Áp dụng bđt: \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)

Trở lại bài toán ta có:

\(C=\left|2000x+2016\right|+\left|2000x-2017\right|\)

\(C=\left|2000x+2016\right|+\left|2017-2000x\right|\)

\(C\ge\left|2000x+2016+2017-2000x\right|=4033\)

Dấu "=" xảy ra khi:

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}2000x+2016\ge0\\2017-2000x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}2000x+2016\le0\\2017-2000x\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2000x\ge-2016\\2000x\le2017\end{matrix}\right.\\loại\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{2016}{2000}\\x\le\dfrac{2017}{2000}\end{matrix}\right.\)

Vậy \(-\dfrac{2016}{2000}\le x\le\dfrac{2017}{2000}\)

đề kiểu sao vậy viết lại đi

\(A=x^{30}-2000x^{29}+2000x^{28}-2000x^{27}+...+2000x^2-2000x+2000\)

Ta có: \(f\left(x\right)=x^{30}-2000x^{29}+2000x^{28}-2000x^{27}+...+2000x^2-2000x+2000\)

\(\Leftrightarrow f\left(2006\right)=2006^{30}-2000.2006^{29}+2000.2006^{28}-2000.2006^{27}+...\)\(+2000.2006^2-2000.2006+2000\)

\(\Rightarrow2006.f\left(2006\right)=2006^{31}-2000.2006^{30}+2000.2006^{29}-2000.2006^{28}+...\)\(+2000.2006^3-2000.2006^2+2000.2006\)

\(\Rightarrow2006.f\left(2006\right)+f\left(2006\right)=2006^{31}-2000.2006^{30}+2000.2006^{29}-2000.2006^{28}+...\)\(+2000.2006^3-2000.2006^2+2000.2006\)\(+2006^{30}-2000.2006^{29}+2000.2006^{28}-2000.2006^{27}+...+2000.2006^2-2000.2006+2000\)

\(\Rightarrow2007.f\left(2006\right)=2006^{31}-2000.2006^{30}+2006^{30}+2000\)

\(\Rightarrow f\left(2006\right)=\frac{2006^{31}-2000.2006^{30}+2006^{30}+2000}{2007}\)

\(\Rightarrow f\left(2006\right)=\frac{2006^{30}\left(2006-2000+1\right)+2000}{2007}\)

\(\Rightarrow f\left(2006\right)=\frac{7.2006^{30}+2000}{2007}\)

Ta có \(\left|2000x+2012\right|+\left|2013-2000x\right|\ge\left|2000x+2012+2013-2000x\right|=\left|4025\right|=4025\)

^.^

thank

Ta có f(1999) = 1999²⁰¹⁵ - 2000.1999²⁰⁰⁴ + 2000.1999²⁰¹³ - 2000.1999²⁰¹² + .... + 2000.1999 - 1

                      = 1999²⁰¹⁵ - 2000(1999²⁰¹⁴ - 1999²⁰¹³ + 1999²⁰¹² - .... - 2000.1999) - 1

         Đặt C = 1999²⁰¹⁴ - 1999²⁰¹³ + 1999²⁰¹² - .... - 2000.1999

  Khi đó : f(1999) = 1999²⁰¹⁵ - 2000C - 1

Ta có : C = 1999²⁰¹⁴ - 1999²⁰¹³ + 1999²⁰¹² - .... - 2000.1999

=> 1999C = 1999²⁰¹⁵ - 1999²⁰¹⁴ + 1999²⁰¹³ - .... - 2000.1999²

Lấy 1999C cộng C theo vế ta có : 

1999C + C = (1999²⁰¹⁵ - 1999²⁰¹⁴ + 1999²⁰¹³ - .... - 2000.1999²) + (1999²⁰¹⁴ - 1999²⁰¹³ + 1999²⁰¹² - .... - 2000.1999)

      2000C = 1999²⁰¹⁵ - 2000.1999

=> f(1999) = 1999²⁰¹⁵ - 1999²⁰¹⁵ +  2000.1999 - 1 = 2000.1999 + 1

Đs 2005x

Thay 2000=x-6 nhé

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