Bài 2: Phân tích đa thức sau thành nhân tử.
a) 3x(x +1) – 5y(x + 1) h) 3x3(2y – 3z) – 15x(2y – 3z)2
b) 3x(x – 6) – 2(x – 6) k) 3x(z + 2) + 5(-x – 2)
c) 4y(x – 1) – (1 – x) l) 18x2(3 + x) + 3(x + 3)
d) (x – 3)3 + 3 – x m) 14x2y – 21xy2 + 28x2y2
e) 7x(x – y) – (y – x) n) 10x(x – y) – 8y(y – x)
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h) \(=3x\left(2y-3z\right)\left[x^2-5\left(2y-3z\right)\right]=3x\left(2y-3z\right)\left(x^2-10y+15z\right)\)
k) \(=\left(x+2\right)\left(3x-5\right)\)
l) \(=\left(18^2+3\right)\left(x+3\right)=327\left(x+3\right)\)
m) \(=7xy\left(2x-3y+4xy\right)\)
n) \(=2\left(x-y\right)\left(5x-4y\right)\)
a,3x3y3-15x2y2=3x2y2(xy-5)
b,2x(x-5y)+8y(5y-x)=2x(x-5y)-8y(x-5y)=(x-5y).(2x-8y)
c,(3x-1)2-16=(3x-1)2-42=(3x-1+4)(3x-1-4)=(3x+3)(3x-5)
d,x3-3x2+3x-1=x3-1-(3x2+3x)=x3-1-3x(x+1)=(x3-1-3x)(x+1)
e,125x3+1=(5x)3+13=(5x+1)(25x2-5x.1+12)
f,x3+6x2y+12xy2+8y3=x3+3.x2.2y+3.x.(2y)2+(2y)3=(x+2y)3
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 )
a, \(x-2y+x^2-4y^2=\left(x-2y\right)+\left(x-2y\right)\left(x+2y\right)=\left(x-2y\right)\left(1+x+2y\right)\)
b, \(x^2-4x^2y^2+y^2+2xy=\left(x+y\right)^2-\left(2xy\right)^2\)
\(=\left(x+y-2xy\right)\left(x+y+2xy\right)\)
c, \(x^6-x^4+2x^3+2x^2=x^6+2x^3+1-x^4+2x^2-1\)
\(=\left(x^3+1\right)^2-\left(x^2-1\right)^2=\left(x^3-x^2+2\right)\left(x^3+x^2\right)\)
\(=x^2\left(x+1\right)\left(x^3-x^2+2\right)\)
d, \(x^3+3x^2+3x+1-8y^3=\left(x+1\right)^3-\left(2y\right)^3=\left(x+1-2y\right)\left(x+1+2y\right)\)
a) Ta có: \(x-2y+x^2-4y^2\)
\(=\left(x-2y\right)+\left(x-2y\right)\left(x+2y\right)\)
\(=\left(x-2y\right)\left(1+x+2y\right)\)
b: Ta có: \(x^2-4x^2y^2+y^2+2xy\)
\(=\left(x+y\right)^2-\left(2xy\right)^2\)
\(=\left(x+y-2xy\right)\left(x+y+2xy\right)\)
\(a,14x^2y-21xy^2+28x^2y^2=7xy\left(x-3y+4xy\right)\\ b,x\left(x+y\right)-5x-5y=x\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(x-5\right)\\ c,10x\left(x-y\right)-8\left(y-x\right)=10x\left(x-y\right)+8\left(x-y\right)=\left(x-y\right)\left(10x+8\right)=2\left(x-y\right)\left(5x+4\right)\)
\(d,\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)=2x\left(4x+2\right)=4x\left(2x+1\right)\)\(e,x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)+3xyz-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
a: \(A=-4x^5y^3-2x^2y^3z^2-2y^4\)
b: \(B=-4x^5y^3-2x^2y^3z^2-2y^4+2x^2y^3z^2-\dfrac{2}{3}y^4+\dfrac{1}{5}x^4y^3=-4x^5y^3+\dfrac{1}{5}x^4y^3-\dfrac{8}{3}y^4\)
a) 3x(x + 1) - 5y(x + 1)
= (x + 1)(3x - 5y)
b) 3x(x - 6) - 2(x - 6)
= (x - 6)(3x - 2)
c) 4y(x - 1) - (1 - x)
= 4y(x - 1) + (x - 1)
= (x - 1)(4y + 1)
d) (x - 3)³ + 3 - x
= (x - 3)³ - (x - 3)
= (x - 3)[(x - 3)² - 1]
= (x - 3)(x - 3 - 1)(x - 3 + 1)
= (x - 3)(x - 4)(x - 2)
e) 7x(x - y) - (y - x)
= 7x(x - y) + (x - y)
= (x - y)(7x + 1)
h) 3x³(2y - 3z) - 15x(2y - 3z)²
= (2y - 3z)[3x³ - 15x(2y - 3x)]
= 3x(2y - 3x)[x² - 5(2y - 3x)]
= 3x(2y - 3x)(x² - 10y + 3x)
= 3x(2y - 3x)(x² + 3x - 10y)
k) 3x(x + 2) + 5(-x - 2)
= 3x(x + 2) - 5(x + 2)
= (x + 2)(3x - 5)
l) 18x²(3 + x) + 3(x + 3)
= (x + 3)(18x² + 3)
= 3(x + 3)(6x² + 1)
m) 7x(x - y) - (y - x)
= 7x(x - y) + (x - y)
= (x - y)(7x + 1)
n) 10x(x - y) - 8y(y - x)
= 10x(x - y) + 8y(x - y)
= (x - y)(10x + 8y)
= 2(x - y)(5x + 4y)