\(x^4\sqrt{x+3}=2x^4-2008x+2008\)
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PT: \(\sqrt{x+3}x^4=2x^4-2008x+2008\)
DK xác định : \(x+3\ge0\Leftrightarrow x\ge-3\)(**)
PT đã cho tương đương:
\(x^4\left(\sqrt{x+3}-2\right)+2008x=2008\)(***)
Nếu :\(x>1\) thì \(x+3>4\Rightarrow x^4\left(\sqrt{x+3}-2\right)+2008x>2008\)
Nếu \(-3\le x\le1\)thì\(0\le x+3< 4\Rightarrow\sqrt{x+3}-2< 0\)và \(x^4\ge0\)
\(\Rightarrow x^4\left(\sqrt{x+3}-2\right)\le0\) Mặt khác : \(2008x< 2008\)
\(\Rightarrow x^4\left(\sqrt{x+3}-2\right)+2008x< 2008\)
* \(x=1\) thỏa mãn (***)
Vậy (***) có nghiệm duy nhất x= 1
KL: Nghiệm của pt đã cho là : x = 1
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x^4+2008x^2+2007x+2008
=x^4+2008x^2+2008x+2008-x
=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)
x4+2008x2+2007x+2008
=(x4+x2+1)+(2007x2+2007x+2007)
=(x4+2x2+1-x2)+2007(x2+x+1)
=(x2+1)2-x2+2007(x2+x+1)
=(x2+1-x)(x2+1+x)+2007(x2+x+1)
=(x2+x+1)(x2+1-x+2007)=(x2+x+1)(x2-x+2008)
Phân tích đa thức thành nhân tử:
a. \(2x^2-5x-7\)
b. \(x^3-5x^2+8x-4\)
c. \(x^4+2008x^2+2007x+2008\)
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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)\(x^4+2008x^2+2007x+2008\)
\(=\left(x^4-x^3+2008x^2\right)+\left(x^3-x^2+2008x\right)+\left(x^2-x+2008\right)\)
\(=\left(x^2-x+2008\right)\left(x^2+x+1\right)\)
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x^4 + 2008x^2 + 2007x + 2008
\(=x^4+x^2+2007x^2+2007x+2007+1\)
\(=x^4+x^2+1+2007\left(x^2+x+1\right)\)
\(=\left(x^2+1\right)^2-x^2+2007\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1\right)+2007\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)