x*2000-x=1999-1997

x*2000-x*1=2

x*(2000-1)=2

x*1999=2

x          =2:1999

x         =2/1999

\(\frac{x+15}{2000}+\frac{x+16}{1999}=\frac{x+17}{1998}+\frac{x+18}{1997}\)

\(\Leftrightarrow\frac{x+15}{2000}+1+\frac{x+16}{1999}+1=\frac{x+17}{1998}+1+\frac{x+18}{1997}+1\)

\(\Leftrightarrow\frac{x+2015}{2000}+\frac{x+2015}{1999}=\frac{x+2015}{1998}+\frac{x+2015}{1997}\)

\(\Leftrightarrow\frac{x+2015}{2000}+\frac{x+2015}{1999}-\frac{x+2015}{1998}-\frac{x+2015}{1997}=0\)

\(\Leftrightarrow\left(x+2015\right)\left(\frac{1}{2000}+\frac{1}{1999}-\frac{1}{1998}-\frac{1}{1997}\right)=0\)

Có: \(\frac{1}{2000}+\frac{1}{1999}-\frac{1}{1998}-\frac{1}{1997}\ne0\)

\(\Rightarrow x+2015=0\Rightarrow x=-2015\)

nhớ là có ai làm rồi mà quên =))

khá dài à =)) bấm máy mỏi :v

Ta có: 1998 ≡ 0 (mod 111) => 1997 ≡ -1 (mod 111) và 1999 ≡ 1 (mod 111)

Nên ta có: 1997^1998 + 1998^1999 +1999^2000 ≡ 2 (mod 111) (1997^1998 + 1998^1999 +1999^2000 )10 ≡ 210 (mod 111)

Mặt khác ta có: 210 = 1024 ≡ 25 (mod 111) Vậy (1997^1998 + 1998^1999 +1999^2000 ) ^ 10 chia cho 111 có số dư là 25

Tại sao 210=1024

2^10 not 210

X.1999.X=1999.1997+1999

X.1999.X=1999.1997+1999.1

X.X.1999=1999.(1997+1)

X.X.1999=1999.1998

X.X=1999.1998:1999=1998

 X × 1999 × X = 1999 × 1997 + 1999

X^2 x 1999 = 1999 × 1997 + 1999 x 1

X^2 x 1999 = 1999 x ( 1997 + 1 )

X^2 x 1999 = 1999 x 1998

X^2 = 1999 x 1998 : 1999

X^2 = 1999 : 1999 x 1998

X^2 = 1 x 1998

X^2 = 1998