\(\frac{x-1}{2000}+\frac{x-3}{1998}+\frac{x-5}{1996}+\frac{x}{667}=6\)

\(\Rightarrow\frac{x-1}{2000}+\frac{x-3}{1998}+\frac{x-5}{1996}+\frac{x}{667}-6=0\)

\(\Rightarrow\left(\frac{x-1}{2000}-1\right)+\left(\frac{x-3}{1998}+1\right)+\left(\frac{x-5}{1996}-1\right)+\left(\frac{x}{667}-3\right)=0\)

\(\Rightarrow\frac{x-1-2000}{2000}+\frac{x-3-1998}{1998}+\frac{x-5-1996}{1996}+\frac{x-3.667}{667}=0\)

\(\Rightarrow\frac{x-2001}{2000}+\frac{x-2001}{1998}+\frac{x-2001}{1996}+\frac{x-2001}{667}=0\)

\(\Rightarrow\left(x-2001\right)\left(\frac{1}{2000}+\frac{1}{1998}+\frac{1}{1996}+\frac{1}{667}\right)=0\)

Ta có: \(\frac{1}{2000}+\frac{1}{1998}+\frac{1}{1996}+\frac{1}{667}\ne0\)

\(\Rightarrow x-2001=0\Rightarrow x=2001\)

Đặt 2020=a; 2021=b; 2022=c, ta có :

\(\frac{bx-ay}{c}=\frac{az-cx}{b}=\frac{cy-bz}{a}\)

Theo tính chất của dãy tỉ số bằng nhau, ta có:

\(\frac{bx-ay}{c}=\frac{az-cx}{b}=\frac{cy-bz}{a}=\frac{bcx-acx}{c^2}=\frac{abz-cbx}{b^2}=\frac{acy-abz}{a^2}=\frac{bcx-acx+abz-bcx+acy-abz}{a^2+b^2+c^2}=0\)

\(\Rightarrow\hept{\begin{cases}\frac{bx-ay}{c}=0\\\frac{az-cx}{b}=0\\\frac{cy-bz}{a}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}bx-ay=0\\az-cx=0\\cy-bz=0\end{cases}}\Leftrightarrow\hept{\begin{cases}ay=bx\\az=cx\\cy=bz\end{cases}}\)

Thay 2020=a; 2021=b; 2022=c, ta có :

\(\hept{\begin{cases}2020y=2021x\\2020z=2022x\\2022y=2021z\end{cases}}\Rightarrow\hept{\begin{cases}\frac{y}{2021}=\frac{x}{2020}\\\frac{z}{2022}=\frac{x}{2020}\\\frac{y}{2021}=\frac{z}{2022}\end{cases}}\)

\(\Rightarrow\frac{x}{2020}=\frac{y}{2021}=\frac{z}{2022}\)

\(C=...=\frac{5\left(2x-3\right)+6}{2x-3}=5+\frac{6}{2x-3}\)

Để C nguyên  \(\Leftrightarrow\frac{6}{2x-3}\) nguyên

---> 6 chia hết cho 2x - 3

---> 2x - 3 \(\in\)( 1 ; -1 ; 2 ; -2 ; 3 ; -3 ; 6 ; -6 )

Giải ra được

---> x \(\in\)( 2 ; 1 ; 3 ; 0 )

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