a) \(x^2-10x+21=0\Rightarrow\left(x-3\right)\left(x-7\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=7\end{matrix}\right.\)

b) ĐK: \(x\ge\dfrac{1}{3}\)

\(\sqrt{3x-1}=2\Rightarrow3x-1=4\Rightarrow3x=5\Rightarrow x=\dfrac{5}{3}\left(TM\right)\)

c) \(\left\{{}\begin{matrix}x^2+y^2=5\\2\left(x+y\right)+xy=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-2xy=5\\xy=-2\left(x+y\right)\end{matrix}\right.\)

\(\Rightarrow\left(x+y\right)^2+4\left(x+y\right)=5\Rightarrow\left(x+y\right)^2+4\left(x+y\right)-5=0\)

\(\Rightarrow\left[{}\begin{matrix}x+y=1\\x+y=-5\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x+y=1\\xy=-2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=1-x\\x\left(1-x\right)+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=1-x\\x^2-x-2=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}y=1-x\\\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2;y=-1\\x=-1;y=2\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}x+y=-5\\xy=10\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=-5-x\\x\left(-5-x\right)-10=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=-5-x\\x^2+5x+10=0\left(L\right)\end{matrix}\right.\)

Vậy PT có 2 nghiệm: \(\left(x;y\right)=\left(2;-1\right);\left(x;y\right)=\left(-1;2\right)\)

\(\dfrac{5}{8}-\dfrac{4}{28}=\dfrac{35}{56}-\dfrac{8}{56}=\dfrac{27}{56}\)