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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 )
d)\(x^2-ax-bx+ab=x\left(x-a\right)-b\left(x-a\right)\)
\(=\left(x-b\right)\left(x-a\right)\)
e)\(x^2y+xy^2-x-y=xy\left(x+y\right)-\left(x+y\right)\)
\(=\left(xy-1\right)\left(x+y\right)\)
f)\(ax^2+ay-bx^2-by=a\left(x^2+y\right)-b\left(x^2+y\right)\)
\(=\left(a-b\right)\left(x^2+y\right)\)
a) \(ax+ay-3x-3y=a\left(x+y\right)-3\left(x+y\right)=\left(a-3\right)\left(x+y\right)\)
b) \(x^3-3x^2+3x-9=x^2\left(x-3\right)+3\left(x-3\right)=\left(x-3\right)\left(x^2+3\right)\)
c) xem lại đề
d) \(9-x^2-2xy-y^2=9-\left(x+y\right)^2=\left(3-x-y\right)\left(3+x+y\right)\)
\(=x^2\left(a+b\right)-6x\left(a+b\right)+9\left(a+b\right)\)
\(=\left(a+b\right)\left(x^2-6x+9\right)\)
\(=\left(a+b\right)\left(x-3\right)^2\)
bn làm sai rồi kìa bn
-6x nhân với b là ra âm 6bx rùi bn
mà đề cho là dương 6bx
a) \(x^2+2x-4y^2-4y=\left(x^2-4y^2\right)+\left(2x-4y\right)=\left(x+2y\right)\left(x-2y\right)+2\left(x-2y\right)\)
\(=\left(x-2y\right).\left(x+2y+2\right)\)
b) \(x^4-6x^3+54x-81=\left(x^4-81\right)-\left(6x^3-54x\right)=\left(x^2-9\right)\left(x^2+9\right)-6x.\left(x^2-9\right)\)
\(=\left(x^2-9\right).\left(x^2+9-6x\right)=\left(x+3\right).\left(x-3\right).\left(x-3\right)^2=\left(x+3\right).\left(x-3\right)^3\)
c) \(ax^2+ax-bx^2-bx-a+b=\left(ax^2-bx^2\right)+\left(ax-bx\right)-\left(a-b\right)\)
\(=x^2.\left(a-b\right)+x.\left(a-b\right)-\left(a-b\right)=\left(a-b\right).\left(x^2+x-1\right)\)
d) \(\left(x^2+y^2-2\right)^2-\left(2xy-2\right)^2=\left(x^2+y^2-2+2xy-2\right).\left(x^2+y^2-2-2xy+2\right)\)
\(=\left(x^2+2xy+y^2-4\right).\left(x^2+y^2-2xy\right)=\left[\left(x+y\right)^2-4\right].\left(x-y\right)^2\)
\(=\left(x+y+2\right).\left(x+y-2\right).\left(x-y\right)^2\)
\(\left(x+a\right)\left(x+b\right)\left(x+c\right)=\left(x^2+bx+ax+ab\right)\left(x+c\right)\)
\(=x^3+cx^2+bx^2+bcx+ax^2+acx+abx+abc\)
\(=x^3+\left(a+b+c\right)x^2+\left(ab+ac+bc\right)x+abc\)
Đồnh nhất đa thức trên với đa thức \(x^3+ax^2+bx+c\),ta đc hệ điều kiện:
\(\hept{\begin{cases}a+b+c=a\left(1\right)\\ab+ac+bc=b\left(2\right)\\abc=c\left(3\right)\end{cases}}\)
Từ \(\left(1\right)a+b+c=a=>b+c=0=>c=-b\)
Thay vào (2),ta đc: \(ab+a.\left(-b\right)+b.\left(-b\right)=b=>ab-ab-b^2=b=>-b^2=b\)
\(=>b^2+b=0=>b\left(b+1\right)=0=>\orbr{\begin{cases}b=0\\b=-1\end{cases}}\)
+b=0 thì từ (1) suy ra c=0 ; a tùy ý
+b=-1 thì từ (1) suy ra c=1
Mà theo (3)\(abc=c=>a=\frac{c}{bc}=\frac{1}{-1}=-1\)
Vậy a=-1 hoặc a tùy ý ;b=0 hoặc b=-1;c=0 hoặc c=1
\(a,ax+by+ay+bx=\left(ax+ay\right)+\left(by+bx\right)=a\left(x+y\right)+b\left(x+y\right)=\left(a+b\right)\left(x+y\right)\)
\(b,x^2y+xy+x+1=xy\left(x+1\right)+\left(x+1\right)=\left(xy+1\right)\left(x+1\right)\)
\(c,x^2-ax-bx+ab=x\left(x-a\right)-b\left(x-a\right)=\left(x-b\right)\left(x-2\right)\)
\(d,x^2y+xy^2-x-y=xy\left(x+y\right)-\left(x+y\right)=\left(xy-1\right)\left(x+y\right)\)
\(e,a\left(x^2+y\right)-b\left(x^2+y\right)=\left(a-b\right)\left(x^2+y\right)\)
\(f,x\left(a-2\right)-a\left(a-2\right)=\left(x-a\right)\left(a-2\right)\)
d) ax2 + ay - bx2 - by
= ( ax2 + ay ) - ( bx2 + by )
= a ( x2 + y ) - b ( x2 + y )
= ( x2 + y )( a - b )
c) x2y + xy2 - x - y
= ( x2y + xy2 ) - ( x + y )
= xy ( x + y ) - ( x+ y )
= ( x + y ) ( xy - 1 )
ax2 - ax + bx2 -bx + a + b
= (ax2+ bx2 ) - (ax + bx) + (a + b)
=x2 (a + b) - x(a + b) + (a + b)
= (x2 - x + 1)(a + b)
ax2 - ax + bx2 - bx + a + b
= ( ax2 + bx2 ) - ( ax + bx ) + ( a + b )
= x2( a + b ) - x( a + b ) + ( a + b )
= ( a + b )( x2 - x + 1 )