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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
\(x^4+2008x^2+2007x+2008\)
\(=x\left[x\left(x^2+2008\right)+2007\right]+2008\)
\(=\left[\left(x-1\right)x+2008\right]\left(x^2+x+1\right)\)
\(=\left(x^2-x+2008\right)\left(x^2+x+1\right)\)
~(‾▿‾~)
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:.\(x^4+2008x^2+2007x+2008\)
\(=x^4+x^2+1+2007x^2+2007x+2007\)
\(=\left(x^4+x^2+1\right)+2007\left(x^2+x+1\right)\)
\(=\left(x^4+x^3+x^2-x^3-x^2-x+x^2+x+1\right)+2007\left(x^2+x+1\right)\)
\(=\left[x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\right]+2007\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1\right)+2007\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)
\(x^8+x^4+1\)
\(=\left(x^4\right)^2+2.x^4+1-x^4\)
\(=\left(x^4+1\right)-\left(x^2\right)^2\)
\(=\left(x^4+1-x^2\right)\left(x^4+x^2+1\right)\)
\(=\left(x^4+1-x^2\right)\left[\left(x^2\right)^2+2x^2+1-x^2\right]\)
\(=\left(x^4+1-x^2\right)\left[\left(x^2+1^2\right)-x^2\right]\)
\(=\left(x^4+1-x^2\right)\left(x^2+x+1\right)\left(x^2-x+1\right)\)
\(x^4+2008x^2+2007x+2008\)
\(=\left(x^4-x\right)+2008\left(x^2+x+1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)+2008\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x-1+2008\right)\)
\(=\left(x^2+x+1\right)\left(x+2007\right)\)
Phân tích đa thức thành nhân tử:
a. \(2x^2-5x-7\)
b. \(x^3-5x^2+8x-4\)
c. \(x^4+2008x^2+2007x+2008\)
a.\(2x^2-5x-7\)
\(=2x^2-7x+2x-7\)
\(=\left(2x^2+2x\right)+\left(-7x-7\right)\)
\(=2x\left(x+1\right)-7\left(x+1\right)\)
\(=\left(2x-7\right)\left(x+1\right)\)
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)\(x^4+2008x^2+2007x+2008\)
\(=\left(x^4-x^3+2008x^2\right)+\left(x^3-x^2+2008x\right)+\left(x^2-x+2008\right)\)
\(=\left(x^2-x+2008\right)\left(x^2+x+1\right)\)
1. x4 + 2008x2 + 2007x + 2008
= (x4 + x2 + 1) + (2007x2 + 2007x + 1)
= (x2 + x + 1)(x2 - x + 1) + 2007(x2 + x + 1)
= (x2 + x + 1)(x2 - x + 2008)
2. x4 - 6x3 + 12x2 - 14x - 3
= x4 - 2x3 + 3x2 - 4x3 + 8x2 - 12x + x2 - 2x + 3
= x2(x2 - 2x + 3) - 4x(x2 - 2x + 3) + (x2 - 2x + 3)
= (x2 - 2x + 3)(x2 - 4x + 1)
bn ơi dòng 2 phải là (x4 + x2 + 1) + (2007x2 + 2007x + 2007 ) ms đúng
Ta có:
\(x^4+2008x^2+2007x+2008\)
\(=\left(x^4+x^2+1\right)+\left(2007x^2+2007x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1\right)+2007\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)
Cách này dễ hơn nè :
\(x^4+2008x^2+2007x+2008\)
= \(x^4-x+2008\left(x^2+x+1\right)\)
=\(x\left(x-1\right)\left(x^2+x+1\right)+2008\left(x^2+x+1\right)\)=\(\left(x^2+x+1\right)\left(x^2-x+2008\right)\)
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) \(x^4+2008x^2+2007x+2008\)
\(=x^4+x^3+x^2-x^3-x^2-x+2008x^2+2008x+2008\)
\(=x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+2008\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)
a )
b)
c) x^5 - x^4 - 1
= x^5 - x^3 - x² - x^4 + x² + x + x^3 - x - 1
= x²( x^3 - x - 1 ) - x( x^3 - x - 1 ) + ( x^3 - x - 1 )
= ( x² - x + 1)( x^3 - x - 1 )
d)
a)\(x^8+2x^4+1-x^4=\left(x^4+1\right)^2-\left(x^2\right)^2\)
\(=\left(x^4+x^2+1\right)\left(x^4-x^2+1\right)\)
\(=\left(x^2-x+1\right)\left(x^2+x+1\right)\left(x^4-x^2+1\right)\)
\(=\left(x^4+x^3+x^2\right)-\left(x^3-2007x^2-2007x-2008\right)\)
\(=x^2\left(x^2+x+1\right)-\left[x\left(x^2+x+1\right)-2008\left(x^2-x-1\right)\right]\)
\(=x^2\left(x^2+x+1\right)-\left(x^2+x+1\right)\left(x-2008\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)