phân tích đa thức sau thành nhân tử x^4-3x^3-x+3
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
6 tháng 11 2021
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
NT
2
26 tháng 6 2021
`a)3x-3a+yx-ya`
`=3(x-a)+y(x-a)`
`=(x-a)(y+3)`
`b)x^2-9-4(x+3)`
`=(x-3)(x+3)-4(x+3)`
`=(x+3)(x-3-4)`
`=(x+3)(x-7)`
NM
3
30 tháng 6 2018
\(=x^5-2x^4+x^3-x^4+2x^3-x^2\)
\(=x^3\left(x^2-2x+1\right)-x^2\left(x^2-2x+1\right)\)
\(=\left(x^2-2x+1\right)\left(x^3-x^2\right)\)
\(=\left(x-1\right)^2x^2\left(x-1\right)=\left(x-1\right)^3x^2\)
30 tháng 6 2018
\(=x^2\left(x^3-1\right)-3x^3\left(x-1\right)\)
\(=x^2\left(x-1\right)\left(x^2+x+1-3x\right)\)
\(=x^2\left(x-1\right)\left(x^2-2x+1\right)\)
\(=x^2\left(x-1\right)\left(x-1\right)^2\)
\(=x^2\left(x-1\right)^3\)
8 tháng 2 2021
\(7xy^5\left(x-1\right)-3x^2y^4\left(1-x\right)+5xy^3\left(x-1\right)\)
\(=7xy^5\left(x-1\right)+3x^2y^4\left(x-1\right)+6xy^3\left(x-1\right)\)
\(=\left(x-1\right)\left(7xy^5+3x^2y^4-6xy^3\right)=xy\left(x-1\right)\left(7y^4+3xy^3-6y^2\right)\)
MK
1
NH
0
H9
HT.Phong (9A5)
CTVHS
10 tháng 6 2023
\(x^4+3x^3+12x-16\)
\(=x^4+4x^3+4x^2+16x-x^3-4x^2-4x-16\)
\(=x\left(x^3+4x^2+4x+16\right)-\left(x^3+4x^2+4x+16\right)\)
\(=\left(x-1\right)\left(x^3+4x^2+4x+16\right)\)
\(=\left(x-1\right)\left[x^2\left(x+4\right)+4\left(x+4\right)\right]\)
\(=\left(x-1\right)\left(x+4\right)\left(x^2+4\right)\)
\(x^4-3x^3-x+3\)
\(=x^4-x^3-2x^3+2x-3x+3\)
\(=\)\(x^3\left(x-1\right)-2x\left(x^2-1\right)-3\left(x-1\right)\)
\(=x^3\left(x-1\right)-2x\left(x-1\right)\left(x+1\right)-3\left(x-1\right)\)
\(=\left[x^3-2x\left(x+1\right)-3\right]\left(x-1\right)\)
\(=\left[x^3-2x^2-2x-3\right]\left(x-1\right)\)
\(=\)\(\left[x^3-3x^2+x^2-3x+x-3\right]\left(x-1\right)\)
\(=\left[x^2\left(x-3\right)+x\left(x-3\right)+\left(x-3\right)\right]\left(x-1\right)\)
\(=\left[\left(x-3\right)\left(x^2+x+1\right)\right]\left(x-1\right)\)
\(x^4-3x^3-x+3\)
= \(x\left(x-3\right)-\left(x-3\right)\)
= \(\left(x-1\right)\left(x-3\right)\)