\(=\dfrac{5y+3}{y\left(y-3\right)}-\dfrac{9-y}{3\left(y-3\right)}=\dfrac{3\left(5y+3\right)-y\left(9-y\right)}{3y\left(y-3\right)}\\ =\dfrac{15y+9-9y+y^2}{3y\left(y-3\right)}=\dfrac{y^2+6y+9}{3y\left(y-3\right)}=\dfrac{\left(y+3\right)^2}{3y\left(y-3\right)}\)
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