;-; nhầm 

Đặt \(A=1+2+2^2+2^3+...+2^{100}\\ 2A=2+2^2+2^3+2^4+...+2^{101}\\ 2A-A=2+2^2+2^3+2^4+...+2^{101}-1-2-2^2-2^3-...-2^{100}\\ A=2^{101}-1\)

\(\Rightarrow2^{100}-\left(2^{100}-1\right)=2^{100}-2^{101}+1\)

\(\dfrac{1}{2}\cdot x+\dfrac{3}{5}\left(x-2\right)=3\\ \dfrac{1}{2}\cdot x+\dfrac{3}{5}\cdot x-\dfrac{13}{5}=3\\ \left(\dfrac{1}{2}+\dfrac{3}{5}\right)x-\dfrac{13}{5}=3\\ \dfrac{11}{10}x-\dfrac{13}{5}=3\\ \dfrac{11}{10}x=\dfrac{28}{5}\\ x=\dfrac{28}{5}:\dfrac{11}{10}\\ x=\dfrac{28}{11}\\ 3-\left(\dfrac{1}{6}-x\right)\cdot\dfrac{2}{3}=\dfrac{2}{3}\\ \left(\dfrac{1}{6}-x\right)\cdot\dfrac{2}{3}=3-\dfrac{2}{3}\\ \left(\dfrac{1}{6}-x\right)\cdot\dfrac{2}{3}=\dfrac{7}{3}\\ \dfrac{1}{6}-x=\dfrac{7}{3}:\dfrac{2}{3}\\ \dfrac{1}{6}-x=\dfrac{7}{2}\\ x=\dfrac{1}{6}-\dfrac{7}{2}\\ x=-\dfrac{10}{3}\)

\(\dfrac{x+1}{x-5}+\dfrac{x-18}{x-5}+\dfrac{x+2}{x-5}\\ =\dfrac{x+1+x-18+x+2}{x-5}\\ =\dfrac{\left(x+x+x\right)+\left(1-18+2\right)}{x-5}\\ =\dfrac{3x-15}{x-5}=\dfrac{3\left(x-5\right)}{x-5}=3\)

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{999\cdot1000}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{999}-\dfrac{1}{1000}\\ =1-\dfrac{1}{1000}=\dfrac{999}{1000}\)

Có `xyz=2023=>2023=xyz` 

Thay vào ta có :

\(\dfrac{xyz\cdot x}{xy+xyz\cdot x+xyz}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{xz+z+1}=1\\ \dfrac{x^2yz}{xy\left(1+xz+z\right)}+\dfrac{y}{y\left(z+1+xz\right)}+\dfrac{z}{xz+z+1}=1\\ \dfrac{xz}{1+xz+z}+\dfrac{1}{z+1+xz}+\dfrac{z}{xz+z+1}=1\\ \dfrac{xz+1+z}{1+xz+z}=1\left(dpcm\right)\)