bài 2:

\(A=\left(a+b+c\right)^3+\left(b+a-c\right)^3+\left(c+a-b\right)^3\)

\(=\left(c+b+a-2c\right)^3+\left(c+a+b-2b\right)^3\)

\(=\left(-2c\right)^3+\left(-2b\right)^3=-8\left(b+c\right)\)

sao nữa nhỉ :v

a. \(\left[{}\begin{matrix}\left\{{}\begin{matrix}a=b=0\\c>0\end{matrix}\right.\\\left\{{}\begin{matrix}a>0\\b^2-4ac< 0\end{matrix}\right.\end{matrix}\right.\)

b. \(\left[{}\begin{matrix}\left\{{}\begin{matrix}a=b=0\\c< 0\end{matrix}\right.\\\left\{{}\begin{matrix}a< 0\\b^2-4ac< 0\end{matrix}\right.\end{matrix}\right.\)

c. \(\left[{}\begin{matrix}\left\{{}\begin{matrix}a=b=0\\c\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}a>0\\b^2-4ac\le0\end{matrix}\right.\end{matrix}\right.\)

d. \(\left[{}\begin{matrix}\left\{{}\begin{matrix}a=b=0\\c\le0\end{matrix}\right.\\\left\{{}\begin{matrix}a< 0\\b^2-4ac\le0\end{matrix}\right.\end{matrix}\right.\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

A = 4acx + 4bcx + 4ax + 4bx ( đã sửa '-' )

= 4x( ac + bc + a + b )

= 4x[ c( a + b ) + ( a + b ) ]

= 4x( a + b )( c + 1 )

B = ax - bx + cx - 3a + 3b - 3c

= x( a - b + c ) - 3( a - b + c )

= ( a - b + c )( x - 3 )

C = 2ax - bx + 3cx - 2a + b - 3c

= x( 2a - b + 3c ) - ( 2a - b + 3c )

= ( 2a - b + 3c )( x - 1 )

D = ax - bx - 2cx - 2a + 2b + 4c

= x( a - b - 2c ) - 2( a - b - 2c )

= ( a - b - 2c )( x - 2 )

E = 3ax² + 3bx² + ax + bx + 5a + 5b

= 3x²( a + b ) + x( a + b ) + 5( a + b )

= ( a + b )( 3x² + x + 5 )

F = ax² - bx² - 2ax + 2bx - 3a + 3b

= x²( a - b ) - 2x( a - b ) - 3( a - b )

= ( a - b )( x² - 2x - 3 )

= ( a - b )( x² + x - 3x - 3 )

= ( a - b )[ x( x + 1 ) - 3( x + 1 ) ]

= ( a - b )( x + 1 )( x - 3 )

\(a,\Leftrightarrow2x^3-x^2+ax+b=\left(x-1\right)\left(x+1\right)\cdot a\left(x\right)\)

Thay \(x=1\Leftrightarrow2-1+a+b=0\Leftrightarrow a+b=-1\)

Thay \(x=-1\Leftrightarrow-2-1-a+b=0\Leftrightarrow b-a=3\)

Từ đó ta được \(\left\{{}\begin{matrix}a+b=-1\\-a+b=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-2\\b=1\end{matrix}\right.\)

\(b,\Leftrightarrow ax^3+bx^2+2x-1=\left(x-1\right)\left(x+6\right)\cdot b\left(x\right)\)

Thay \(x=1\Leftrightarrow a+b+2-1=0\Leftrightarrow a+b=-1\)

Thay \(x=-6\Leftrightarrow-216a+36b+12-1=0\Leftrightarrow216a-36b=11\)

Từ đó ta được \(\left\{{}\begin{matrix}a+b=-1\\216a-36b=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{25}{252}\\b=-\dfrac{227}{252}\end{matrix}\right.\)

\(c,\Leftrightarrow ax^4+bx^3+1=\left(x+1\right)^2\cdot c\left(x\right)\)

Thay \(x=-1\Leftrightarrow a-b+1=0\Leftrightarrow b=a+1\)

\(\Leftrightarrow ax^4+\left(a+1\right)x^3+1⋮\left(x+1\right)\\ \Leftrightarrow ax^4+ax^3+x^3+1⋮\left(x+1\right)\\ \Leftrightarrow ax^3\left(x+1\right)+\left(x+1\right)\left(x^2-x+1\right)⋮\left(x+1\right)\\ \Leftrightarrow\left(x+1\right)\left(ax^3+x^2-x+1\right)⋮\left(x+1\right)\\ \Leftrightarrow ax^3+x^2-x+1⋮\left(x+1\right)\)

Thay \(x=-1\Leftrightarrow-a+1+1+1=0\Leftrightarrow a=3\Leftrightarrow b=4\)