a) TA có :

\(\left(x^2+cx+2\right)\left(ax+b\right)=ax^3+bx^2+acx^2+bcx+2ax+2b\)

\(=ax^3+x^2\left(b+ac\right)+x\left(bc+2a\right)+2b\) = \(=x^3-x^2-2\)

=> a = 1 

=>\(2b=-2\Rightarrow b=-1\)

=> b + ac = -1 => -1 + 1.c = -1 => -1 + c = -1 => c = -1 + 1 = 0 

VẬy a = 1 ; b = -1 ; c = 0 

\(a,xy+1-x-y\)

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

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

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

\(b,ax+ay-3x-3y\)

\(=a\left(x+y\right)-3\left(x+y\right)\)

\(=\left(x+y\right)\left(a-3\right)\)

\(c,x^3-2x^2+2x-4\)

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

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

\(d,x^2+ab+ax+bx\)

\(=\left(x^2+ax\right)+\left(ab+bx\right)\)

\(=x\left(a+x\right)+b\left(a+x\right)\)

\(=\left(a+x\right)\left(b+x\right)\)

\(e,16-x^2+2xy-y^2\)

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

\(=4^2-\left(x-y\right)^2\)

\(=\left(4-x+y\right)\left(4+x-y\right)\)

\(f,ax^2+ax-bx^2-bx-a+b\)

\(=\left(ax^2-bx^2\right)+\left(ax-bx\right)-\left(a-b\right)\)

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

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

a) \(xy+1-x-y\)

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

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

b) \(ax+ay-3x-3y\)

\(=a\left(x+y\right)-3\left(x+y\right)\)

\(=\left(x+y\right)\left(a-3\right)\)

c) \(x^3-2x^2+2x-4\)

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

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

d) \(x^2+ab+ax+bx\)

\(=x\left(b+x\right)+a\left(b+x\right)\)

\(=\left(b+x\right)\left(a+x\right)\)

e) \(16-x^2+2xy-y^2\)

\(=16-\left(x^2-2xy+y^2\right)\)

\(=4^2-\left(x-y\right)^2\)

\(=\left(4-x+y\right)\left(4+x-y\right)\)

f) \(ax^2+ax-bx^2-bx-a+b\)

\(=\left(ax^2+ax-a\right)-\left(bx^2+bx-b\right)\)

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

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

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

b) \(ax+ay-3x-3y=a\left(x+y\right)-3\left(x-y\right)=\left(a-3\right)\left(x+y\right)\)

c) \(x^3-2x^2+2x-4=x^2\left(x-2\right)+2\left(x-2\right)=\left(x^2+2\right)\left(x-2\right)\)

d) \(x^2+ab+ax+bx=\left(x^2+ax\right)+\left(ab+bx\right)=x\left(x+a\right)+b\left(a+x\right)=\left(x+b\right)\left(x+a\right)\)

e) \(16-x^2+2xy-y^2=16-\left(x-y\right)^2=\left(4-x+y\right)\left(4+x-y\right)\)

f) \(ax^2+ax-bx^2-bx-a+b=\left(a-b\right)x^2+\left(a-b\right)x-a+b=\left(a-b\right)\left(x^2+x-1\right)\)

làm mẫu 1 phần thôi men còn lại tự làm 

giải

a) 

 

Để \(A\left(x\right)⋮B\left(x\right)\)\(\Leftrightarrow\hept{\begin{cases}b-3a+16a=0\\24-12a=0\end{cases}}\)

                                    \(\Leftrightarrow\hept{\begin{cases}b+13.2=0\\a=2\end{cases}}\)

                                     \(\Leftrightarrow\hept{\begin{cases}b=-26\\a=2\end{cases}}\)

Tiểu biểu một câu thôi, mấy câu còn lại tương tự. 

Tư tưởng là phân tích vế trái để sử dụng đồng nhất hệ số.

b) \(\left(ax+b\right)\left(x^2-x-1\right)=ax^3+cx^2-1\)

\(\Leftrightarrow ax^3-ax^2+bx^2-ax-bx-b=ax^3+cx^2-1\)

\(\Leftrightarrow ax^3+x^2\left(-a+b\right)-x\left(a+b\right)-b=ax^3+c\cdot x^2-0\cdot x-1\)

Đồng nhất hệ số:

\(\hept{\begin{cases}-a+b=c\\a+b=0\\b=1\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a=-1\\b=1\\c=2\end{cases}}\)

Các câu còn lại tương tự.

dài thế