Question

cần lời giải chi tiết

 

Answer 1

\(a,=\dfrac{2y\left(3x-1\right)^2}{3x^2}\\ b,=\dfrac{\left(3x+2-x-2\right)\left(3x+2+x+2\right)}{x^2\left(x-1\right)}\\ =\dfrac{2x\left(4x+4\right)}{x^2\left(x-1\right)}=\dfrac{8\left(x+1\right)}{x\left(x-1\right)}\\ c,=\dfrac{x^2+3x+4x+12}{x^2+2x+3x+6}\\ =\dfrac{\left(x+3\right)\left(x+4\right)}{\left(x+2\right)\left(x+3\right)}=\dfrac{x+4}{x+2}\)