Question

PLEASE HELP ASAP!!!!!!

Answer 1

Answer:

Step-by-step explanation:

$\dfrac{xy}{a^2+a^3}\times \dfrac{a+a^2}{x^2y^2}$

Apply the fraction rule $\dfrac{a}{b}\times \dfrac{c}{d}=\dfrac{a \times c}{b \times d}$ :

$\implies \dfrac{xy(a+a^2)}{x^2y^2(a^2+a^3)}$

Cancel the common factor $xy$ :

$\implies \dfrac{(a+a^2)}{xy(a^2+a^3)}$

Factor $(a+a^2)=a(1+a) \ \ \textsf{and} \ \ a^2+a^3=a^2(1+a)$ :

$\implies \dfrac{a(1+a)}{xy \cdot a^2(1+a)}$

Cancel the common factor $a(1+a)$ :

$\implies \dfrac{1}{axy}$

I think the "if" part is a, x and y cannot equal zero, as otherwise the expression will be undefined.

So I would put: $x\neq 0; y\neq 0, a\neq 0$

Answer 2

Answer:

$\frac{1}{axy}$

Step-by-step explanation:

$\frac{xy}{a^2+a^3}$ × $\frac{a+a^2}{x^2y^2}$ ( factorise denominator and numerator of 2 fractions )

= $\frac{xy}{a^2(1+a)}$ × $\frac{a(1+a)}{x^2y^2}$

Cancel (1 + a) from both fractions

= $\frac{xy}{a^2}$ × $\frac{a}{x^2y^2}$

Cancel a and xy from both fractions

= $\frac{1}{a}$ × $\frac{1}{xy}$

= $\frac{1}{axy}$

a ≠ 0 , x ≠ 0 , y ≠ 0

as this would would make the function undefined