Phân tích các đa thức sau thành nhân tử:
\(A=4x^2+6x\). \(B=\left(2x+3\right)^2-x\left(2x+3\right)\). \(C=\left(9x^2-1\right)-\left(3x-1\right)^2\).
\(D=x^3-16x\). \(E=4x^2-25y^2\). \(G=\left(2x+3\right)^2-\left(2x-3\right)^2\).
\(A=4x^2+6x=2x\left(2x+3\right)\)
\(B=\left(2x+3\right)^2-x\left(2x+3\right)=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\)
\(C=\left(9x^2-1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1-3x+1\right)=2\left(3x+1\right)\)
\(D=x^3-16x=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\)
\(E=4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)
\(G=\left(2x+3\right)^2-\left(2x-3\right)^2=\left(2x+3-2x+3\right)\left(2x+3+3x-3\right)=6.4x=24x\)
\(A=2x\left(2x+3\right)\\ B=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\\ C=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2\\ =\left(3x-1\right)\left(3x+1-3x+1\right)\\ =2\left(3x-1\right)\\ D=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\\ E=\left(2x-5y\right)\left(2x+5y\right)\\ G=\left(2x+3-2x+3\right)\left(2x+3+2x-3\right)\\ =24x\)
phân tích đa thức thành nhân tử:
a. \(ax^2-a^2x-x+a\)
b. \(18x^3-12x^2+2x\)
c. \(x^3-5x^2-4x+20\)
d. \(\left(x+7\right)\left(x+15\right)+15\)
\(a.\) \(ax^2-a^2x-x+a\)
\(=\left(ax^2-a^2x\right)-\left(x-a\right)\)
\(=ax\left(x-a\right)-\left(x-a\right)\)
\(=\left(ax-1\right)\left(x-a\right)\)
\(b.\) \(18x^3-12x^2+2x\)
\(=2x\left(9x^2-6x+1\right)\)
\(=2x\left(3x-1\right)^2\)
\(c.\) \(x^3-5x^2-4x+20\)
\(=\left(x^3-5x^2\right)-\left(4x-20\right)\)
\(=x^2\left(x-5\right)-4\left(x-5\right)\)
\(=\left(x^2-4\right)\left(x-5\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(x-5\right)\)
\(d.\) \(\left(x+7\right)\left(x+15\right)+15\)
\(=x^2+15x+7x+105+15\)
\(=x^2+22x+120\)
\(=\left(x+10\right)\left(x+12\right)\)
\(\left(x^2+2x\right)^2-2x^2-4x-3\)
Phân tích đa thức thành nhân tử
\(\left(x^2+2x\right)^2-2x^2-4x-3\)
\(=\left(x^2+2x\right)^2-2.\left(x^2+2x\right)-3\)
\(=\left(x^2+2x\right)^2-2\left(x^2+2x\right)+1-4\)
\(=\left(x^2+2x+1\right)^2-4\)
\(=\left(x^2+2x-2\right)\left(x^2+2x+2\right)\)
Phân tích đa thức thành nhân tử
\(a,2x^2-xy+4x-2y\)
\(b,\left(a^2-a+2012\right)\left(a^2-a+2014\right)-3\)
a) 2x² - xy + 4x - 2y
<=> (2x² + 4x)-(xy + 2y)
<=> 2x(x + 2) - y(x + 2)
<=> (x + 2)(2x - y)
b) (a²−a+2012)(a²−a+2014)−3
Đặt a²−a+2012 là x , ta có :
x(x + 2) - 3
<=> x² + 2x - 3
<=> x² + 3x - x - 3
<=> x(x + 3) - (x + 3)
<=> (x +3)(x - 1)
Thay x = a²−a+2012 , ta được :
(a²−a+2015)(a²−a+2011)
Phân tích đa thức thành nhân tử :
\(f\left(x\right)-x^5-2x^4-3x^3-4x^2+2\)
Phân tích đa thức thành nhân tử:
\(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\)
đặt y=x2+4x+8 ta được
y2+3xy+2x2=y2+xy+2xy+2x2=y(y+x)+2x(y+x)
=(y+x)(y+2x)
thay y=x2+4x+8 ta được
(x2+5x+8)(x2+7x+8)
=(x^2+4x+8)2+2x(x^2+4x+8)+(x^2+4x+8)+2x^2
=(x^2+5x+8)(x^2+6x+8)
(x^2+4x+8)^2+3x(x^2+4x+8)+2x^2
dat x^2+4x+8=y
ta co:y^2+3xy+2x^2
=y^2+xy+2xy+2x^2
=y(y+x)+2x(y+x)
=(y+2x)(y+x)
=(x^2+4x+8+2x)(x^2+4x+8+x)
=(x^2+6x+8)(x^2+5x+8)
KL:......................
phân tích đa thức thành nhân tử
a. \(\left(2x-y\right)\left(x-y\right)-\left(3y-4x\right)^2+\left(y-2x\right)\left(2y-3x\right)\)
\(\left(2x-y\right)\left(x-y\right)-\left(3y-4x\right)^2+\left(y-2x\right)\left(2y-3x\right)\)
=(2x-y)(x-y)-(2x-y)(2y-3x)-(4x-3y)2
=(2x-3y)(x-y-2y+3x)-(4x-3y)2
=(2x-3y)(4x-3y)-(4x-3y)2
=(4x-3y)(2x-3y-4x+3y)
=(4x-3y))(-2x)
Phân tích đa thức thành nhân tử :\(\left(x^2-2x+3\right)\left(x^2-2x+5\right)-8\)
Đặt \(x^2-2x+4=a\)
Khi đó \(\left(x^2-2x+3\right)\left(x^2-2x+5\right)-8=\left(a-1\right)\left(a+1\right)-8\)
\(=a^2-1-8\)
\(=a^2-9\)
\(=\left(a-3\right)\left(a+3\right)\)
\(=\left(x^2-2x+4-3\right)\left(x^2-2x+4+3\right)\)
\(=\left(x^2-2x+1\right)\left(x^2-2x+7\right)\)
\(=\left(x-1\right)^2\left(x^2-2x+7\right)\)
1 phân tích đa thức thành nhân tử
a,\(3x^2-6xy+3y^2\)
b,\(\left(x-y\right)^2-4x^2\)
2.tìm x biết
a,2x(x-3)-x+3=0
b,\(x^2+5x+6=0\)
`1)`
`a)3x^2-6xy+3y^2=3(x^2-2xy+y^2)=3(x-y)^2`
`b)(x-y)^2-4x^2=(x-y-2x)(x-y+2x)=(-x-y)(3x-y)`
`2)`
`a)2x(x-3)-x+3=0`
`<=>2x(x-3)-(x-3)=0`
`<=>(x-3)(2x-1)=0`
`<=>[(x=3),(x=1/2):}`
`b)x^2+5x+6=0`
`<=>x^2+2x+3x+6=0`
`<=>(x+2)(x+3)=0`
`<=>[(x=-2),(x=-3):}`
Phân tích đa thức thành nhân tử:
\(x^3-8+2x\left(x-2\right)\)
\(x^3-8+2x\left(x-2\right)\\ =\left(x-2\right)\left(x^2+2x+4\right)+2x\left(x-2\right)\\ =\left(x-2\right)\left(x^2+2x+4+2x\right)=\left(x-2\right)\left(x^2+4x+4\right)\\ =\left(x-2\right)\left(x+2\right)^2\)
=\(\left(x-2\right)\left(x^2+2x+4\right)+2x\left(x-2\right)\)
=\(\left(x-2\right)\left(x^2+4x+4\right)\)
=\(\left(x-2\right)\left(x+2\right)^2\)