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

Ta có: x=1999

nên x+1=2020

Ta có: \(f\left(x\right)=x^{17}-2020\cdot x^{16}+2020\cdot x^{15}-2020\cdot x^{14}+...+2000x-1\)

\(=x^{17}-x^{16}\left(x+1\right)+x^{15}\left(x+1\right)-x^{14}\left(x+1\right)+...+x\left(x+1\right)-1\)

\(=x^{17}-x^{17}-x^{16}+x^{16}+x^{15}-x^{15}-x^{14}+...+x^2+x-1\)

\(=x-1\)

\(=1999-1=1998\)

f(x) = x^17 - 2000x^16 + 2000x^15 - 2000x^14 + ... + 2000x - 1

⇒ f(1999) = 1999^17 - 2000.1999^16 + 2000.1999^15 - 2000.1999^14 + ... + 2000.1999 - 1

⇒ 1999. f(1999) = 1999^18 - 1999.1999^17 + 2000.1999^16 - 2000.1999^15 + ... + 2000.1999^2 - 1999

⇒ 1999. f(1999) + f(1999) =(1999^18 - 1999.1999^17 + 2000.1999^16 - 2000.1999^15 + ... + 2000.1999^2 - 1999) + (1999^17 - 2000.1999^16 + 2000.1999^15 - 2000.1999^14 + ... + 2000.1999 - 1)

⇒ 2000. f(1999) = 19992−1

⇔ f(1999) =1999^2-1/2000(ghi dưới dạng phân số nha)

Từ :\(\hept{\begin{cases}x+y+z=3\\x^4+y^4+z^4=3xyz\end{cases}}\)\(\Rightarrow x^4+y^4+z^4=\left(x+y+z\right)xyz=x^2yz+xy^2z+xyz^2\)

Áp dụng AM - GM ta có :

\(x^2yz=x.x.y.z\le\frac{x^4+x^4+y^4+z^4}{4}=\frac{2x^4+y^4+z^4}{4}\)

\(xy^2z=x.y.y.z\le\frac{x^4+y^4+y^4+z^4}{4}=\frac{x^4+2y^4+z^4}{4}\)

\(xyz^2=x.y.z.z\le\frac{x^4+y^4+z^4+z^4}{4}=\frac{x^4+y^4+2z^4}{4}\)

\(\Rightarrow x^2yz+xy^2z+xyz^2\le\frac{4\left(x^4+y^4+z^4\right)}{4}=x^4+y^4+z^4\)

Mà đề lại cho \(x^4+y^4+z^4=x^2yz+xy^2z+xyz^2\) \(\Rightarrow x=y=z\)

Kết hợp với x + y + z = 3 \(\Rightarrow x=y=z=1\)

Thay vào M ta được : \(M=2000.1^{2016}+1^{2016}+1^{2016}=2002\)

Thanks bạn

Ta có : x = 1999

 \(\Leftrightarrow\)x + 1 = 2000

Thay 2000 = x + 1 vào biểu thức A ta được :

A = x²⁰⁰⁰ - ( x + 1 )x¹⁹⁹⁹ + ( x + 1 )x¹⁹⁹⁸ - ( x + 1 )x¹⁹⁹⁷ + ... - ( x + 1 )x² + ( x + 1 )x + 727

A = x²⁰⁰⁰ - x²⁰⁰⁰ - x¹⁹⁹⁹ + x¹⁹⁹⁹ + x¹⁹⁹⁸ - x¹⁹⁹⁸ - x¹⁹⁹⁷ + ... - x³ - x² + x² + x + 727

A = x + 727

Thay x = 1999 vào A ta được :

A = 1999 + 727 = 2726

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

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

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

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

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

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

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

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

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

a)

\(\frac{x}{10}=\frac{y}{15}=>\frac{x}{2}=\frac{y}{3}\)

\(x=\frac{z}{2}=>\frac{x}{2}=\frac{z}{4}\)

=> x/2=y/3=z/4=\(\frac{x+2y-3z}{2+4-12}=\frac{-24}{-6}=4\)

x=4x2=8

y=4x3=12

z=4x4=16