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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 )
\(1,2x^2-6xy+5x-15y\)
\(=2x\left(x-3y\right)+5\left(x-3y\right)\)
\(=\left(x-3y\right)\left(2x+5\right)\)
\(2,ax^{2\:}-3axy+bx-3by\)
\(=ax\left(x-3y\right)+b\left(x-3y\right)\)
\(=\left(x-3y\right)\left(ax+b\right)\)
\(3,5ax^2-3axy+3ay^2-3axy\) ( Đề sai )
Sửa : \(3ax^2-3axy+3ay^2-3axy\)
\(=3ax\left(x-y\right)+3ay\left(y-x\right)\)
\(=3ax\left(x-y\right)-3ay\left(x-y\right)\)
\(=3a\left(x-y\right)^2\)
\(4,4acx+4bcx+4ax+4bx\)
\(=4cx\left(a+b\right)+4x\left(a+b\right)\)
\(=4x\left(a+b\right)\left(c+1\right)\)
\(6,ax^{2\:}y-bx^2y-ax+bx+2a-2b\)
\(=x^2y\left(a-b\right)-x\left(a-b\right)+2\left(a-b\right)\)
\(=\left(a-b\right)\left(x^2y-x+2\right)\)
\(7,ax^{2\:}-bx^2-2ax+2bx-3a+3b\)
\(=x^2\left(a-b\right)-2x\left(a-b\right)-3\left(a-b\right)\)
\(=\left(a-b\right)\left(x^2-2x-3\right)\)
\(8,ax^{2\:}-5x^2-ax+5x+a-5\)
\(=x^2\left(a-5\right)-x\left(a-5\right)+\left(a-5\right)\)
\(=\left(a-5\right)\left(x^2-x+1\right)\)
\(9,ax+bx+cx-2a-2b+2c\) Đề sai
Sửa :\(ax+bx+cx-2a-2b-2c\)
\(=x\left(a+b+c\right)-2\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(x-2\right)\)
\(10,2ax-bx+3cx-2a+b-3c\)
\(=\left(2ax-2a\right)-\left(bx-b\right)+\left(3cx-3c\right)\)
\(=2a\left(x-1\right)-b\left(x-1\right)+3c\left(x-1\right)\)
\(=\left(x-1\right)\left(2a-b+3c\right)\)
Mấy câu đề sai mk sửa chỗ nào ko đúng thì nói mk nha !
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 )
a) \(x^2-x-y^2+y\)
\(=x\left(x-1\right)-y\left(y-1\right)\)
\(=\left(x-1\right)\left(x-y\right)\)
b) \(x^2-2xy-z^2+y^2\)
\(=\left(x^2-2xy+y^2\right)-z^2\)
\(=\left(x-y\right)^2-z^2\)
\(=\left(x-y-z\right)\left(x-y+z\right)\)
c) \(5x-5y+ax-ay\)
\(=5\left(x-y\right)+a\left(x-y\right)\)
\(=\left(x-y\right)\left(5+a\right)\)
a) ax2 - 2bxy + 2bx2 - axy
= ( ax2 - axy ) + ( 2bx2 - 2bxy )
= ax( x - y ) + 2bx( x - y )
= ( x - y )( ax + 2bx )
= x( x - y )( a + 2b )
b) x2 + 2x - 4y2 + 8y - 3 < đã sửa >
= ( x2 + 2x + 1 ) - ( 4y2 - 8y + 4 )
= ( x + 1 )2 - ( 2y - 2 )2
= [ ( x + 1 ) - ( 2y - 2 ) ][ ( x + 1 ) + ( 2y - 2 ) ]
= ( x + 1 - 2y + 2 )( x + 1 + 2y - 2 )
= ( x - 2y + 3 )( x + 2y - 1 )
c) x4 + 5x3 + 20x - 16
= x4 + 5x3 + 4x2 - 4x2 + 20x - 16
= ( x4 + 5x3 - 4x2 ) + ( 4x2 + 20x - 16 )
= x2( x2 + 5x - 4 ) + 4( x2 + 5x - 4 )
= ( x2 + 5x - 4 )( x2 + 4 )
a) \(x^3-2x^2+2x-1^3\)
\(=x\left(x^2-2x+1\right)+x-1\)
\(=x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x+1\right)\left(x-1\right)\)
b) \(x^2y+xy+x+1\)
\(=xy\left(x+1\right)+\left(x+1\right)\)
\(=\left(xy+1\right)\left(x+1\right)\)
c) \(ax+by+ay+bx\)
\(=a\left(x+y\right)+b\left(x+y\right)\)
\(=\left(a+b\right)\left(x+y\right)\)
d) \(x^2-\left(a+b\right)x+ab\)
\(=x^2-ax-bx+ab\)
\(=\left(x^2-ax\right)-\left(bx-ab\right)\)
\(=x\left(x-a\right)-b\left(x-a\right)\)
\(=\left(x-b\right)\left(x-a\right)\)
e) Ko biết làm
f) \(ax^2+ay-bx^2-by\)
\(=\left(ax^2+ay\right)-\left(bx^2+by\right)\)
\(=a\left(x^2+y\right)-b\left(x^2+y\right)\)
\(=\left(a-b\right)\left(x^2+y\right)\)
\(ax+bx+ay+by\)
\(=x\left(a+b\right)+y\left(a+b\right)\)
\(=\left(x+y\right)\left(a+b\right)\)
\(xy+1-x-y\)
\(=x\left(y-1\right)-\left(y-1\right)\)
\(=\left(x-1\right)\left(y-1\right)\)
ax2-5x2-ax+5x+a-5
=x^2(a-5)-x(a-5)+(a-5)
=(a-5)(x^2-x+1)
cậu ghi sai đề rồi phải là
3ax2+3bx2+ax+bx+5a+5b
=3x^2(a+b)+x(a+b)+5(a+b)
=(a+b)(3x^2+x+5)
Đề sai nhé .Sửu lại
\(x^2-4x^2y^2+4+4x\)
\(=\left(x^2+4x+4\right)-4x^2y^2\)
\(=\left(x+2\right)^2-\left(2xy\right)^2\)
\(=\left(x+2+2xy\right)\left(x+2-2xy\right)\)
Bài 2: Phân tích các đa thức sau thành nhân tử ( phương pháp nhóm hạng tử ) .
a) ax2 - 5x2 - ax + 5x + a - 5;
= ( ax2 - 5x2) - (ax - 5x) + (a - 5)
= x2(a - 5) - x(a - 5) + (a - 5)
= (a - 5) (x2 - x + 1)
b) ax - bx + cx - 3a + 3b - 3c
= ( ax - bx + cx) - (3a - 3b + 3c)
= x(a - b + c) - 3(a - b + c)
= (a - b + c)(x - 3)