Question

Answer 1

mọi người giải giùm em ạ, em cảm ơn nhiều

 

Answer 2

\(a,=\dfrac{4x^2y+3-3+5xy^2}{3xy}=\dfrac{xy\left(4x+5y\right)}{3xy}=\dfrac{4x+5y}{3}\\ b,=\dfrac{3}{\left(3x+1\right)^2}+\dfrac{3x}{\left(1-3x\right)\left(3x+1\right)}\\ =\dfrac{3-9x+9x^2+3x}{\left(3x+1\right)^2\left(1-3x\right)}=\dfrac{9x^2-6x+3}{\left(3x+1\right)^2\left(1-3x\right)}\\ c,=\dfrac{5x-5+2x^2+2x}{\left(x-1\right)\left(x+1\right)^2}\cdot\dfrac{3\left(x-1\right)}{2x^2+7x-5}\\ =\dfrac{2x^2+7x-5}{\left(x+1\right)^2}\cdot\dfrac{3}{2x^2+7x-5}=\dfrac{3}{\left(x+1\right)^2}\)