\(\frac{1}{a}+\frac{1}{b}=2-\frac{1}{c}\Rightarrow\left(\frac{1}{a}+\frac{1}{b}\right)^2=\left(2-\frac{1}{c}\right)^2\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{ab}=\frac{1}{c^2}-\frac{4}{c}+4\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}=-\frac{4}{c}\Rightarrow\frac{1}{c}=-\frac{1}{4a^2}-\frac{1}{4b^2}\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}-\frac{1}{4a^2}-\frac{1}{4b^2}=2\Rightarrow\frac{1}{4a^2}-\frac{1}{a}+1+\frac{1}{4b^2}-\frac{1}{b}+1=0\)
\(\Rightarrow\left(\frac{1}{2a}-1\right)^2+\left(\frac{1}{2b}-1\right)^2=0\Rightarrow\left\{{}\begin{matrix}\frac{1}{2a}=1\\\frac{1}{2b}=1\end{matrix}\right.\)
\(\Rightarrow a=b=\frac{1}{2}\Rightarrow\frac{1}{c}=2-\left(\frac{1}{a}+\frac{1}{b}\right)=-2\Rightarrow c=-\frac{1}{2}\)
\(\Rightarrow\) Đề bài câu a sai, đề đúng phải là \(\left(\frac{1}{a}+\frac{1}{c}\right)^2+\left(\frac{1}{b}+\frac{1}{c}\right)^2=0\)
\(Q=\left(\frac{1}{2}+\frac{1}{2}-1\right)^{2019}=0\)
