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\(ax^2-3axy+bx-3by\\ =x\left(ax+b\right)-3y\left(ax+b\right)\\ =\left(x-3y\right)\left(ax+b\right)\)
\(5x^2y+5xy^2-a^2x-a^2y\\ =5xy\left(x+y\right)-a^2\left(x+y\right)\\ =\left(5xy-a^2\right)\left(x+y\right)\)
\(2ax^3+6ax^2+6ax+18a\\ =2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\)
\(10xy^2-5by^2+2ax-ab\\ =5y^2\left(2x-b\right)+a\left(2x-b\right)\\ =\left(5y^2+a\right)\left(2x-b\right)\)
\(ax-bx+cx-3a+3b-3c\\ =x\left(a-b+c\right)-3\left(a-b+c\right)\\ =\left(x-3\right)\left(a-b+c\right)\)
a) \(2a^2x-5by-5a^2y+2by\)
\(=3\left(\frac{2}{3}a^2x-\frac{5}{3}a^2y\right)-3by\)
\(=3\left(\frac{2}{3}a^2x-\frac{5}{3}a^2y-by\right)\)
Bài 2:
a) x(x - 3)- y(3 - x)
= x(x - 3) + y(x - 3)
= (x - 3)(x + y) (1)
Thay x = \(\frac{1}{3}\); y = \(\frac{8}{3}\)vào (1)
Ta có: (\(\frac{1}{3}\)- 3)(\(\frac{1}{3}\)+ \(\frac{8}{3}\))
= \(\frac{-8}{3}\). 3
= -8
\(2ax-bx+3cx-2a+b-3c\\ =x\left(2a-b+3c\right)-\left(2a-b+3c\right)\\ =\left(x-1\right)\left(2a-b+3c\right)\)
\(ax-bx-2cx-2a+2b+4c\\ =x\left(a-b-2c\right)-2\left(a-b-2c\right)\\ =\left(x-2\right)\left(a-b-2c\right)\)
\(3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\)
\(ax^2-bx^2-2ax+2bx-3a+3b\\ =x^2\left(a-b\right)-2x\left(a-b\right)-3\left(a+b\right)\\ =\left(x^2-2x-3\right)\left(a+b\right)\\ =\left(x+1\right)\left(x-3\right)\left(a+b\right)\)
a, x(a - b) + (a - b)
= (x + 1)(a - b)
b, x(a + b) - a - b
= x(a + b) - (a + b)
= (x - 1)(a + b)
c, 10ax - 5ay - 2x + y
= 5a(2x - y) - (2x - y)
= (5a - 1)(2x - y)
d, 2a^2x - 5by - 5a^2y + 2bx
= 2x(a^2 + b) - 5y(b + a^2)
= (2a - 5y)(a^2 + b)
làm tiếp:
2ax2 - bx2 - 2ax +bx +4a-2b
= x2(2a-b) - x(2a-b) +2(2a-b)
=(2a-b)(x2-x+2)
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 = 10ax - 5ay - 2x + y
= ( 10ax - 5ay ) - ( 2x - y )
= 5a( 2x - y ) - ( 2x - y )
= ( 2x - y )( 5a - 1 )
B = 2x2 - 6xy + 5x - 15y
= 2x( x - 3y ) + 5( x - 3y )
= ( x - 3y )( 2x + 5 )
C = ax2 - 3axy + bx - 3by
= ( ax2 + bx ) - ( 3axy + 3by )
= x( ax + b ) - 3y( ax + b )
= ( ax + b )( x - 3y )
D = 2ax3 + 6ax2 + 6ax + 18a
= 2ax2( x + 3 ) + 6a( x + 3 )
= ( x + 3 )( 2ax2 + 6a )
= ( x + 3 )2a( x2 + 3 )
E = 5x2y + 5xy2 + a2x + a2y ( đã sửa 1 dấu '-' )
= 5xy( x + y ) + a2( x + y )
= ( x + y )( 5xy + a2 )
F = 10xy2 - 5by2 + 2a2x - aby ( xem lại đề chứ không phân tích được :)) )
a) Ta có: \(4\left(2-x\right)^2+xy-2y\)
\(=4\left(x-2\right)^2+y\left(x-2\right)\)
\(=\left(x-2\right)\left[4\left(x-2\right)+y\right]\)
\(=\left(x-2\right)\left(4x-8+y\right)\)
b) Ta có: \(3a^2x-3a^2y+abx-aby\)
\(=3a^2\left(x-y\right)+ab\left(x-y\right)\)
\(=\left(x-y\right)\left(3a^2+ab\right)\)
\(=a\left(x-y\right)\left(3a+b\right)\)
c) Ta có: \(x\left(x-y\right)^3-y\left(y-x\right)^2-y^2\left(x-y\right)\)
\(=x\left(x-y\right)^3-y\left(x-y\right)^2-y^2\left(x-y\right)\)
\(=\left(x-y\right)\left[x\left(x-y\right)^2-y\left(x-y\right)-y^2\right]\)
\(=\left(x-y\right)\left[x\left(x^2-2xy+y^2\right)-yx+y^2-y^2\right]\)
\(=\left(x-y\right)\left(x^3-2x^2y+xy^2-xy\right)\)
d) Ta có: \(2ax^3+6ax^2+6ax+18a\)
\(=2ax^2\left(x+3\right)+6a\left(x+3\right)\)
\(=\left(x+3\right)\left(2ax^3+6a\right)\)
\(=2a\left(x+3\right)\left(x^3+3\right)\)
e) Ta có: \(x^2y-xy^2-3x+3y\)
\(=xy\left(x-y\right)-3\left(x-y\right)\)
\(=\left(x-y\right)\left(xy-3\right)\)
