:D mong được giải tuần 1 lần trong cuộc đời 

\(a,\dfrac{1}{2}-\dfrac{1}{3}-\left(-\dfrac{5}{4}\right)=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{5}{4}=\dfrac{1\times6-1\times4+5\times3}{12}=\dfrac{6-4+15}{12}=\dfrac{17}{12}\\ b,\dfrac{5}{4}-\dfrac{1}{2}-\dfrac{7}{8}=\dfrac{5\times2-1\times4-7}{8}=\dfrac{10-4-7}{8}=-\dfrac{1}{8}\\ c,\dfrac{1}{5}-\dfrac{1}{2}+\dfrac{9}{10}=\dfrac{1\times2-1\times5+9}{10}=\dfrac{2-5+9}{10}=\dfrac{6}{10}=\dfrac{3}{5}\\ d,\dfrac{5}{4}-\dfrac{1}{3}+\dfrac{7}{6}=\dfrac{5\times3-1\times4+7\times2}{12}=\dfrac{15-4+14}{12}=\dfrac{25}{12}\)

\(\left(3x-1\right)^2=44\\ \Rightarrow\left[{}\begin{matrix}3x-1=2\sqrt{11}\\3x-1=-2\sqrt{11}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{2\sqrt{11}+1}{3}\\x=\dfrac{-2\sqrt{11}+1}{3}\end{matrix}\right.\)

3

Vậy ABCD là hình vuông hả 

:v ơ sửa đề nhanh vậy 

\(\left(3x-1\right)^2=144\\\Rightarrow\left[{}\begin{matrix}\left(3x-1\right)^2=12^2\\\left(3x-1\right)^2=\left(-12\right)^2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}3x-1=12\\3x-1=-12\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}3x=13\\3x=-11\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{13}{3}\\x=-\dfrac{11}{3}\end{matrix}\right.\)