K
Khách
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NN
2
NC
1 tháng 6 2021
a.\(x^2+7x+6\)
\(=x^2+x+6x+6\)
\(=x\left(x+1\right)+6\left(x+1\right)\)
\(=\left(x+1\right)\left(x+6\right)\)
Sửa đề:.\(x^4+2008x^2+2007x+2008\)
\(=x^4+x^2+1+2007x^2+2007x+2007\)
\(=\left(x^4+x^2+1\right)+2007\left(x^2+x+1\right)\)
\(=\left(x^4+x^3+x^2-x^3-x^2-x+x^2+x+1\right)+2007\left(x^2+x+1\right)\)
\(=\left[x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\right]+2007\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1\right)+2007\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)
DD
6
NV
3
T
24 tháng 3 2019
a)\(x^8+2x^4+1-x^4=\left(x^4+1\right)^2-\left(x^2\right)^2\)
\(=\left(x^4+x^2+1\right)\left(x^4-x^2+1\right)\)
\(=\left(x^2-x+1\right)\left(x^2+x+1\right)\left(x^4-x^2+1\right)\)
T
24 tháng 3 2019
\(=\left(x^4+x^3+x^2\right)-\left(x^3-2007x^2-2007x-2008\right)\)
\(=x^2\left(x^2+x+1\right)-\left[x\left(x^2+x+1\right)-2008\left(x^2-x-1\right)\right]\)
\(=x^2\left(x^2+x+1\right)-\left(x^2+x+1\right)\left(x-2008\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)
NH
4
30 tháng 5 2017
giải phương trình:
- Nếu \(x\ge1\)phương trình trở thành : \(x^2-3x+2=x-1\Leftrightarrow x^2-4x+3=0\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}TM}\)
- Nếu \(x< 1\)\(\Rightarrow x^2-3x+2=1-x\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1L\)VẬY NGHIỆM PHƯƠNG TRÌNH LÀ : x=1 hoặc x=3
KH
30 tháng 5 2017
\(x^4+2008x^2+2007x+2008\)
\(=x\left[x\left(x^2+2008\right)+2007\right]+2008\)
\(=\left[\left(x-1\right)x+2008\right]\left(x^2+x+1\right)\)
\(=\left(x^2-x+2008\right)\left(x^2+x+1\right)\)
~(‾▿‾~)
PN
2
MA
7 tháng 10 2016
a) \(x^8+x+1\)
\(=x^8+x^7+x^6-x^7-x^6-x^5+x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1\)
\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)
b) \(x^4+2008x^2+2007x+2008\)
\(=x^4+x^3+x^2-x^3-x^2-x+2008x^2+2008x+2008\)
\(=x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+2008\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)
M
1
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 4 2019
\(x^4-x+2008x^2+2008x+2008\)
\(=x\left(x^3-1\right)+2008\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2008\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)
NT
1
23 tháng 3 2019
\(x^8+x^4+1\)
\(=\left(x^4\right)^2+2.x^4+1-x^4\)
\(=\left(x^4+1\right)-\left(x^2\right)^2\)
\(=\left(x^4+1-x^2\right)\left(x^4+x^2+1\right)\)
\(=\left(x^4+1-x^2\right)\left[\left(x^2\right)^2+2x^2+1-x^2\right]\)
\(=\left(x^4+1-x^2\right)\left[\left(x^2+1^2\right)-x^2\right]\)
\(=\left(x^4+1-x^2\right)\left(x^2+x+1\right)\left(x^2-x+1\right)\)
\(x^4+2008x^2+2007x+2008\)
\(=\left(x^4-x\right)+2008\left(x^2+x+1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)+2008\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x-1+2008\right)\)
\(=\left(x^2+x+1\right)\left(x+2007\right)\)
KA
Phân tích đa thức thành nhân tử:
a. \(2x^2-5x-7\)
b. \(x^3-5x^2+8x-4\)
c. \(x^4+2008x^2+2007x+2008\)
2
TT
1 tháng 12 2017
a.\(2x^2-5x-7\)
\(=2x^2-7x+2x-7\)
\(=\left(2x^2+2x\right)+\left(-7x-7\right)\)
\(=2x\left(x+1\right)-7\left(x+1\right)\)
\(=\left(2x-7\right)\left(x+1\right)\)
TC
1 tháng 12 2017
a)\(2x^2-5x-7\)
\(=\left(2x^2+2x\right)-\left(7x+7\right)\)
\(=\left(x+1\right)\left(2x-7\right)\)
b) \(x^3-5x^2+8x-4\)
\(=\left(x^3-x^2\right)-\left(4x^2-4x\right)+\left(4x-4\right)\)
\(=\left(x-1\right)\left(x^2-4x+4\right)\)
\(=\left(x-1\right)\left(x-2\right)^2\)
c)\(x^4+2008x^2+2007x+2008\)
\(=\left(x^4-x^3+2008x^2\right)+\left(x^3-x^2+2008x\right)+\left(x^2-x+2008\right)\)
\(=\left(x^2-x+2008\right)\left(x^2+x+1\right)\)
a) \(x^2+7x+6\)
\(=x^2+x+6x+6\)
\(=x\left(x+1\right)+6\left(x+1\right)\)
\(=\left(x+1\right)\left(x+6\right)\)
b) \(x^4 +2008.x^2+2007.x+2008\)
\(= x^4 +2008x^2+2008x-x+2008\)
\(= x(x^3-1)+2008(x^2+x+1) \)
\(= x(x-1)(x^2+x+1)+2008(x^2+x+1) \)
\(= (x^2+x+1)(x^2-x+2008) \)