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n thuộc N
B=x^2 +2x +1 =(x+1)^2
\(A=x^{4n+2}+2.x^{2n+1}+1=\left(x^{2n+1}\right)^2+2.\left(x^{2n+1}\right)+1=\left(x^{2n+1}+1\right)^2\)
\(\dfrac{A}{B}=\left(\dfrac{x^{2n+1}+1}{x+1}\right)^2\)
với n =0 đúng
n >0 =>2n+1 >=3
=> x^(2n+1) =(x+1).g(x) => dpcm
Ta có :
\(x^{4n+2}+2x^{2n+1}+1=\left(x^{2n+1}\right)^2+2x^{2n+1}+1==\left(x^{2n+1}+1\right)^2\)
Vì \(x^{2n+1}+1⋮x+1\forall x;n\in Z\) nên \(\left(x^{2n+1}+1\right)^2⋮\left(x+1\right)^2=\forall x;n\in Z\)
Hay \(x^{4n+2}+2x^{2n+1}+1⋮x^2+2x+1\)
Đặt \(A=x^{20}+x^{10}+1\)
\(x^{50}+x^{10}+1\)
\(=x^{50}-x^{20}+A\)
\(=x^{20}\left(x^{30}-1\right)+A\)
\(=x^{20}\left(x^{10}-1\right)A+A\)
\(=\left(x^{30}-x^{20}+1\right)A\)
mà \(\left(x^{30}-x^{20}+1\right)A⋮A\)
\(\Rightarrow\left(x^{50}+x^{10}+1\right)⋮\left(x^{20}+x^{10}+1\right)\)
Đề đúng : Chứng minh : \(\frac{x^4+4}{x\left(x^2+2\right)-2x^2-\left(x-1\right)^2-1}=\frac{x^2+2x+2}{x-1}\)
Điều kiện : \(x\ne1\)
Phân tích : \(x^4+4=\left(x^4+4x^2+4\right)-4x^2=\left(x^2+2\right)^2-\left(2x\right)^2=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
\(x\left(x^2+2\right)-2x^2-\left(x-1\right)^2-1=x^3+2x-2x^2-\left(x^2-2x+1\right)-1\)
\(=x^3-3x^2+4x-2=\left(x^3-3x^2+3x-1\right)+\left(x-1\right)=\left(x-1\right)^3+\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-2x+2\right)\)
Suy ra : \(\frac{x^4+4}{x\left(x^2+2\right)-2x^2-\left(x-1\right)^2-1}=\frac{\left(x^2-2x+2\right)\left(x^2+2x+2\right)}{\left(x-1\right)\left(x^2-2x+2\right)}=\frac{x^2+2x+2}{x-1}\)
