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e) \(E=0,7.2\frac{2}{3}.20.0,375.\frac{5}{28}\)
\(=\left(0,7.20\right)\left(2\frac{2}{3}.0,375\right)\frac{5}{28}\)
\(=14.1.\frac{5}{28}\)
\(=14.\frac{5}{28}\)
\(=\frac{5}{2}\)
f) \(F=\left(9,75.21\frac{3}{7}+\frac{39}{4}.18\frac{4}{7}\right)\frac{15}{78}\)
\(=\left(\frac{39}{4}.21\frac{3}{7}+\frac{39}{4}.18\frac{4}{7}\right)\frac{15}{78}\)
\(=\frac{39}{4}\left(21\frac{3}{7}+18\frac{4}{7}\right)\frac{15}{78}\)
\(=\frac{39}{4}.40.\frac{15}{78}\)
\(=390.\frac{15}{78}\)
\(=78\)
Chúc bạn học tốt môn Toán!!!
Mình ko ghi đề đâu
\(A=49\frac{8}{23}-5\frac{7}{32}-14\frac{8}{23}=\left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}=35-5\frac{7}{32}=29\frac{25}{32}\)
\(B=71\frac{38}{45}-43\frac{8}{45}+1\frac{17}{57}=\left(71\frac{38}{45}-43\frac{8}{45}\right)+1\frac{17}{57}=28\frac{2}{3}+1\frac{17}{57}=29\frac{55}{57}\)
\(C=-\frac{3}{7}.\left(\frac{5}{9}+\frac{4}{9}\right)+2\frac{3}{7}=-\frac{3}{7}.1+2\frac{3}{7}=-\frac{3}{7}+2\frac{3}{7}=2\)
Cây a, bạn nhân cả 2 vế với 3
Lấy vế nhân với 3 trừ đi ban đầu tất cả chia 2
b) Tính như bình thường
Câu c hình như sai đề
Bài đây tính nhanh nhé ミ★ʟuғғʏ☆мũ☆ʀơм★彡 chứ không phải quy đồng lên đâu :)
a) \(A=49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)
\(A=49\frac{8}{23}-5\frac{7}{32}-14\frac{8}{23}\)
\(A=\left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}=35-5\frac{7}{32}=35-\frac{167}{32}=\frac{953}{32}\)
b) \(B=\frac{-3}{7}\cdot\frac{5}{9}+\frac{4}{9}:\frac{-7}{3}+2\frac{3}{7}\)
\(B=\frac{-3}{7}\cdot\frac{5}{9}+\frac{4}{9}\cdot\frac{-3}{7}+2\frac{3}{7}\)
\(B=\frac{-3}{7}\left(\frac{5}{9}+\frac{4}{9}\right)+2\frac{3}{7}\)
\(B=\frac{-3}{7}+\frac{17}{7}=\frac{14}{7}=2\)
c) \(C=\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right)\cdot\frac{4}{5}\)
\(C=\left[\left(19\frac{5}{8}-13\frac{1}{4}\right):\frac{7}{12}\right]\cdot\frac{4}{5}\)
\(C=\left[\left(19\frac{5}{8}-13\frac{2}{8}\right):\frac{7}{12}\right]\cdot\frac{4}{5}\)
\(C=6\frac{3}{8}\cdot\frac{4}{5}=\frac{51}{8}\cdot\frac{4}{5}=\frac{51}{2}\cdot\frac{1}{5}=\frac{51}{10}\)
d) \(D=\frac{54\cdot107-53}{53\cdot107+54}=\frac{\left(53+1\right)\cdot107-53}{53\cdot107+54}=\frac{53\cdot107+107-53}{53\cdot107+54}=\frac{53\cdot107+54}{53\cdot107+54}=1\)
\(=\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}:\frac{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right):\frac{919191}{808080}\)
\(=\left(\frac{1}{2}:4\right):\frac{919191}{808080}=\frac{1}{8}\cdot\frac{808080}{919191}=\frac{10}{91}\)
Bài giải
\(\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}}\text{ : }\frac{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right)\text{ : }\frac{919191}{808080}\)
\(=\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}\text{ : }\frac{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right)\text{ : }\frac{91}{80}\)
\(=\left(\frac{1}{2}\text{ : }\frac{4}{1}\right)\text{ : }\frac{91}{80}=\frac{1}{8}\text{ : }\frac{91}{80}=\frac{10}{91}\)
a,\(\frac{3}{7}.\frac{4}{9}+\frac{3}{7}.\frac{5}{9}+\frac{5}{14}\)
\(=\frac{3}{7}.\left(\frac{4}{9}+\frac{5}{9}\right)+\frac{5}{14}\)
\(=\frac{3}{7}.1+\frac{5}{14}\)
\(=\frac{3}{7}+\frac{5}{14}=\frac{6}{14}+\frac{5}{14}=\frac{11}{14}\)
b,\(\frac{-11}{23}.\frac{6}{7}+\frac{8}{9}.\frac{-11}{23}-\frac{1}{23}\)
\(=\)\(\frac{-11}{23}.\left(\frac{6}{7}+\frac{8}{9}\right)-\frac{1}{23}\)
\(=\frac{-11}{23}.\frac{110}{63}-\frac{1}{23}\)
=\(\frac{-1210}{1449}\)-\(\frac{1}{23}\)
\(=\frac{-1273}{1449}\)
