Câu 1 : Phân tích đa thức thành nhân tử : x4 - 2x2y +x2 +y2 - 2y +1
Câu 2: a) CMR: ∀ x ∈ Z+ thì ta luôn có:
\(A=\dfrac{x^5}{120}+\dfrac{x^4}{12}+\dfrac{7x^3}{24}+\dfrac{5x^2}{12}+\dfrac{x}{5}\) ∈ Z+
b) Rút gọn biểu thức : \(B=\dfrac{x^{2004}+x^{2000}+x^{1996}+...+x^4+1}{x^{2006}+x^{2004}+x^{2002}+...+x^2+1}\)
1. Sửa đề
\(x^4-2x^2y+x^2+y^2-2y+1\)
\(=x^2\left(x^2-2y+1\right)+\left(x^2-2y+1\right)\)
\(=\left(x^2-2y+1\right)\left(x^2+1\right)\)
2.
a. \(A=\dfrac{x^5}{120}+\dfrac{x^4}{12}+\dfrac{7x^3}{24}+\dfrac{5x^2}{12}+\dfrac{x}{5}\)
\(=\dfrac{x^5+10x^4+35x^3+50x^2+24x}{120}\)
Ta có: \(x^5+10x^4+35x^3+50x^2+24x\)
\(=x\left(x^4+10x^3+35x^2+50x+24\right)\)
\(=x\left(x^4+x^3+9x^3+9x^2+26x^2+26x+24x+24\right)\)
\(=x\left[x^3\left(x+1\right)+9x^2\left(x+1\right)+26x\left(x+1\right)+24\left(x+1\right)\right]\)
\(=x\left(x+1\right)\left(x^3+9x^2+26x+24\right)\)
\(=x\left(x+1\right)\left(x^3+2x^2+7x^2+14x+12x+24\right)\)
\(=x\left(x+1\right)\left[x^2\left(x+2\right)+7x\left(x+2\right)+12\left(x+2\right)\right]\)
\(=x\left(x+1\right)\left(x+2\right)\left(x^2+7x+12\right)\)
\(=x\left(x+1\right)\left(x+2\right)\left(x^2+3x+4x+12\right)\)
\(=x\left(x+1\right)\left(x+2\right)\left[x\left(x+3\right)+4\left(x+3\right)\right]\)
\(=x\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)⋮\left(1\cdot2\cdot3\cdot4\cdot5\right)=120\)
\(\Rightarrow\dfrac{x^5+10x^4+35x^3+50x^2+24x}{120}\in Z\)
b.
\(B=\dfrac{x^{2004}+x^{2000}+x^{1996}+...+x^4+1}{x^{2006}+x^{2004}+x^{2002}+...+x^2+1}\)
\(=\dfrac{x^{2004}+x^{2000}+x^{1996}+...+x^4+1}{\left(x^2+1\right)\left(x^{2004}+x^{2000}+...+1\right)}=\dfrac{1}{x^2+1}\)
