a/

\(x^3\left(x-4\right)+x\left(x-4\right)=\left(x-4\right)\left(x^3+x\right)=\)

\(=x\left(x^2+1\right)\left(x-4\right)\)

b/

\(\left(x-1\right)\left(x^2-9\right)-2x\left(x-3\right)=\)

\(=\left(x-1\right)\left(x-3\right)\left(x+3\right)-2x\left(x-3\right)=\)

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

\(=\left(x-3\right)\left(x^2-3\right)\)

\(\left(a\right):=x\left(y+z\right)+y\left(y+z\right)=\left(y+z\right)\left(x+y\right)\\ \left(b\right):=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\\ \left(c\right):=4^2-\left(x+y\right)^2=\left(4+x+y\right)\left(4-x-y\right)\\ \left(d\right):=x\left(x+y+z\right)-\left(x+y+z\right)=\left(x+y+z\right)\left(x-1\right)\)

Em nên gõ công thức trực quan để được hỗ trợ tốt nhất nhé

Em nên gõ công thức trực quan để được hỗ trợ tốt nhất nhé

\(x^2+y^2+z^2=xy+3x+2z-4\)

\(\Leftrightarrow x^2+y^2+z^2-xy-3x-2z+4=0\)

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

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

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x=y\\z=1\\\dfrac{1}{2}x=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\z=1\\x=2\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)=\left(2;1;1\right)\)

=>x^2+y^2+z^2-xy-3y-2z+4=0
=>x^2-xy+1/4y^2+3/4y^2-3y+3+z^2-2z+1=0
=>(x-1/2y)^2+3/4(y-2)^2+(z-1)^2=0
=>(x-1/2y)^2=0 (y-2)^2=0 (z-1)^2=0 
=>x=1/2y y=2 z=1
=>x=1,y=2,z=1

     (x+1)² + (x-2)(x+3) -4x

=    x² + 2x + 1 + x²+3x - 2x   -6 - 4x

=2 x² -x  - 5