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6) \(9x^3y^2+3x^2y^2=3x^2y^2\left(3x+1\right)\)
7) \(x^3+2x^2+3x=x\left(x^2+2x+3\right)\)
8) \(6x^2y+4xy^2+2xy=2xy\left(3x+2y+1\right)\)
9) \(5x^2\left(x-2y\right)-15x\left(x-2y\right)=5x\left(x-2y\right)\left(x-3\right)\)
10) \(3\left(x-y\right)-5x\left(y-x\right)=\left(x-y\right)\left(3+5x\right)\)
6) 9x3y2 + 3x2y2 = 3x2y2( 3x + 1 )
7) x3 + 2x2 + 3x = x( x2 + 2x + 3 )
8) 6x2y + 4xy2 + 2xy = 2xy( 3x + 2y + 1 )
9) 5x2( x - 2y ) - 15x( x - 2y ) = 5x( x - 2y )( x - 3 )
10 3( x - y ) - 5x( y - x ) = 3( x - y ) + 5x( x - y ) = ( x - y )( 3 + 5x )
16) 2x + 2y - x2 - xy = ( 2x + 2y ) - ( x2 + xy ) = 2( x + y ) - x( x + y ) = ( x + y )( 2 - x )
17) x2 - 2x - 4y2 - 4y = ( x2 - 4y2 ) - ( 2x + 4y ) = ( x - 2y )( x + 2y ) - 2( x + 2y ) = ( x + 2y )( x - 2y - 2 )
18) x2y - x3 - 9y + 9x = ( x2y - x3 ) - ( 9y - 9x ) = x2( y - x ) - 9( y - x ) = ( y - x )( x2 - 9 ) = ( y - x )( x - 3 )( x + 3 )
19) x2( x - 1 ) + 16( 1 - x ) = x2( x - 1 ) - 16( x - 1 ) = ( x - 1 )( x2 - 16 ) = ( x - 1 )( x - 4 )( x + 4 )
20) 2x2 + 3x - 2xy - 3y = ( 2x2 - 2xy ) + ( 3x - 3y ) = 2x( x - y ) + 3( x - y ) = ( x - y )( 2x + 3 )
20, \(2x^2+3x-2xy-3y=2x\left(x-y\right)+3\left(x-y\right)=\left(2x+3\right)\left(x-y\right)\)
16, \(2x+2y-x^2-xy=2\left(x+y\right)-x\left(x+y\right)=\left(2-x\right)\left(x+y\right)\)
17, \(x^2-2x-4y^2-4y=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)=\left(x-2y-2\right)\left(x+2y\right)\)
18, \(x^2y-x^3-9y+9x=-x\left(x^2-9\right)+y\left(x^2-9\right)=\left(-x-y\right)\left(x^2-9\right)=\left(y-x\right)\left(x-3\right)\left(x+3\right)\)
19, \(x^2\left(x-1\right)+16\left(1-x\right)=x^2\left(x-1\right)-16\left(x-1\right)=\left(x^2-16\right)\left(x-1\right)=\left(x-4\right)\left(x+4\right)\left(x-1\right)\)
1
a, 2x2+4x+2-2y2 = 2(x2+2x+1-y2)= 2[(x+1)2-y2 ] = 2(x-y+1)(x+y+1)
b, 2x - 2y - x2 + 2xy - y2= 2(x -y) - (x2 - 2xy + y2) = 2(x-y)-(x-y)2=(x-y)(2-x+y)
c, x2-y2-2y-1=x2-(y2+2y+1)=x2-(y+1)2=(x-y-1)(x+y+1)
d, x2-4x-2xy-4y+y2= x2-2xy+y2-4x-4y=(x-y)
2.
a, x2-3x+2=x2-x-2x+2=x(x-1)-2(x-1)=(x-2)(x-1)
b, x2+5x+6=x2+2x+3x+6=x(x+2)+3(x+2)=(x+3)(x+2)
c, x2+6x-6=
a) Ta có: \(4x\left(2y-z\right)+7y\left(z-2y\right)\)
\(=4x\left(2y-z\right)-7y\left(2y-z\right)\)
\(=\left(4x-7y\right)\left(2y-z\right)\)
b) Ta có: \(2x\left(x+3\right)+\left(3+x\right)\)
\(=\left(2x+1\right)\left(x+3\right)\)
\(7xy^5\left(x-1\right)-3x^2y^4\left(1-x\right)+5xy^3\left(x-1\right)\)
\(=7xy^5\left(x-1\right)+3x^2y^4\left(x-1\right)+6xy^3\left(x-1\right)\)
\(=\left(x-1\right)\left(7xy^5+3x^2y^4-6xy^3\right)=xy\left(x-1\right)\left(7y^4+3xy^3-6y^2\right)\)
11) \(3x\left(x-1\right)+5\left(1-x\right)=\left(3x-5\right)\left(x-1\right)\)
12) \(2\left(2x-1\right)+3\left(1-2x\right)=1-2x\)
13) \(10x\left(x-y\right)-8y\left(y-x\right)=2\left(x-y\right)\left(5x+4y\right)\)
14) \(3x\left(y+2\right)-3\left(y+2\right)=3\left(x-1\right)\left(y+2\right)\)
15) \(x^2-y^2-2x+2y\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)=\left(x-y\right)\left(x+y-2\right)\)
không cần phương pháp đó đâu, mik có cách này hay hơn nè
tìm nghiệm của đthức trên
nếu nghiệm là số dương thì khi phân tích xong sẽ có 1 tsố là (x-1)
nếu nghiệm là số âm thì...........................................1..........(x+1)
VD: phân tích thành nhân tử: 2x^2+5x-3
Nghiệm của đa thức trên là 3
=> 2x^2+6x-x-3
=> 2x(x+3)-1(x+3)
=> (2x-1)(x+3)
ĐÓ, KICK MIK NHA
8: \(=\left(x-2y\right)\cdot x\cdot\left(x+3\right)\)
9: \(=\left(5x+2\right)\left(x-3\right)-x\left(x-3\right)\)
\(=\left(x-3\right)\left(4x+2\right)\)
=2(2x+1)(x-3)
3: \(=2\left(x+2\right)\left(25x-15-x\right)\)
\(=2\left(x+2\right)\left(24x-15\right)\)
=6(x+2)(8x-5)