Question

The expression: $\frac{\left(\frac{x^2+3x+2}{x+2}\right)+3x-\frac{x(x^3+1)}{(x+1)(x^2-x+1)}{\log }_{2}8}{(x-1)\left({\log }_{2}4\right)\left({\log }_{4}5\right)\left({\log }_{5}2\right)}$ reduces to

• A. $\frac{x+1}{x-1}$
• B. $\frac{x^2+3x+2}{\left({\log }_{2}5\right)x-1}$
• C. $\frac{3x}{x-1}$
• D. $x$

Answer 1

The correct option is A $\frac{x+1}{x-1}$

Given:$\frac{\left(\frac{x^2+3x+2}{x+2}\right)+3x-\frac{x(x^3+1)}{(x+1)(x^2-x+1)}{\log }_{2}8}{(x-1)\left({\log }_{2}4\right)\left({\log }_{4}5\right)\left({\log }_{5}2\right)}$

$⟹\frac{\left(\frac{x^2+2x+x+2}{x+2}\right)+3x-\frac{x(x^3+1)}{x^3+1}{\log }_2{2}^3}{(x-1)\times \frac{\log 4}{\log 2}\times \frac{\log 5}{\log 4}\times \frac{\log 2}{\log 5}}$

$⟹\frac{\frac{x(x+2)+1(x+2)}{x+2}+3x-x\times 3{\log }_{2}2}{(x-1)}\because$ (common terms are cut)

$⟹\frac{\frac{(x+1)(x+2)}{x+2}+3x-3x}{x-1}\because {\log }_{2}2=1$

$⟹\frac{x+1}{x-1}$