
\(0,2.\frac{15}{36}-\left(\frac{2}{5}+\frac{2}{3}\right):1\frac{1}{5}\)
13) Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. a) \(\frac{-3}{7}+\frac{15}{26}-\left(\frac{2}{13}-\frac{3}{7}\right)=\frac{-3}{7}+\frac{15}{26}-\frac{2}{13}+\frac{3}{7}=\left(\frac{-3}{7}+\frac{3}{7}\right)+\left(\frac{15}{26}-\frac{2}{13}\right)\) \(=\frac{15-4}{26}=\frac{11}{26}\) c) \(\frac{-11}{23}.\frac{6}{7}+\frac{8}{7}.\frac{-11}{23}-\frac{1}{23}=\frac{-11}{23}.\left(\frac{6}{7}+\frac{8}{7}\right)-\frac{1}{23}\) \(=\frac{-11}{23}.2-\frac{1}{23}=\frac{-22-1}{23}=\frac{-23}{23}=-1\) a. \(\frac{-3}{7}+\frac{15}{26}-\left(\frac{2}{13}-\frac{3}{7}\right)=\frac{-3}{7}+\frac{15}{26}-\frac{2}{13}+\frac{3}{7}=\frac{15}{13.2}-\frac{2}{13}=\frac{15}{13.2}-\frac{2.2}{13.2}=\frac{15-4}{26}=\frac{11}{26}\) C. \(\frac{-11}{23}.\frac{6}{7}+\frac{8}{7}.\frac{-11}{23}-\frac{1}{23}=\frac{1}{23}.\left(-11.\frac{6}{7}-11.\frac{8}{7}-1\right)=\frac{1}{23}.\left(-22-1\right)=\frac{1}{23}.\left(-23\right)=-1\) \(1)A=\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{9}-\frac{2}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{9}-\frac{4}{11}}\) \(=\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}\) \(=\frac{2}{4}=\frac{1}{2}\) \(2)B=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}\) \(=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}.\frac{4.4}{4.5}\) \(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}\) \(=\frac{1.2.3.4}{2.3.4.5}=\frac{1}{5}\) \(3)C=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}\) \(=\frac{2.2.3.3.4.4.5.5}{1.3.2.4.3.5.4.6}\) \(=\frac{2.5}{1.6}=\frac{2.5}{1.3.2}=\frac{5}{3}\) \(4)D=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right)\left(\frac{1}{5}-\frac{1}{6}-\frac{1}{30}\right)\) \(=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right)\left(\frac{6}{30}-\frac{5}{30}-\frac{1}{30}\right)\) \(=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right).0=0\) \(5)M=8\frac{2}{7}-\left(3\frac{4}{9}+3\frac{9}{7}\right)\) \(N=\left(10\frac{2}{9}+2\frac{3}{5}\right)-6\frac{2}{9}\) \(=\frac{58}{7}-\left(\frac{31}{9}+\frac{30}{7}\right)\) \(=\left(\frac{92}{9}+\frac{13}{5}\right)-\frac{56}{9}\) \(=\frac{58}{7}-\left(\frac{217}{63}+\frac{270}{63}\right)\) \(=\left(\frac{460}{45}+\frac{117}{45}\right)-\frac{280}{45}\) \(=\frac{58}{7}-\frac{487}{63}\) \(=\frac{577}{45}-\frac{280}{45}\) \(=\frac{522}{63}-\frac{487}{63}=\frac{5}{9}\) \(=\frac{33}{5}\) \(P=M-N\) \(\Rightarrow P=\frac{5}{9}-\frac{33}{5}\) \(\Rightarrow P=\frac{25}{45}-\frac{297}{45}\) \(\Rightarrow P=\frac{-272}{45}\) Vậy P = \(\frac{-272}{45}\) \(6)E=10101\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{3.7.11.13.37}\right)\) \(=\frac{5}{11}+\frac{5}{22}-\left(10101.\frac{4}{111111}\right)\) \(=\frac{10}{22}+\frac{5}{22}-\frac{4}{11}\) \(=\frac{15}{22}-\frac{8}{22}=\frac{7}{22}\) \(7)F=\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}+\frac{2}{7}-\frac{2}{13}}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{256}+\frac{3}{64}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\) \(=\frac{1\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{13}\right)}{2\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{13}\right)}.\frac{3\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{256}+\frac{1}{64}\right)}{1\left(1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}\right)}+\frac{5}{8}\) \(=\frac{1}{2}.\frac{3\left(\frac{16}{64}-\frac{4}{64}+\frac{1}{64}-\frac{1}{256}\right)}{1\left(\frac{64}{64}-\frac{16}{64}+\frac{4}{64}-\frac{1}{64}\right)}+\frac{5}{8}\) \(=\frac{1}{2}.\frac{3\left(\frac{13}{64}-\frac{1}{256}\right)}{1.\frac{51}{64}}+\frac{5}{8}\) \(=\frac{1}{2}.\frac{3\left(\frac{52}{256}-\frac{1}{256}\right)}{\frac{51}{64}}+\frac{5}{8}\) \(=\frac{1}{2}.\frac{3\left(\frac{51}{256}\right)}{\frac{51}{64}}+\frac{5}{8}\) \(=\frac{1}{2}.\frac{\frac{153}{256}}{\frac{51}{64}}+\frac{5}{8}\) \(=\frac{1}{2}.\frac{153}{256}:\frac{51}{64}+\frac{5}{8}\) \(=\frac{1}{2}.\frac{3}{4}+\frac{5}{8}\) \(=\frac{3}{8}+\frac{5}{8}=1\) Xin lỗi tớ đã làm hết buổi tối mà chỉ có 7 bài mong bạn thông cảm cho mình nhé ! \(A=\left(\frac{3}{8}+\frac{-3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\) \(A=\left(\frac{3}{8}+\frac{-6}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\) \(A=\left(\frac{-3}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\) \(A=\left(\frac{-36}{24}+\frac{56}{24}\right):\frac{5}{6}+\frac{1}{2}\) \(A=\frac{5}{6}:\frac{5}{6}+\frac{1}{2}\) \(A=\frac{5}{6}\times\frac{6}{5}+\frac{1}{2}\) \(A=1+\frac{1}{2}\) \(A=\frac{1}{1}+\frac{1}{2}=\frac{2}{2}+\frac{1}{2}\) \(A=\frac{3}{2}\) \(a)\frac{5}{6}+\left(\frac{-1}{2}\right)+\frac{3}{4}\) \(=\frac{1}{3}+\frac{3}{4}\) \(=\frac{13}{12}\) \(b)\left(0,75-\frac{1}{3}\right):\frac{7}{15}\) \(=\left(\frac{3}{4}-\frac{1}{3}\right).\frac{15}{7}\) \(=\frac{5}{12}.\frac{15}{7}\) \(=\frac{25}{28}\) \(c)\frac{7}{12}-\frac{3}{4}.\frac{5}{6}\) \(=\frac{7}{12}-\frac{5}{8}\) \(=\frac{-1}{24}\) \(d)\left(2\frac{1}{3}+1\frac{3}{4}\right).\frac{12}{13}\) \(=\left(\frac{7}{3}+\frac{7}{4}\right).\frac{12}{13}\) \(=\frac{49}{12}.\frac{12}{13}\) \(=\frac{49}{13}\) a)\(\frac{5}{6}+\frac{-1}{2}+\frac{3}{4}=\frac{10}{12}-\frac{6}{12}+\frac{9}{12}=\frac{10-6+9}{12}=\frac{13}{12}\) b)\(\left\{0,75-\frac{1}{3}\right\}:\frac{7}{15}=\left\{\frac{3}{4}-\frac{1}{3}\right\}.\frac{15}{7}=\left\{\frac{9}{12}-\frac{4}{12}\right\}.\frac{15}{7}=\frac{5}{12}.\frac{15}{7}=\frac{75}{84}\) c)\(\frac{7}{12}-\frac{3}{4}.\frac{5}{6}=\frac{7}{12}-\frac{5}{8}=\frac{14}{24}-\frac{15}{24}=\frac{-1}{24}\) d)\(\left\{2\frac{1}{3}+1\frac{3}{4}\right\}.\frac{12}{13}=\left\{\frac{7}{3}+\frac{7}{4}\right\}.\frac{12}{13}=\left\{\frac{28}{12}+\frac{21}{12}\right\}.\frac{12}{13}=\frac{39}{12}.\frac{12}{13}=3\)