Question

(3xy/3x^2-12)*(x^2+3x+2/xy+y)

Answer 1

Value of expression $(3xy/3x^2-12)*(x^2+3x+2/xy+y)$ is  $\frac{x}{x-2}$ .

Step-by-step explanation:

Here we need to evaluate expression : (3xy/3x^2-12)*(x^2+3x+2/xy+y)  or ,

$(3xy/3x^2-12)*(x^2+3x+2/xy+y)$

Let's simplify this

⇒ $(\frac{3xy}{3x^2-12})(\frac{x^2+3x+2}{xy+y})$

Factorizing the terms we get:

⇒ $(\frac{3xy}{3(x^2-4)})(\frac{x^2+2x+x+2}{y(x+1)})$              { $a^2-b^2=(a+b)(a-b)$      }

⇒ $(\frac{xy}{(x-2)(x+2)})(\frac{x(x+2)+1(x+2)}{y(x+1)})$

⇒ $(\frac{xy}{(x-2)(x+2)})(\frac{(x+1)(x+2)}{y(x+1)})$

Cancelling similar terms we get:

⇒ $\frac{x}{x-2}$

Therefore , Value of expression $(3xy/3x^2-12)*(x^2+3x+2/xy+y)$ is  $\frac{x}{x-2}$ .