\(x^3+y^3=5+x^2y+xy^2\Rightarrow x^3+y^3-\left(x^2y+xy^2\right)=5\)

\(\Rightarrow\left(x+y\right)\left(x^2-xy+y^2\right)-xy\left(x+y\right)=5\)

\(\Rightarrow\left(x+y\right)\left(x-y\right)^2=5\)

Vì \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\\5>0\end{matrix}\right.\Rightarrow x+y>0\)

Lại có \(\left\{{}\begin{matrix}\left(x-y\right)^2\in N\\\left(x-y\right)^2< 5\end{matrix}\right.\) và \(\left(x-y\right)^2\) là số chính phương

\(\Rightarrow\left(x-y\right)^2=1\Rightarrow\left\{{}\begin{matrix}x+y=5\\x-y=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)

mn giúp mk vs ạ

\(\Leftrightarrow\left(x^2+\dfrac{y^2}{4}+\dfrac{9}{4}+xy-3x-\dfrac{3y}{2}\right)+\dfrac{3}{4}\left(y^2-2y+1\right)=0\)

\(\Leftrightarrow\left(x+\dfrac{y}{2}-\dfrac{3}{2}\right)^2+\dfrac{3}{4}\left(y-1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{y}{2}-\dfrac{3}{2}=0\\y-1=0\end{matrix}\right.\)

\(\Rightarrow x=y=1\)

<=> \(x^2+y^2+\left(xy-3x\right)-\left(3y-3\right)=0\)

\(\Leftrightarrow x^2+y^2+3x^2\left(y-1\right)-3\left(y-1\right)=0\)

<=> \(x^2+y^2\left(3x^2-3\right)\left(y-1\right)=0\)

\(\Leftrightarrow x^2+y^2.3.\left(x-1\right)\left(x+1\right)\left(y-1\right)=0\)

=> x = 0 ; 1 ; -1  . y =0  ; 1

P/s : ngủ được ròi:3

a) x(x-1)=0+12

    x(x-1)=12

    x(x-1)=4.3

=>x=4

a, \(x^2-x-12=0\)

\(x^2+\left(-x\right)+\left(-12\right)=0\)

\(\Delta=-1^2-4.1.\left(-12\right)=1+48=49>0\)

Nên pt có 2 nghiệm phân biệt 

\(x_1=\frac{1-\sqrt{49}}{2.1}=\frac{1-7}{2}=-\frac{6}{2}=-3\)

\(x_2=\frac{1+\sqrt{49}}{2.1}=\frac{1+7}{2}=\frac{8}{2}=4\)