a) Hình như nhầm đề thì phải :v
\(P=\dfrac{\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}}{\dfrac{5}{12}+1-\dfrac{6}{11}}\)
\(=\dfrac{\dfrac{5}{12}+\dfrac{5}{11}}{\dfrac{5}{12}+\dfrac{5}{11}}=1\)
b) \(Q=\dfrac{0,125-\dfrac{1}{5}+\dfrac{1}{7}}{0,375-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0,2}{\dfrac{3}{4}+0,5-\dfrac{3}{10}}\)
\(Q=\dfrac{0,125-\dfrac{1}{5}+\dfrac{1}{7}}{3\left(0,125-\dfrac{1}{5}+\dfrac{1}{7}\right)}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0,2}{\dfrac{3}{4}+0,5-\dfrac{3}{10}}\)
\(Q=\dfrac{1}{3}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0,2}{\dfrac{3}{4}+0,5-\dfrac{3}{10}}\)
\(Q=\dfrac{1}{3}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{2}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{3}\right)}=\dfrac{1}{3}+\dfrac{1}{\dfrac{3}{2}}\)
\(Q=\dfrac{1}{3}+\dfrac{2}{3}=1\)
a,\(P=\dfrac{\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}}{\dfrac{5}{12}+1-\dfrac{7}{11}}=\dfrac{\left(\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}\right).132}{\left(\dfrac{5}{12}+1-\dfrac{7}{11}\right).132}=\dfrac{88-33+60}{55+132-84}=\dfrac{115}{103}\)
b, Ta có : 0,125 = \(\dfrac{1}{8}\) ; 0,375 = \(\dfrac{3}{8}\) ; 0,2 = \(\dfrac{1}{5}\) ; 0,5 = \(\dfrac{3}{6}\)
\(Q=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{\dfrac{3}{8}-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{4}+\dfrac{3}{6}-\dfrac{3}{10}}\)
\(Q=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{3\cdot\left(\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}\right)}+\dfrac{2\cdot\left(\dfrac{1}{4}+\dfrac{1}{6}-\dfrac{1}{10}\right)}{3\cdot\left(\dfrac{1}{4}+\dfrac{1}{6}-\dfrac{1}{10}\right)}\)
\(Q=\dfrac{1}{3}+\dfrac{2}{3}=1\)