• C. ${\left(\frac{y}{x}\right)}^{\left(a+b\right)-\left(p+q\right)}$
• D. ${\left(\frac{x}{y}\right)}^{\left(a+b\right)-\left(p+q\right)}$

Answer 1

Answer: (B)

${\left(\frac{y}{x}\right)}^{\left(a+b\right)+\left(p+q\right)}$

Let $a=\frac{{\left(\frac{1}{x}+y\right)}^{\left(a+b\right)}{\left(\frac{1}{y}-x\right)}^{-\left(p+q\right)}}{{\left(\frac{1}{x}-y\right)}^{-\left(p+q\right)}{\left(x+\frac{1}{y}\right)}^{\left(a+b\right)}}$

$={\left(\frac{\frac{1}{x}+y}{x+\frac{1}{y}}\right)}^{a+b}{\left(\frac{\frac{1}{y}-x}{\frac{1}{x}-y}\right)}^{-(p+q)}$

$={\left(\frac{y}{x}\right)}^{a+b}{\left(\frac{x}{y}\right)}^{-(p+q)}$

$={\left(\frac{y}{x}\right)}^{a+b}{\left(\frac{y}{x}\right)}^{p+q}$

$a={\left(\frac{y}{x}\right)}^{(a+b)+(p+q)}$

$\therefore \frac{{\left(\frac{1}{x}+y\right)}^{\left(a+b\right)}{\left(\frac{1}{y}-x\right)}^{-\left(p+q\right)}}{{\left(\frac{1}{x}-y\right)}^{-\left(p+q\right)}{\left(x+\frac{1}{y}\right)}^{\left(a+b\right)}}={\left(\frac{y}{x}\right)}^{(a+b)+(p+q)}$