\(x^2-2x+y^2-8y+17=0\)

\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2-8y+16\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2+\left(y-4\right)^2=0\)

Vì \(\left(x-1\right)^2\ge0\)

\(\left(y-4\right)^2\ge0\)

\(\Rightarrow\left(x-1\right)^2=\left(y-4\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y-4\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=4\end{matrix}\right.\)

Vậy phương trình có nghiệm \(\left(x,y\right)=\left(1;4\right)\)

c) Ta có: \(\dfrac{2-x}{2017}-1=\dfrac{1-x}{2018}-\dfrac{x}{2019}\)

\(\Leftrightarrow\dfrac{2-x}{2017}+1=\dfrac{1-x}{2018}+1-\dfrac{x}{2019}+1\)

\(\Leftrightarrow\dfrac{2-x}{2017}+1=\left(\dfrac{1-x}{2018}+1\right)-\left(\dfrac{x}{2019}-1\right)\)

\(\Leftrightarrow\dfrac{2-x+2017}{2017}=\dfrac{1-x+2018}{2018}-\dfrac{x-2019}{2019}\)

\(\Leftrightarrow\dfrac{2019-x}{2017}=\dfrac{2019-x}{2018}+\dfrac{2019-x}{2019}\)

\(\Leftrightarrow\dfrac{2019-x}{2017}-\dfrac{2019-x}{2018}-\dfrac{2019-x}{2019}=0\)

\(\Leftrightarrow\left(2019-x\right)\left(\dfrac{1}{2017}-\dfrac{1}{2018}-\dfrac{1}{2019}\right)=0\)

\(\Leftrightarrow2019-x=0\)(vì \(\dfrac{1}{2017}-\dfrac{1}{2018}-\dfrac{1}{2019}\ne0\))

\(\Leftrightarrow x=2019\)

Vậy nghiệm của phương trình là \(x=2019\)

\(\frac{92}{23}=\frac{4}{1}=\frac{4x6}{1x6}=\frac{24}{6}=>Sốcầnđiềnlà6\)

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