Phân tích đa thức thành nhân tử
a)x3-9x2+6x+16
b)x3-6x2-x+30
c)3x2+22xy+11x+37y+7y2+10
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1) \(x^2-7x+6=x^3+1-7x-7=\left(x^3+1\right)-7\left(x+1\right)=\left(x+1\right)\left(x^2-x-6\right)\)
2) \(x^3-9x^2+6x+16\)
\(\left(x^3+1\right)-\left[\left(9x^2-6x+1\right)-16\right]\)
\(=\left(x^3+1\right)-\left[\left(3x-1\right)^2-16\right]=\left(x^3+1\right)-\left(3x-1+4\right)\left(3x-1-4\right)\)\(=\left(x^3+1\right)-3\left(3x-5\right)\left(x+1\right)\)\(=\left(x+1\right)\left[x^2-x+1-9x+15\right]=\left(x+1\right)\left(x^2-10x+16\right)\)
\(=\left(x+1\right)\left[x\left(x-2\right)-8\left(x-2\right)\right]\)\(\left(x+1\right)\left(x-2\right)\left(x-8\right)\)
3) \(x^3-6x^2-x+30\)
\(=x^3-5x^2-x^2+5x-6x+30\)
\(=x^2\left(x-5\right)-x\left(x-5\right)-6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2-x-1\right)\)
4) \(2x^3-x^2+5x+3=\left(2x^3+x^2\right)-\left(2x^2+x\right)+\left(6x+3\right)\)
\(=x^2\left(2x+1\right)-x\left(2x+1\right)+3\left(2x+1\right)\)
\(=\left(2x+1\right)\left(x^2-x+3\right)\)
5) \(27x^3-27x^2+18x-4=\left(27x^3-1\right)-\left(27x^2-18x+3\right)\)
\(=\left(3x-1\right)\left(9x^2+3x+1\right)-3\left(9x^2-6x+1\right)\)
\(=\left(3x-1\right)\left(9x^2+3x+1\right)-3\left(3x-1\right)^2\)
\(=\left(3x-1\right)\left(9x^2+3x+1-9x+3\right)=\left(3x-1\right)\left(9x^2-6x+4\right)\)
gửi phần này trước còn lại làm sau !!! tk mk nka !!!
a,\(x^3-7x+6\)
\(=x^3-2x^2+2x^2-4x-3x+6\)
\(=\left(x^3-2x^2\right)+\left(2x^2-4x\right)-\left(3x-6\right)\)
\(=x^2.\left(x-2\right)+2x.\left(x-2\right)-3.\left(x-2\right)\)
\(=\left(x-2\right).\left(x^2+2x-3\right)\)
\(=\left(x-2\right).\left(x^2-x+3x-3\right)\)
\(=\left(x-2\right).\left[\left(x^2-x\right)+\left(3x-3\right)\right]\)
\(=\left(x-2\right).\left[x.\left(x-1\right)+3.\left(x-1\right)\right]\)
\(=\left(x-2\right).\left(x-1\right).\left(x+3\right)\)
b,\(x^3-9x^2+6x+16\)
\(=x^3-8x^2-x^2+8x-2x+16\)
\(=\left(x^3-8x^2\right)-\left(x^2-8x\right)-\left(2x-16\right)\)
\(=x^2.\left(x-8\right)-x.\left(x-8\right)-2.\left(x-8\right)\)
\(=\left(x-8\right).\left(x^2-x-2\right)\)
\(=\left(x-8\right).\left(x^2+x-2x-2\right)\)
\(=\left(x-8\right).\left[\left(x^2+x\right)-\left(2x+2\right)\right]\)
\(=\left(x-8\right).\left[x.\left(x+1\right)-2.\left(x+1\right)\right]\)
\(=\left(x-8\right).\left(x+1\right).\left(x-2\right)\)
c,\(x^3-6x^2-x+30\)
\(=x^3-5x^2-x^2+5x-6x+30\)
\(=\left(x^3-5x^2\right)-\left(x^2-5x\right)-\left(6x-30\right)\)
\(=x^2.\left(x-5\right)-x.\left(x-5\right)-6.\left(x-5\right)\)
\(=\left(x-5\right).\left(x^2-x-6\right)\)
\(=\left(x-5\right).\left(x^2+2x-3x-6\right)\)
\(=\left(x-5\right).\left[\left(x^2+2x\right)-\left(3x+6\right)\right]\)
\(=\left(x-5\right).\left[x.\left(x+2\right)-3.\left(x+2\right)\right]\)
\(=\left(x-5\right).\left(x+2\right).\left(x-3\right)\)
Chúc bạn học tốt!!!
d,\(2x^3-x^2+5x+3\)
\(=2x^3+x^2-2x^2-x+6x+3\)
\(=\left(2x^3+x^2\right)-\left(2x^2+x\right)+\left(6x+3\right)\)
\(=x^2.\left(2x+1\right)-x.\left(2x+1\right)+3.\left(2x+1\right)\)
\(=\left(2x+1\right).\left(x^2-x+3\right)\)
e, \(27x^3-27x^2+18x-4\)
\(=27x^3-9x^2-18x^2+6x+12x-4\)
\(=\left(27x^2-9x^2\right)-\left(18x^2-6x\right)+\left(12x-4\right)\)
\(=9x^2.\left(3x-1\right)-6x.\left(3x-1\right)+4.\left(3x-1\right)\)
\(=\left(3x-1\right).\left(9x^2-6x+4\right)\)
Chúc bạn học tốt!!!
11) Ta có: \(a^6+a^4+a^2b^2+b^4-b^6\)
\(=a^6-b^6+a^4+a^2b^2+b^4\)
\(=\left(a^2-b^2\right)\left(a^4+a^2b^2+b^4\right)+\left(a^4+a^2b^2+b^4\right)\)
\(=\left(a^4+a^2b^2+b^4\right)\left(a^2-b^2+1\right)\)
12) Ta có: \(x^3+3xy+y^3-1\)
\(=\left(x^3+3x^2y+3xy^2+y^3-1\right)-3x^2y-3xy^2+3xy\)
\(=\left[\left(x+y\right)^3-1\right]-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left[x^2+2xy+y^2+x+y+1\right]-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left(x^2-xy+y^2+x+y+1\right)\)
14) Ta có: \(x^8+x+1\)
\(=x^8+x^7-x^7-x^6+x^6+x^5-x^5-x^4+x^4+x^3-x^3+x^2-x^2+x+1\)
\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)
15) Ta có: \(x^8+3x^4+4\)
\(=x^8+4x^4+4-x^4\)
\(=\left(x^4+2\right)^2-\left(x^2\right)^2\)
\(=\left(x^4-x^2+2\right)\left(x^4+x^2+2\right)\)
`1)x^3-7x+6`
`=x^3-x-6x+6`
`=x(x-1)(x+1)-6(x-1)`
`=(x-1)(x^2+x-6)`
`=(x-1)(x^2-2x+3x-6)`
`=(x-1)[x(x-2)+3(x-2)]`
`=(x-1)(x-2)(x+3)`
`2)x^3-9x^2+6x+16`
`=x^3-2x^2-7x^2+14x-8x+16`
`=x^2(x-2)-7x(x-2)-8(x-2)`
`=(x-2)(x^2-7x-8)`
`=(x-2)(x^2-8x+x-8)`
`=(x-2)[x(x-8)+x-8]`
`=(x-2)(x-8)(x+1)`
`3)x^3-6x^2-x+30`
`=x^3+2x^2-8x^2-16x+15x+30`
`=x^2(x+2)-8x(x+2)+15(x+2)`
`=(x+2)(x^2-8x+15)`
`=(x+2)(x^2-3x-5x+15)`
`=(x+2)[x(x-3)-5(x-3)]`
`=(x+2)(x-3)(x-5)`
`4)2x^3-x^2+5x+3`
`=2x^3+x^2-2x^2-x+6x+3`
`=x^2(2x+1)-x(2x+1)+3(2x+1)`
`=(2x+1)(x^2-x+3)`
`5)27x^3-27x^2+18x-4`
`=27x^3-9x^2-18x^2+6x+12x-4`
`=9x^2(3x-1)-6x(3x-1)+4(3x-1)`
`=(3x-1)(9x^2-6x+4)`
1) Ta có: \(x^3-7x+6\)
\(=x^3-x-6x+6\)
\(=x\left(x^2-1\right)-6\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x-6\right)\)
\(=\left(x-1\right)\left(x+3\right)\left(x-2\right)\)
2) Ta có: \(x^3-9x^2+6x+16\)
\(=x^3-2x^2-7x^2+14x-8x+16\)
\(=x^2\left(x-2\right)-7x\left(x-2\right)-8\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2-7x-8\right)\)
\(=\left(x-2\right)\left(x-8\right)\left(x+1\right)\)
3) Ta có: \(x^3-6x^2-x+30\)
\(=x^3+2x^2-8x^2-16x+15x+30\)
\(=x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-8x+15\right)\)
\(=\left(x+2\right)\left(x-3\right)\left(x-5\right)\)
Bài 2:
a: =x(x^2-25)
=x(x-5)(x+5)
b: =x(x-2y)+3(x-2y)
=(x-2y)(x+3)
c: =(2x-3)(4x^2+6x+9)+2x(2x-3)
=(2x-3)(4x^2+8x+9)
Chia nhỏ ra cậu ơi :v
Cậu đặt câu hỏi free nên đặt nhỏ ra thì mới có người làm nha để như này dày cộp không ai dám làm đou =(((
\(a,=x\left(y^2-25\right)=x\left(y-5\right)\left(y+5\right)\\ b,=x\left(x-y\right)+2\left(x-y\right)=\left(x+2\right)\left(x-y\right)\\ c,=x^2\left(x-3\right)-4\left(x-3\right)\\ =\left(x-2\right)\left(x+2\right)\left(x-3\right)\)
trên gg có kết quả kìa