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\(a,x^4-4x^3+x^2-4x=0\)
\(\Rightarrow\left(x^4-4x^3\right)+\left(x^2-4x\right)=0\)
\(\Rightarrow x^3\left(x-4\right)+x\left(x-4\right)=0\)
\(\Rightarrow\left(x-4\right)\left(x^2+x\right)=0\)
\(\Rightarrow x\left(x-4\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-1\end{matrix}\right.\)
\(b,x^3-5x^2+4x-20=0\)
\(\Rightarrow\left(x^3-5x^2\right)+\left(4x-20\right)=0\)
\(\Rightarrow x^2\left(x-5\right)+4\left(x-5\right)=0\)
\(\Rightarrow\left(x-5\right)\left(x^2+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\x^2+4=0\end{matrix}\right.\)
\(\Rightarrow x=5\)
a) \(x^4-4x^3+x^2-4x=0\)
\(\Leftrightarrow\left(x^4-4x^3\right)+\left(x^2-4x\right)=0\)
\(\Leftrightarrow x^3\left(x-4\right)+x\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x^3+x\right)=0\)
\(\Leftrightarrow x\left(x-4\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x^2+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x^2=-1\left(loai\right)\end{matrix}\right.\)
Vậy x=0; x=4
b) \(x^3-5x^2+4x-20=0\)
\(\Leftrightarrow\left(x^3-5x^2\right)+\left(4x-20\right)=0\)
\(\Leftrightarrow x^2\left(x-5\right)+4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x^2+4=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=5\\x^2=-4\left(loai\right)\end{matrix}\right.\)
Vậy x=5
a,\(x^5-x^4-x^4+x^3+2x^3-2x^2-2x^2+2\)2x-2x+2\(x^4\left(x-1\right)-x^3\left(x-1\right)+2x^2\left(x-1\right)-2x\left(x-1\right)+2\left(x-1\right)\)
=\(\left(x^4-x^3+2x^2-2x+2\right)\left(x-1\right)\)
b,
x=2016 =>x+1=2017
Thay 2007=x+1 vào A ................................................. tự típ
2: \(=a^2\left(a+3\right)+4\left(a+3\right)=\left(a+3\right)\left(a^2+4\right)\)
3: \(=\left(2a-1\right)^2-4b^2\)
\(=\left(2a-1-2b\right)\left(2a-1+2b\right)\)
4: \(=-\left(x^2+x-2\right)=-\left(x+2\right)\left(x-1\right)\)
5: \(=7\left(x^2-2xy^2+y^4\right)=7\left(x-y^2\right)^2\)
6: \(=\left(x+2\right)^2-y^2=\left(x+2+y\right)\left(x+2-y\right)\)
Dễ thầy 2017=2016+1=x+1
Thay vào ta có:
\(x^{10}-2017x^9+2017x^8-.....+2017x^2-2017x+2017\)
\(=x^{10}-\left(x+1\right)x^9+\left(x+1\right)x^8-....+\left(x+1\right)x^2-\left(x+1\right)x+2017\)
\(=x^{10}-x^{10}-x^9+x^9+x^8-....+x^3+x^2-x^2-x+2017=-x+2017=-2016+2017=1\)
Vậy..........
2017 = 2016 + 1 = x + 1
suy ra 2017x15 = x16 + x15
2017x14 = x15 + x14
....
từ đó ta dễ tính ra A