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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 !
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) \(ax^2+ay-bx^2-by\)
\(=\left(ax^2+ay\right)-\left(bx^2+by\right)\)
\(=a\left(x^2+y\right)-b\left(x^2+y\right)\)
\(=\left(a-b\right)\left(x^2+y\right)\)
a,\(\left(a-b\right)\left(a+2b\right)-\left(b-a\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)
\(=\left(a-b\right)\left(a+2b\right)+\left(a-b\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)
\(=\left(a-b\right)\left(a+2b+2a-b-a-3b\right)\)
\(=\left(a-b\right)\left(2a-2b\right)\)
\(=\left(a-b\right)2\left(a-b\right)\)
\(=2\left(a-b\right)^2\)
b,\(\left(x+y\right)\left(2x-y\right)+\left(2x-y\right)\left(3x-y\right)-\left(y-2x\right)\)
\(=\left(x+y\right)\left(2x-y\right)+\left(2x-y\right)\left(3x-y\right)+\left(2x-y\right)\)
\(=\left(2x-y\right)\left(x+y+3x-y+1\right)\)
\(=\left(2x-y\right)\left(4x+1\right)\)
c,\(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\)
\(=x^2y-x^2z+y^2z-y^2x+z^2\left(x-y\right)\)
\(=x^2y-y^2x-x^2z+y^2z+z^2\left(x-y\right)\)
\(=xy\left(x-y\right)-z\left(x^2-y^2\right)+z^2\left(x-y\right)\)
\(=xy\left(x-y\right)-z\left(x-y\right)\left(x+y\right)+z^2\left(x-y\right)\)
\(=\left(x-y\right)\left(xy-zx-zy+z^2\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)
a) ax - 2x - a2 + 2a
= ( ax - 2x ) - ( a2 - 2a )
= x ( a - 2 ) - a ( a - 2 )
= ( a - 2 ) ( x - a )
b) x2 + x - ax - a
= ( x2 + x ) - ( ax + a )
= x ( x + 1 ) - a ( x + 1 )
= ( x + 1 ) ( x - a )
Hok Tốt!!!
a) ax -2x- a2+ 2a
= (ax -2x ) -(a2 -2a )
= x(a-2) -a ( a-2 )
= (x-a) (a-2)
b) x2 +x -ax -a
=( x2 +x ) - ( ax +a )
= x( x+1 ) -a ( x+1 )
= ( x-a ) (x+ 1)
c) 2x2 +4ax +x +2a
=( 2x2 + 4ax ) + ( x+ 2a )
= 2x ( x+ 2a ) + ( x+2a )
= ( 2x +1 ) (x+2a )
d) 2xy -ax +x2 - 2ay
= (2xy -2ay ) + ( -ax + x2 )
= 2y( x-a ) + x ( x-a)
= ( 2y +x ) ( x -a )
Bài làm:
a) \(x^2-2xy+y^2-zx+yz\)
\(=\left(x-y\right)^2-z\left(x-y\right)\)
\(\left(x-y\right)\left(x-y-z\right)\)
a/ \(x^2-2xy+y^2-zx+yz.\)
\(=\left(x-y\right)^2-z\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-z\right)\)
c/ \(x^2-y^2-2x-2y.\)
\(=x^2-2x+1-y^2-2y-1\)
\(=\left(x^2-2x+1\right)-\left(y^2+2y+1\right)\)
\(=\left(x-1\right)^2-\left(y+1\right)^2\)
\(=\left(x-1+y+1\right)\left(x-1-y-1\right)\)
\(=\left(x+y\right)\left(x-y-2\right)\)
Lời giải:
31.
\(2a^2x-5by-6a^2y+2bx=(2a^2x+2bx)-(5by+5a^2y)\)
\(=2x(a^2+b)-5y(b+a^2)=(a^2+b)(2x-5y)\)
34.
\(4acx+4bcx+4ax+4bx=4x(ac+bc+a+b)\)
\(=4x[(ac+bc)+(a+b)]=4x[c(a+b)+(a+b)]=4x(c+1)(a+b)\)
37. Sửa đề:
\(2ax^2-bx^2-2ax+bx+4a-2b\)
\(=(2ax^2-bx^2)-(2ax-bx)+(4a-2b)\)
\(=x^2(2a-b)-x(2a-b)+2(2a-b)=(2a-b)(x^2-x+2)\)
Câu 31:
\(2a^2x-5by-5a^2y+2bx\)
\(=2x\left(a^2+b\right)-5y\left(a^2+b\right)\)
\(=\left(a^2+b\right)\left(2x-5y\right)\)
Câu 34:
\(4acx+4bcx+4ax+4bx\)
\(=4cx\left(a+b\right)+4x\left(a+b\right)\)
\(=\left(a+b\right)\left(4cx+4x\right)\)
\(=4x\left(a+b\right)\left(c+1\right)\)
Câu 37:
\(2ax^2-bx^2-2ax+bx+4a-2b\)
\(=x^2\left(2a-b\right)-x\left(2a-b\right)+2\left(2x-b\right)\)
\(=\left(2a-b\right)\left(x^2-x+2\right)\)
\(=\left(2a-b\right)\left(x^2-x+2\right)\)