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a) x2 + x - 12 = x2 - 3x + 4x - 12 = x( x - 3 ) + 4( x - 3 ) = ( x - 3 )( x + 4 )
b) x2 - 4x - 5 = x2 + x - 5x - 5 = x( x + 1 ) - 5( x + 1 ) = ( x + 1 )( x - 5 )
c) x2 - 2x - 3 = x2 + x - 3x - 3 = x( x + 1 ) - 3( x + 1 ) = ( x + 1 )( x - 3 )
d) x2 - 2x - 8 = x2 + 2x - 4x - 8 = x( x + 2 ) - 4( x + 2 ) = ( x + 2 )( x - 4 )
e) x2 - 5x - 6 = x2 + x - 6x - 6 = x( x + 1 ) - 6( x + 1 ) = ( x + 1 )( x - 6 )
f) x2 - 6x + 8 = x2 - 2x - 4x + 8 = x( x - 2 ) - 4( x - 2 ) = ( x - 2 )( x - 4 )
g) x2 + 4x + 3 = x2 + x + 3x + 3 = x( x + 1 ) + 3( x + 1 ) = ( x + 1 )( x + 3 )
h) x2 - 2x - 15 = x2 + 3x - 5x - 15 = x( x + 3 ) - 5( x + 3 ) = ( x + 3 )( x - 5 )
i) x2 + 7x + 12 = x2 + 3x + 4x + 12 = x( x + 3 ) + 4( x + 3 ) = ( x + 3 )( x + 4 )
j) x2 - 5x - 14 = x2 + 2x - 7x - 14 = x( x + 2 ) - 7( x + 2 ) = ( x + 2 )( x - 7 )
a)x2+5x-6
=x2+6x-x-6
=x(x+6)-(x+6)
=(x-1)(x+6)
b)7x-6x2-2
=-6x2+3x+4x-2
=-3x(2x-1)+2(2x-1)
=(2x-1)(-3x+2)
c)x2+4x+3
=x2+3x+x+3
=x(x+3)+(x+3)
=(x+1)(x+3)
d)2x2+3x-5
=2x2+5x-2x-5
=x(2x+5)-(2x+5)
=(x-1)(2x+5)
\(x^4+3x^2+36\)
\(=\left(x^2\right)^2+2.x^2.6+6^2-9x^2\)
\(=\left(x^2+6\right)^2-\left(3x\right)^2=\left(x^2-3x+6\right)\left(x^2+3x+6\right)\)
\(2x^4-3x^3-7x^2+6x+8\)
\(=2x^4+2x^3-5x^3-5x^2-2x^2-2x+8x+8\)
\(=2x^3\left(x+1\right)-5x^2\left(x+1\right)-2x\left(x+1\right)+8\left(x+1\right)\)
\(=\left(x+1\right)\left(2x^3-5x^2-2x+8\right)\)
\(=\left(x+1\right)\left[2x^2\left(x-2\right)-x\left(x-2\right)-4\left(x-2\right)\right]\)
\(=\left(x+1\right)\left(x-2\right)\left(2x^2-x-4\right)\)
Chúc bạn học tốt.
Đổi dấu – (4yx2 + yz2)(z – y2) = (4yx2 + yz2)( y2 – z), ta có thừa số
(y2 – z) chung:
C = (y2 – z)(2x2y – yz) – (4yx2 + yz2)(z – y2) + 6x2z(y2 – z)
= (y2 – z)(2x2y – yz) + (4yx2 + yz2)( y2 – z) + 6x2z(y2 – z)
= (y2 – z)[( 2x2y – yz ) + (4yx2 + yz2) + 6x2z]
= (y2 – z)[ 2x2y + 4yx2 + 6x2z]
= (y2 – z)[ 2xy2 + 4yx2 + 6x2z]
= (y2 – z)[ 2x2(y + 2y + 3z)]
= (y2 – z)[ 2x2(3y + 3z)]
= (y2 – z) 2x2 .3(y + z)
= 6x2(y2 – z)(y + z).
a) 7x2 - 4x
= x ( 7x - 4 )
b) 5x2 - 2x + 10 xy - 4y
= x ( 5x - 2 ) + 2y ( 5x - 2 )
= ( x + 2y ) ( 5x - 2 )
Ta nhân thấy nghiệm của f(x) nếu có thì x = 
Cách 1:
x3 – x2 – 4 =(x3-2x2)+(x2-2x)+(2x-4)=x2(x-2)+x(x-2)+2(x-2)=(x-2)(x2+x+2)
Cách 2:
(x-2)[(x2+2x+4)-(x+2)]=(x-2)(x2+x+2)
x3-x2-4=x3-8-x2+4=(x3-8)-(x2-4)=(x-2)(x2+2x+4)-(x-2)(x+2)
a: \(2x^2+9x-35\)
\(=2x^2+14x-5x-35\)
\(=\left(x+7\right)\left(2x-5\right)\)
b: \(-6x^2+x+12\)
\(=-6x^2+9x-8x+12\)
\(=-3x\left(2x-3\right)-4\left(2x-3\right)\)
\(=\left(2x-3\right)\left(-3x-4\right)\)
a, \(3x^2-6x=3x\left(x-2\right)\)
b, \(5x\left(x-3\right)-2\left(x-3\right)=\left(5x-2\right)\left(x-3\right)\)
c, \(x^2-25+y^2+2xy=\left(x+y\right)^2-25=\left(x+y-5\right)\left(x+y+5\right)\)
d, \(2x^2+7x-15=2x^2+10x-3x-15=2x\left(x+5\right)-3\left(x+5\right)=\left(2x-3\right)\left(x+5\right)\)
- a] =-[6x^2 - 7x +2] = - [ 6x^2 - 3x - 4x + 2 ] = -[ 3x [ 2x-1] - 2 [2x - 1]] = - [ 2x - 1] [3x - 2 ]
1.
7x(2x-1)=14x2-7x
2
a. x2+2x=x(x+2)
b.x2-xy+3x-3y
=x(x-y)+3(x-y)
=(x+3)(x-y)
Câu 2:
1. 2x/2x-5 - 5/2x-5
=2x-5/2x-5
=1
2. (6x3-7x2-x+2) : (x-1)=6x2-x-2
Phân tích đa thức thành nhân tử:
a) x2 (2x - 5) + 6x - 15 = x2 (2x - 5) + 3(2x - 5) =( x2 + 3 )(2x - 5)
b)x2 + 7x + 12 = x2 + 3x + 4x + 12 = (x + 3 )( x + 4)