Phân thức đối của \(\dfrac{-3}{x-y}\) là \(\dfrac{3}{x-y}\)

\(\dfrac{1}{3}x^2y^5\left(\dfrac{-3}{5}x^3y\right)+x^5y^6=\dfrac{-1}{5}x^5y^6+x^5y^6=\dfrac{4}{5}x^5y^6\)

\(B=\dfrac{3x^2+6x+9+1}{x^2+2x+3}=\dfrac{3\left(x^2+2x+3\right)+1}{x^2+2x+3}=3+\dfrac{1}{x^2+2x+3}\)

Vậy B đạt GTLN khi \(\dfrac{1}{x^2+2x+3}\) đạt GTLN. Vậy \(x^2+2x+3\) phải đạt GTNN

\(x^2+2x+1+2=\left(x+1\right)^2+2\ge2\)

Vậy GTNN của B là 2 khi x=-1

b) \(\dfrac{5x+4-3\left(x-2\right)}{3\left(x+5\right)}=\dfrac{5x+4-3x+6}{3\left(x+5\right)}=\dfrac{2x+10}{3\left(x+5\right)}=\dfrac{2\left(x+5\right)}{3\left(x+5\right)}=\dfrac{2}{3}\)

a) \(\dfrac{x^2+2x}{x+2}=\dfrac{x\left(x+2\right)}{x+2}=x\)