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7 tháng 2 2018

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7 tháng 2 2018

  x^4 + 2008x^2 + 2007x + 2008

\(=x^4+x^2+2007x^2+2007x+2007+1\)

\(=x^4+x^2+1+2007\left(x^2+x+1\right)\)

\(=\left(x^2+1\right)^2-x^2+2007\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+1\right)+2007\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)

14 tháng 3 2022

2x^4 hay x^4

14 tháng 2 2016

x^4+2008x^2+2007x+2008

=x^4+2008x^2+2008x-x+2008

=(x^4-x)+(2008x^2+2008x+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)

18 tháng 6 2018

       x4+2008x2+2007x+2008

<=> x4-x+2008x2+2008x+2008

<=> x(x3-1)+2008(x2+x+1)

<=> x(x-1)(x2+x+1)+2008(x2+x+1)

<=> (x2+x+1)(x2-x+2008)

14 tháng 3 2015

\(\left(x^4+x^2+1\right)+\left(2007x^2+2007x+2007\right)\)

=\(\left(x^2+x+1\right)\left(x^2-x+1\right)+2007\left(x^2+x+1\right)\)

=\(\left(x^2+x+1\right)\left(x^2-x+2008\right)\)

29 tháng 12 2017

x4_x+2008(x2+x+1)=x(x-1)(x2+x+1)+2008(x2+x+1)=(x2-x+2008)(x2+x+1)

5 tháng 12 2017

=x4+2008x2+2008x-x+2008

=(x4-x)+(2008x2+2008x+2008)

=x(x3-1)+2008(x2+x+1)

=x(x-1)(x2+x+1)+2008(x2+x+1)

=(x2+x++1)(x2-x+2008)

24 tháng 3 2019

a)\(x^8+2x^4+1-x^4=\left(x^4+1\right)^2-\left(x^2\right)^2\)

\(=\left(x^4+x^2+1\right)\left(x^4-x^2+1\right)\)

\(=\left(x^2-x+1\right)\left(x^2+x+1\right)\left(x^4-x^2+1\right)\)

24 tháng 3 2019

\(=\left(x^4+x^3+x^2\right)-\left(x^3-2007x^2-2007x-2008\right)\)

\(=x^2\left(x^2+x+1\right)-\left[x\left(x^2+x+1\right)-2008\left(x^2-x-1\right)\right]\)

\(=x^2\left(x^2+x+1\right)-\left(x^2+x+1\right)\left(x-2008\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)

30 tháng 5 2017

giải phương trình:

  1. Nếu \(x\ge1\)phương trình trở thành : \(x^2-3x+2=x-1\Leftrightarrow x^2-4x+3=0\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}TM}\)
  2. 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
30 tháng 5 2017

   \(x^4+2008x^2+2007x+2008\)

\(=x\left[x\left(x^2+2008\right)+2007\right]+2008\)

\(=\left[\left(x-1\right)x+2008\right]\left(x^2+x+1\right)\)

\(=\left(x^2-x+2008\right)\left(x^2+x+1\right)\)

~(‾▿‾~)

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:.\(x^4+2008x^2+2007x+2008\)

\(=x^4+x^2+1+2007x^2+2007x+2007\)

\(=\left(x^4+x^2+1\right)+2007\left(x^2+x+1\right)\)

\(=\left(x^4+x^3+x^2-x^3-x^2-x+x^2+x+1\right)+2007\left(x^2+x+1\right)\)

\(=\left[x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\right]+2007\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+1\right)+2007\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)

1 tháng 6 2021

Trả lời:

a, x2 + 7x + 6

= x2 + x + 6x + 6

= ( x2 + x ) + ( 6x + 6 )

= x ( x + 1 ) + 6 ( x + 1 )

= ( x + 6 ) ( x + 1 )