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

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

Vì \(\left(x^2-1\right)\left(x+1\right)⋮\left(x+1\right)\)\(\Rightarrow\)Để \(x^3+x^2+a-x\)chia hết cho \(x+1\)thì \(a+1=0\)

\(\Rightarrow a=-1\)

Vậy \(a=-1\)

\(x^3y-xy^3-2xy^2-xy\)

\(=xy\left(x^2-y^2-2y-1\right)\)

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

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

\(=xy\left(x-y-1\right)\left(x+y+1\right)\)

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

x có điều kiện gì ko bạn

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

Vì \(2\left(x-1\right)^2\ge0\forall x\)\(\Rightarrow2x^2-4x\ge-2\forall x\)

Dấu " = " xảy ra \(\Leftrightarrow x-1=0\)\(\Leftrightarrow x=1\)

Vậy GTNN của biểu thức là \(-2\)\(\Leftrightarrow x=1\)

mn giúp mình vs,đang cần gấp.

1a) 5x - 5y + 3x (x - y)

= (5x - 5y) + 3x (x - y)

= 5 (x - y) + 3x (x - y)

= (5 + 3x) (x - y)

b) x² + 2xy + y² - 4

= (x² + 2xy + y²) - 2²

= (x + y)² - 2²

= [(x + y) + 2] [(x + y) - 2]

= (x + y + 2) (x + y - 2)

#Học tốt!!!

~NTTH~