a) \(\dfrac{x+1}{2020}+\dfrac{x+2}{2019}=\dfrac{x+3}{2018}+\dfrac{x}{2021}\)
\(\Leftrightarrow\dfrac{x+1}{2020}+1+\dfrac{x+2}{2019}+1=\dfrac{x+3}{2021}+1+\dfrac{x}{2021}+1\)
\(\Leftrightarrow\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}=\dfrac{x+2021}{2018}+\dfrac{x+2021}{2021}\)
\(\Leftrightarrow\left(x+2021\right)\left(\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2021}\right)=0\)
mà \(\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2021}\ne0\)
\(\Rightarrow x+2021=0\Leftrightarrow x=-2021\)
Vậy \(x=-2021\)
b) \(\dfrac{10-x}{100}+\dfrac{20-x}{110}+\dfrac{30-x}{120}=3\)
\(\Leftrightarrow\dfrac{10-x}{100}-1+\dfrac{20-x}{110}-1+\dfrac{30-x}{120}-1=0\)
\(\Leftrightarrow\dfrac{-90-x}{100}+\dfrac{-90-x}{110}+\dfrac{-90-x}{120}=0\)
\(\Leftrightarrow\left(-90-x\right)\left(\dfrac{1}{100}+\dfrac{1}{110}+\dfrac{1}{120}\right)=0\)
mà \(\dfrac{1}{100}+\dfrac{1}{110}+\dfrac{1}{120}>0\)
\(\Rightarrow-90-x=0\Leftrightarrow x=-90\)
Vậy \(x=-90\)
c) \(\dfrac{x+2}{13}+\dfrac{2x+45}{15}=\dfrac{3x+8}{37}+\dfrac{4x+69}{9}\)
\(\Leftrightarrow\dfrac{x+2}{13}+1+\dfrac{2x+45}{15}-1=\dfrac{3x+8}{37}+1+\dfrac{4x+69}{9}-1\)
\(\Leftrightarrow\dfrac{x+15}{13}+\dfrac{2\left(x+15\right)}{15}=\dfrac{3\left(x+15\right)}{37}+\dfrac{4\left(x+15\right)}{9}\)
\(\Leftrightarrow\left(x+15\right)\left(\dfrac{1}{13}+\dfrac{1}{15}-\dfrac{1}{37}-\dfrac{1}{9}-5\right)=0\)
mà \(\dfrac{1}{13}+\dfrac{1}{15}-\dfrac{1}{37}-\dfrac{1}{9}-5\ne0\)
\(\Rightarrow x+15=0\Leftrightarrow x=-15\)
Vậy \(x=-15\)
d) \(\dfrac{x+1}{2}+\dfrac{x+5}{3}+\dfrac{x+11}{4}+\dfrac{x+19}{5}=10\)
\(\Leftrightarrow\dfrac{x+1}{2}-1+\dfrac{x+5}{3}-2+\dfrac{x+11}{4}-3+\dfrac{x+19}{5}-4=0\)
\(\Leftrightarrow\dfrac{x-1}{2}+\dfrac{x-1}{3}+\dfrac{x-1}{4}+\dfrac{x-1}{5}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}\right)=0\)
mà \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}>0\)
\(\Rightarrow x-1=0\Leftrightarrow x=1\)
Vậy \(x=1\)
