Phân tích các đa thức sau thành nhân tử:
\(g,4x^3-13x^2+9x-18\)
\(h,x^2-4x+3+4y-4y^2\)
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\(g.4x^3-13x^2+9x-18=4x^3-12x^2-x^2+3x+6x-18=4x^2\left(x-3\right)-x\left(x-3\right)+6\left(x-3\right)=\left(x-3\right)\left(4x^2-x+6\right)\)
\(h.x^2-4x+3+4y-4y^2=x^2-4x+4-4y^2+4y-1=\left(x-2\right)^2-\left(2y-1\right)^2=\left(x-2+2y-1\right)\left(x-2-2y+1\right)=\left(x+2y-3\right)\left(x-2y-1\right)\)
f) = x2( x - 4 ) - 9( x - 4 ) = ( x - 4 )( x - 3 )( x + 3 )
g) = 4( x - y ) + ( x - y )2 = ( x - y )( x - y + 4 )
h) = x3( x + 1 ) + ( x - 1 )( x + 1 ) = ( x + 1 )( x3 + x - 1 )
i) = ( x - y )( x + y ) - 4( x + y ) = ( x + y )( x - y - 4 )
j) = ( x - y )( x2 + xy + y2 ) - 3( x - y ) = ( x - y )( x2 + xy + y2 - 3 )
Trả lời:
f, x3 - 4x2 - 9x + 36 = ( x3 - 4x2 ) - ( 9x - 36 ) = x2 ( x - 4 ) - 9 ( x - 4 ) = ( x - 4 )( x2 - 9 ) = ( x - 4 )( x - 3 )( x + 3 )
g, 4x - 4y + x2 - 2xy + y2 = ( 4x - 4y ) + ( x2 - 2xy + y2 ) = 4 ( x - y ) + ( x - y )2 = ( x - y ) ( 4 + x - y )
h, x4 + x3 + x2 - 1 = ( x4 + x3 ) + ( x2 - 1 ) = x3 ( x + 1 ) + ( x - 1 )( x + 1 ) = ( x + 1 )( x3 + x - 1 )
i, x2 - y2 - 4x - 4y = ( x2 - y2 ) - ( 4x + 4y ) = ( x - y )( x + y ) - 4 ( x + y ) = ( x + y )( x - y - 4 )
j, x3 - y3 - 3x + 3y = ( x3 - y3 ) - ( 3x - 3y ) = ( x - y )( x2 + xy + y2 ) - 3 ( x - y ) = ( x - y )( x2 + xy + y2 - 3 )
1. \(x^3+2x^2-6x-27=\left(x-3\right)\left(x^2+5x+9\right)\)
2. \(9x^2+6x-4y^2-4y=\left(9x^2-4y^2\right)+\left(6x-4y\right)\)
\(=\left(3x-2y\right)\left(3x+2y\right)+2\left(3x-2y\right)=\left(3x-2y\right)\left(3x+2y+2\right)\)
3. \(12x^3+4x^2-27x-9=4x^2\left(3x+1\right)-9\left(3x+1\right)\)
\(=\left(3x+1\right)\left(x^2-\dfrac{9}{4}\right)=\left(x+\dfrac{1}{3}\right)\left(x+\dfrac{3}{2}\right)\left(x-\dfrac{3}{2}\right)\)
1) Ta có: \(x^3+2x^2-6x-27\)
\(=\left(x-3\right)\left(x^2+3x+9\right)+2x\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2+5x+9\right)\)
2: Ta có: \(9x^2+6x-4y^2-4y\)
\(=\left(3x-2y\right)\left(3x+2y\right)+2\left(3x-2y\right)\)
\(=\left(3x-2y\right)\left(3x+2y+2\right)\)
a: 2x^2y-50xy=2xy(x-25)
b: 5x^2-10x=5x(x-2)
c: 5x^3-5x=5x(x^2-1)=5x(x-1)(x+1)
d: \(x^2-xy+x=x\left(x-y+1\right)\)
e: x(x-y)-2(y-x)
=x(x-y)+2(x-y)
=(x-y)(x+2)
f: 4x^2-4xy-8y^2
=4(x^2-xy-2y^2)
=4(x^2-2xy+xy-2y^2)
=4[x(x-2y)+y(x-2y)]
=4(x-2y)(x+y)
f1: x^2ỹ-y^2+y
=(x-y)(x+y)+(x+y)
=(x+y)(x-y+1)
a) $x^3-3x^2y+4x-12y$
$=(x^3-3x^2y)+(4x-12y)$
$=x^2(x-3y)+4(x-3y)$
$=(x-3y)(x^2+4)$
b) $4x^2-y^2+4y-4$
$=4x^2-(y^2-4y+4)$
$=(2x)^2-(y^2-2\cdot y\cdot2+2^2)$
$=(2x)^2-(y-2)^2$
$=[2x-(y-2)][2x+(y-2)]$
$=(2x-y+2)(2x+y-2)$
c) $9x^2-6x-y^2+2y$
$=(9x^2-y^2)-(6x-2y)$
$=[(3x)^2-y^2]-2(3x-y)$
$=(3x-y)(3x+y)-2(3x-y)$
$=(3x-y)(3x+y-2)$
$\text{#}Toru$
a) \(x-xy+y-y^2=x\left(1-y\right)+y\left(1-y\right)=\left(x+y\right)\left(1-y\right)\)
b) \(x^2-2x-y^2+1=\left(x^2-2x+1\right)-y^2=\left(x-1\right)^2-y^2=\left(x-y-1\right)\left(x+y-1\right)\)
c) \(4x^2-4xy+y^2=\left(2x\right)^2-2.2x.y+y^2=\left(2x-y\right)^2\)
d) \(9x^3-9x^2y-4x+4y=9x^2\left(x-y\right)-4\left(x-y\right)=\left(9x^2-4\right)\left(x-y\right)=\left(3x-2\right)\left(3x+2\right)\left(x-y\right)\)
e) \(x^3+2+3\left(x^3-2\right)=x^3+2+3x^3-6=4x^3-4=4\left(x^3-1\right)=4\left(x-1\right)\left(x^2+x+1\right)\)
a) Xem lại đề
b) x³ - 4x²y + 4xy² - 9x
= x(x² - 4xy + 4y² - 9)
= x[(x² - 4xy + 4y² - 3²]
= x[(x - 2y)² - 3²]
= x(x - 2y - 3)(x - 2y + 3)
c) x³ - y³ + x - y
= (x³ - y³) + (x - y)
= (x - y)(x² + xy + y²) + (x - y)
= (x - y)(x² + xy + y² + 1)
d) 4x² - 4xy + 2x - y + y²
= (4x² - 4xy + y²) + (2x - y)
= (2x - y)² + (2x - y)
= (2x - y)(2x - y + 1)
e) 9x² - 3x + 2y - 4y²
= (9x² - 4y²) - (3x - 2y)
= (3x - 2y)(3x + 2y) - (3x - 2y)
= (3x - 2y)(3x + 2y - 1)
f) 3x² - 6xy + 3y² - 5x + 5y
= (3x² - 6xy + 3y²) - (5x - 5y)
= 3(x² - 2xy + y²) - 5(x - y)
= 3(x - y)² - 5(x - y)
= (x - y)[(3(x - y) - 5]
= (x - y)(3x - 3y - 5)
\(4x^3-13x^2+9x-18 \)
\(=4x^2\left(x-3\right)-x\left(x-3\right)+6\left(x-3\right)\)
\(=\left(x-3\right)\left(4x^2-x+6\right)\)