Question

3(x+4)-x^2-4x=0

x^3-x+3x^2y +y^3-y

Answer 1

\(a,3\left(x+4\right)-x\left(x+4\right)=0\\ \left(3-x\right)\left(x+4\right)=0\\ \left[{}\begin{matrix}3-x=0\\x+4=0\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\)

b, \(Sửađề\) 

\(x^3+3x^2y+3xy^2+y^3-x-y\\ =\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\\ =\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)

Answer 2

a: =>(x+4)(3-x)=0

=>x=3 hoặc x=-4