Question

Answer 1

a)

$\dfrac{4x}{25y^3}\cdot\dfrac{2y^2}{x^2}$

$=\dfrac{4x\cdot2y^2}{25y^3\cdot x^2}$

$=\dfrac{8xy^2}{25x^2y^3}$

$=\dfrac{8}{25xy}$

b)

$\dfrac{9a^2}{a+3}\cdot\dfrac{a^2-9}{6a^3}$

$=\dfrac{9a^2}{a+3}\cdot\dfrac{(a-3)(a+3)}{6a^3}$

$=\dfrac{9a^2(a-3)}{6a^3}$

$=\dfrac{3(a-3)}{2a}$

c)

$\dfrac{4n^2}{17m^4}\cdot\left(-\dfrac{7m^2}{12n}\right)$

$=-\dfrac{4n^2\cdot7m^2}{17m^4\cdot12n}$

$=-\dfrac{28n}{204m^2}$

$=-\dfrac{7n}{51m^2}$

d)

$\dfrac{3b+6}{(b-9)^2}\cdot\dfrac{2b-18}{(b+2)^2}$

$=\dfrac{3(b+2)\cdot2(b-9)}{(b-9)^2(b+2)^2}$

$=\dfrac{6}{(b-9)(b+2)}$

e)

$\dfrac{x+3}{x}\cdot\dfrac{x^2y}{x^2+6x+9}$

$=\dfrac{x+3}{x}\cdot\dfrac{x^2y}{(x+3)^2}$

$=\dfrac{x^2y}{x(x+3)}$

$=\dfrac{xy}{x+3}$

g)

$(x^2+2x+1)\cdot\dfrac{2x+3}{x^2-1}$

$=\dfrac{(x+1)^2(2x+3)}{(x-1)(x+1)}$

$=\dfrac{(x+1)(2x+3)}{x-1}$

Answer 2

\(a,\frac{4x}{25y^3}.\frac{2y^2}{x^2}=\frac{4x.2y^2}{25y^3.x^2}=\frac{4.2}{25y.x}=\frac{8}{25xy}\)
\(b,\frac{9a^2}{a+3}.\frac{a^2-9}{6a^3}=\frac{9a^2\left(a-3\right)\left(a+3\right)}{\left(a+3\right).6a^3}=\frac{3\left(a-3\right)}{2a}=\frac{3a-9}{2a}\)
\(c,\frac{4n^2}{17m^4}.\left(-\frac{7m^2}{12m}\right)=\frac{4n^2.\left(-7m^2\right)}{17m^4.12m}=\frac{n.\left(-7\right)}{17m^2.3}=\frac{-7n}{51m^2}\)
\(d,\frac{3b+6}{\left(b-9\right)^2}.\frac{2b-18}{\left(b+2\right)^2}=\frac{3\left(b+2\right).2\left(b-9\right)}{\left(b-9\right)^2.\left(b+2\right)^2}=\frac{3.2}{\left(b-9\right)\left(b+2\right)}=\frac{6}{\left(b-9\right)\left(b+2\right)}\)
\(e,\frac{x+3}{x}.\frac{x^2y}{x^2+6x+9}=\frac{\left(x+3\right).x^2y}{x.\left(x^2+6x+9\right)}=\frac{\left(x+3\right).x^2y}{x.\left(x+3\right)^2}=\frac{xy}{x+3}\)
\(\)