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\(x=2009\)
\(\Rightarrow x-1=2008\left(1\right)\)
Thay (1) vào A ta được:
\(A=x^{2009}-2008x^{2008}-2008x^{2007}-...-2008x+1\)
\(A=x^{2009}-\left(x-1\right)x^{2008}-...-\left(x-1\right)x+1\)
\(A=x^{2009}-x^{2009}+x^{2008}-...-x^2-x+1\)
\(A=-x+1\)
\(A=-2009+1\)
\(A=-2008\)
x=2009x=2009
⇒x−1=2008(1)⇒x−1=2008(1)
Thay (1) vào A ta được:
A=x^2009−2008x^2008−2008x^2007−...−2008x+1
A=x^2009−(x−1)x^2008−...−(x−1)x+1
A=x^2009−x^2009+x^2008−...−x^2−x+1
A=−x+1
A=−2009+1
A=−2008
\(x=2009\Leftrightarrow x-1=2008\\ \Leftrightarrow A=x^x-\left(x-1\right)x^{x-1}-\left(x-1\right)x^{x-2}-...-\left(x-1\right)x+1\\ \Leftrightarrow A=x^x-x^x+x^{x-1}-x^{x-1}+x^{x-2}-...-x^2-x+1\\ \Leftrightarrow A=1-x=1-2009=-2008\)
Để PT có nghiệm khi \(2009y^{2010}\) lẻ \(\Rightarrow y^{2010}\)lẻ Hay \(y\) lẻ
\(\Rightarrow y^2\equiv1\left(mod4\right)\)\(\Rightarrow2009y^{2010}\equiv1\left(mod4\right)\)
Mà \(2008x^{2009}\equiv0\left(mod4\right)\) nên \(2008x^{2009}+2009y^{2010}\equiv1\left(mod4\right)\)
Mà \(2011\equiv3\left(mod4\right)\)
\(\Rightarrow2008x^{2009}+2009y^{2010}\ne2011\forall x;y\in Z\)
Vậy PT vô nghiệm nguyên
a)\(x^8+2x^4+1-x^4=\left(x^4+1\right)^2-\left(x^2\right)^2\)
\(=\left(x^4+x^2+1\right)\left(x^4-x^2+1\right)\)
\(=\left(x^2-x+1\right)\left(x^2+x+1\right)\left(x^4-x^2+1\right)\)
\(=\left(x^4+x^3+x^2\right)-\left(x^3-2007x^2-2007x-2008\right)\)
\(=x^2\left(x^2+x+1\right)-\left[x\left(x^2+x+1\right)-2008\left(x^2-x-1\right)\right]\)
\(=x^2\left(x^2+x+1\right)-\left(x^2+x+1\right)\left(x-2008\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)
Câu 1:
a) \(2x^2+5x-3=\left(2x^2+6x\right)-\left(x+3\right)\)
\(=2x\left(x+3\right)-\left(x+3\right)=\left(x+3\right)\left(2x-1\right)\)
b) \(x^4+2009x^2+2008x+2009\)
\(=\left(x^4-x\right)+\left(2009x^2+2009x+2009\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2009\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2009\right)\)
c) \(\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]=-16\) (đã sửa đề)
\(\Leftrightarrow\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16=0\)
\(\Leftrightarrow\left(x^2+10x+20\right)^2-16+16=0\)
\(\Leftrightarrow\left(x^2+10x+20\right)^2=0\)
\(\Leftrightarrow\left(x+5\right)^2-5=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-5-\sqrt{5}\\x=-5+\sqrt{5}\end{cases}}\)
Câu 1.
a) 2x2 + 5x - 3 = 2x2 + 6x - x - 3 = 2x( x + 3 ) - ( x + 3 ) = ( x + 3 )( 2x - 1 )
b) x4 + 2009x2 + 2008x + 2009
= x4 + 2009x2 + 2009x - x + 2009
= ( x4 - x ) + ( 2009x2 + 2009x + 2009 )
= x( x3 - 1 ) + 2009( x2 + x + 1 )
= x( x - 1 )( x2 + x + 1 ) + 2009( x2 + x + 1 )
= ( x2 + x + 1 )[ x( x - 1 ) + 2009 ]
= ( x2 + x + 1 )( x2 - x + 2009 )
c) ( x + 2 )( x + 4 )( x + 6 )( x + 8 ) = 16 ( xem lại đi chứ không phân tích được :v )
Câu 2.
3x2 + x - 6 - √2 = 0
<=> ( 3x2 - 6 ) + ( x - √2 ) = 0
<=> 3( x2 - 2 ) + ( x - √2 ) = 0
<=> 3( x - √2 )( x + √2 ) + ( x - √2 ) = 0
<=> ( x - √2 )[ 3( x + √2 ) + 1 ] = 0
<=> \(\orbr{\begin{cases}x-\sqrt{2}=0\\3\left(x+\sqrt{2}\right)+1=0\end{cases}}\)
+) x - √2 = 0 => x = √2
+) 3( x + √2 ) + 1 = 0
<=> 3( x + √2 ) = -1
<=> x + √2 = -1/3
<=> x = -1/3 - √2
Vậy S = { √2 ; -1/3 - √2 }
Câu 3.
A = x( x + 1 )( x2 + x - 4 )
= ( x2 + x )( x2 + x - 4 )
Đặt t = x2 + x
A = t( t - 4 ) = t2 - 4t = ( t2 - 4t + 4 ) - 4 = ( t - 2 )2 - 4 ≥ -4 ∀ t
Dấu "=" xảy ra khi t = 2
=> x2 + x = 2
=> x2 + x - 2 = 0
=> x2 - x + 2x - 2 = 0
=> x( x - 1 ) + 2( x - 1 ) = 0
=> ( x - 1 )( x + 2 ) = 0
=> x = 1 hoặc x = -2
=> MinA = -4 <=> x = 1 hoặc x = -2
\(x^8+x^4+1\)
\(=\left(x^4\right)^2+2.x^4+1-x^4\)
\(=\left(x^4+1\right)-\left(x^2\right)^2\)
\(=\left(x^4+1-x^2\right)\left(x^4+x^2+1\right)\)
\(=\left(x^4+1-x^2\right)\left[\left(x^2\right)^2+2x^2+1-x^2\right]\)
\(=\left(x^4+1-x^2\right)\left[\left(x^2+1^2\right)-x^2\right]\)
\(=\left(x^4+1-x^2\right)\left(x^2+x+1\right)\left(x^2-x+1\right)\)
\(x^4+2008x^2+2007x+2008\)
\(=\left(x^4-x\right)+2008\left(x^2+x+1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)+2008\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x-1+2008\right)\)
\(=\left(x^2+x+1\right)\left(x+2007\right)\)
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:.\(x^4+2008x^2+2007x+2008\)
\(=x^4+x^2+1+2007x^2+2007x+2007\)
\(=\left(x^4+x^2+1\right)+2007\left(x^2+x+1\right)\)
\(=\left(x^4+x^3+x^2-x^3-x^2-x+x^2+x+1\right)+2007\left(x^2+x+1\right)\)
\(=\left[x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\right]+2007\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1\right)+2007\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)
Có ai giúp mình làm ko?
Lộn đề
\(A=x^{2009}-2008x^{2008}-2008x^{2007}-...-2008x+1\)1