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\(x^2-x-1=x^2-2.\frac{1}{2}x+\frac{1}{4}-\frac{5}{4}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2-\frac{5}{4}=0\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=\frac{5}{4}\)
Khai căn hai vế ra ta được
\(x-\frac{1}{2}=\frac{\sqrt{5}}{2}\)
\(\Leftrightarrow x=\frac{\sqrt{5}}{2}+\frac{1}{2}\)
\(\Leftrightarrow x=1,618033989\)
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\(\left(1-x\right)\left(1+x+x^2+...+x^{31}\right)=1-x^{32}\)
\(\left(1-x\right)\left(1+x\right)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)\)
\(=\left(1-x^2\right)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)\)
\(=\left(1-x^4\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)\)
\(=\left(1-x^8\right)\left(1+x^8\right)\left(1+x^{16}\right)\)
\(=\left(1-x^{16}\right)\left(1+x^{16}\right)\)
\(=1-x^{32}\)
Ta có đpcm.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(1-x\right)\left(x^{31}+x^{30}+...+x+1\right)=\left(1-x\right)\left(1+x\right)\left(1+x^2\right).....\left(1+x^{16}\right)\)
VP = 1 - x32
Đặt \(A=x^{31}+x^{30}+..+x+1\Leftrightarrow xA=x^{32}+x^{31}+.......+x^2+x\)
VT = \(A-xA=\left(1-x\right)A=1-x^{32}\)= VP (dpcm)