Câu 1:
\(a,A=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\right)\cdot\dfrac{2019}{2018}\\ A=\left[\dfrac{2\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}-\dfrac{\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{2}\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}\right)}\right]\cdot\dfrac{2019}{2018}\\ A=\left(\dfrac{2}{7}-\dfrac{2}{7}\right)\cdot\dfrac{2019}{2018}=0\)
\(b,\Leftrightarrow\left\{{}\begin{matrix}\left|2x-1\right|=0\\\left(x+2y\right)^{24}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\x+2y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{1}{4}\end{matrix}\right.\)
\(c,\dfrac{x}{2}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{6}\Leftrightarrow\dfrac{x}{10}=\dfrac{y}{20}=\dfrac{z}{24}=\dfrac{2x+3y+4z}{20+60+96}=\dfrac{3x+4y+5z}{30+80+120}\\ \Leftrightarrow A=\dfrac{20+60+96}{30+80+120}=\dfrac{176}{230}=\dfrac{88}{115}\)