Tham khảo
`a)` Có `x+y+1=0⇒x+y=-1`
`⇒N=x^2(x+y)-y^2(x+y)+(x^2-y^2)+2(-1)+3`
`⇒N=(x+y)(x^2-y^2)+(x^2-y^2)+1`
`⇒N=(x^2-y^2)(x+y+1)+1`
`⇒N=(x^2-y^2).0+1`
`⇒N=1`
`b)` Có `\frac{x}{14}=\frac{y}{15}⇔x=\frac{14y}{15}`
Do đó $\dfrac{5x-4y}{3x-2y}=\dfrac{5.\dfrac{14y}{15}-4y}{3.\dfrac{14y}{15}-2y}$
$=\dfrac{\dfrac{14y}{3}-4y}{\dfrac{14}{5}-2y}$
$=\dfrac{\dfrac{2y}{3}}{\dfrac{4y}{5}}$
$=\dfrac{\dfrac{2}{3}}{\dfrac{4}{5}}$
`=\frac{5}{6}`
`\text{©CBT}`