Phân tích đa thức thành nhân tử:
a) x2 - 5x +6
b) 3x2 + 9x - 30
c) x2 - 3x + 2
d) x2 - 9x + 18
e) x2 - 6x + 8
f) x2 - 5x - 14
g) x2 - 6x + 5
h) x2 - 7x + 12
i) x2 - 7x + 10
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CM
19 tháng 1 2017
a) (x - 2)(x - 3). b) 3(x - 2)(x + 5).
c) (x - 2)(3x + 1). d) (x-2y)(x - 5y).
e) (x + l)(x + 2)(x - 3). g) (x-1)(x + 3)( x 2 + 3).
h) (x + y - 3)(x - y + 1).
8 tháng 10 2021
\(a,=\left(x-2\right)^2-y^2=\left(x-y-2\right)\left(x+y-2\right)\\ b,=4x^2\left(x^2+2x+1\right)=4x^2\left(x+1\right)^2\\ c,=xy^2\left(x^2-2xy+y^2\right)=xy^2\left(x-y\right)^2\\ d,=\left(x-y\right)\left(x+y\right)-7\left(x-y\right)=\left(x-y\right)\left(x+y-7\right)\\ e,=\left(5x-2y\right)\left(5x+2y\right)\\ f,=x^2+3x+4x+12=\left(x+3\right)\left(x+4\right)\\ i,=x^2+2x-7x-14=\left(x+2\right)\left(x-7\right)\)
HT
3
26 tháng 7 2021
a, \(x^2-5x+6=x^2+x-6x+6=x\left(x-1\right)-6\left(x-1\right)=\left(x-1\right)\left(x-6\right)\)
b, \(3x^2+9x-30=3\left(x^2+3x-10\right)=3\left(x^2-2x+5x-10\right)\)
\(=3\left[x\left(x-2\right)+5\left(x-2\right)\right]=3\left(x-2\right)\left(x+5\right)\)
c, \(x^2+7x+10=x^2+2x+5x+10=x\left(x+2\right)+5\left(x+2\right)=\left(x+2\right)\left(x+5\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 12 2020
a.
\(1-4x^2=\left(1-2x\right)\left(1+2x\right)\)
b.
\(8-27x^3=\left(2\right)^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)
c.
\(27+27x+9x^2+x^3=x^3+3.x^2.3+3.3^2.x+3^3\)
\(=\left(x+3\right)^3\)
d.
\(2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)^2\)
e.
\(x^2-y^2-5x+5y=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-5\right)\)
f.
\(x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)
CM
31 tháng 1 2017
a) (x - 1)(x - 2). b) 4(x - 2)(x - 7).
c) (x + 2)(2x +1). d) (x - l)(2x - 7).
e) (2x + 3y - 3)(2x - 3y +1). g) (x - 3)( x 3 + x 2 - x +1).
h) (x + y)(x + y-l)(x + y + l).
28 tháng 10 2017
a, \(x^4+2x^2+1-x^2\)
= \(\left(x^2+1\right)^2-x^2\)
= \(\left(x^2+x+1\right)\left(x^2-x+1\right)\)
b, \(x^4+x^2+1\)
= \(x^4+2x^2+1-x^2\)
= .. ( như phần a )
c, \(y^4+64\)
= \(\left(y^2+8\right)\left(y^2-8\right)\)
d, \(4xy+3z-12y-xz\)
\(=4y\left(x-3\right)-z\left(x-3\right)\)
\(=\left(x-3\right)\left(4y-z\right)\)
e, \(x^2-4xy+4y^2-z^2+6z-9\)
\(=\left(x-2y\right)^2-\left(z-3\right)^2\)
g, \(x^2-4xy+5x+4y^2-10y\)
\(=\left(x^2-4xy+4y^2\right)+\left(5x-10y\right)\)
\(=\left(x-2y\right)^2+5\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x-2y+5\right)\)
h, \(x^2-7x+6\)
\(=x^2-6x-x+6\)
\(=x\left(x-6\right)-\left(x-6\right)\)
\(=\left(x-6\right)\left(x-1\right)\)
i, \(x^3+5x^2+6x+2\)
\(=x^3+x^2+4x^2+4x+2x+2\)
\(=x^2\left(x+1\right)+4x\left(x+1\right)+2\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+4x+2\right)\)
2 tháng 9 2021
Bài 2:
a) \(3x^2-7x-10=\left(x+1\right)\left(3x-10\right)\)
b) \(x^2+6x+9-4y^2=\left(x+3\right)^2-\left(2y\right)^2=\left(x+3-2y\right)\left(x+3+2y\right)\)
c) \(x^2-2xy+y^2-5x+5y=\left(x-y\right)^2-5\left(x-y\right)=\left(x-y\right)\left(x-y-5\right)\)
d) \(4x^2-y^2-6x+3y=\left(2x-y\right)\left(2x+y\right)-3\left(2x-y\right)=\left(2x-y\right)\left(2x+y-3\right)\)
e) \(1-2a+2bc+a^2-b^2-c^2=\left(a-1\right)^2-\left(b-c\right)^2=\left(a-1-b+c\right)\left(a-1+b-c\right)\)
f) \(x^3-3x^2-4x+12=\left(x+2\right)\left(x-3\right)\left(x-2\right)\)
g) \(x^4+64=\left(x^2+8\right)^2-16x^2=\left(x^2+8-4x\right)\left(x^2+6+4x\right)\)h) \(x^4-5x^2+4=\left(x+2\right)\left(x+1\right)\left(x-1\right)\left(x-2\right)\)
i) \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+16=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+16=\left(x^2+8x+7\right)^2+8\left(x^2+8x+7\right)+16=\left(x^2+8x+11\right)^2\)
2 tháng 9 2021
a: \(3x^2-7x-10\)
\(=3x^2+3x-10x-10\)
\(=\left(x+1\right)\left(3x-10\right)\)
b: \(x^2+6x+9-4y^2\)
\(=\left(x+3\right)^2-4y^2\)
\(=\left(x+3-2y\right)\left(x+3+2y\right)\)
c: \(x^2-2xy+y^2-5x+5y\)
\(=\left(x-y\right)^2-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-5\right)\)
16 tháng 8 2021
a: \(x^2-6x+5=\left(x-5\right)\left(x-1\right)\)
b: \(x^2-x-12=\left(x-4\right)\left(x+3\right)\)
c: \(x^2+8x+15=\left(x+5\right)\left(x+3\right)\)
d: \(2x^2-5x-12=\left(x-4\right)\left(2x+3\right)\)
e: \(x^2-13x+36=\left(x-9\right)\left(x-4\right)\)
27 tháng 10 2021
\(a,=\left(x+1\right)\left(x+3\right)\\ b,=-5x^2+15x+x-3=\left(x-3\right)\left(1-5x\right)\\ c,=2x^2+2x+5x+5=\left(2x+5\right)\left(x+1\right)\\ d,=2x^2-2x+5x-5=\left(x-1\right)\left(2x+5\right)\\ e,=x^3+x^2-4x^2-4x+x+1=\left(x+1\right)\left(x^2-4x+1\right)\\ f,=x^2+x-5x-5=\left(x+1\right)\left(x-5\right)\)
a/\(x^2-5x+6=x^2-2x-3x+6=x\left(x-2\right)-3\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)b/
\(3x^2+9x-30=3\left(x^2+3x-10\right)\)
c/
\(x^2-3x+2=x^2-x-2x+2=x\left(x-1\right)-2\left(x-1\right)=\left(x-1\right)\left(x-2\right)\)
d/\(x^2-9x+18=x^2-3x-6x+18=x\left(x-3\right)-6\left(x-3\right)=\left(x-3\right)\left(x-6\right)\)e/
\(x^2-6x+8=x^2-2x-4x+8=x\left(x-2\right)-4\left(x-2\right)=\left(x-2\right)\left(x-4\right)\)f/\(x^2-5x-14=x^2+2x-7x-14=x\left(x+2\right)-7\left(x+2\right)=\left(x+2\right)\left(x-7\right)\)
g/
\(x^2-6x+5=x^2-x-5x+5=x\left(x-1\right)-5\left(x-1\right)=\left(x-1\right)\left(x-5\right)\)
h/
\(x^2-7x+12=x^2-4x-3x+12=x\left(x-4\right)-3\left(x-4\right)=\left(x-4\right)\left(x-3\right)\)i/\(x^2-7x+10=x^2-2x-5x+10=x\left(x-2\right)-5\left(x-2\right)=\left(x-2\right)\left(x-5\right)\)
a) Ta có: \(x^2-5x+6\)
\(=x^2-2x-3x+6\)
\(=x\left(x-2\right)-3\left(x-2\right)\)
\(=\left(x-2\right)\left(x-3\right)\)
b) Ta có: \(3x^2+9x-30\)
\(=3\left(x^2+3x-10\right)\)
\(=3\left(x^2+5x-2x-10\right)\)
\(=3\left[x\left(x+5\right)-2\left(x+5\right)\right]\)
\(=3\left(x+5\right)\left(x-2\right)\)
c) Ta có: \(x^2-3x+2\)
\(=x^2-x-2x+2\)
\(=x\left(x-1\right)-2\left(x-1\right)\)
\(=\left(x-1\right)\left(x-2\right)\)
d) Ta có: \(x^2-9x+18\)
\(=x^2-3x-6x+18\)
\(=x\left(x-3\right)-6\left(x-3\right)\)
\(=\left(x-3\right)\left(x-6\right)\)
e) Ta có: \(x^2-6x+8\)
\(=x^2-4x-2x+8\)
\(=x\left(x-4\right)-2\left(x-4\right)\)
\(=\left(x-4\right)\left(x-2\right)\)
f) Ta có: \(x^2-5x-14\)
\(=x^2-7x+2x-14\)
\(=x\left(x-7\right)+2\left(x-7\right)\)
\(=\left(x-7\right)\left(x+2\right)\)
g) Ta có: \(x^2-6x+5\)
\(=x^2-x-5x+5\)
\(=x\left(x-1\right)-5\left(x-1\right)\)
\(=\left(x-1\right)\left(x-5\right)\)
h) Ta có: \(x^2-7x+12\)
\(=x^2-3x-4x+12\)
\(=x\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x-4\right)\)
i) Ta có: \(x^2-7x+10\)
\(=x^2-2x-5x+10\)
\(=x\left(x-2\right)-5\left(x-2\right)\)
\(=\left(x-2\right)\left(x-5\right)\)