Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{7}{10.11}+\frac{7}{11.12}+...+\frac{7}{69.70}\)
\(=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{69.70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(=7.\frac{3}{35}\)
\(=\frac{3}{5}\)
Chúc bạn học tốt !
\(\frac{7}{10\cdot11}+\frac{7}{11\cdot12}+...+\frac{7}{69\cdot70}\)
\(=7\cdot\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(=7\cdot\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(=7\cdot\frac{3}{35}\)
\(=\frac{3}{5}\)
\(B=\frac{7}{10.11}+\frac{7}{11.12}+...+\frac{7}{69.70}\)
\(B=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{69.70}\right)\)
\(B=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(B=7.\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(B=7.\left(\frac{7}{70}-\frac{1}{70}\right)\)
\(B=7.\frac{6}{70}\)
\(B=\frac{3}{5}\)
Chúc bạn học tốt !!!
\(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(=7.\frac{1}{10.11}+7.\frac{1}{11.12}+7.\frac{1}{12.13}+...+7.\frac{1}{69.70}\)
\(=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{70}\right)=7.\frac{3}{35}=\frac{3}{5}\)
\(A=7.\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(=\frac{6}{70}\)
\(=\frac{3}{35}\)
\(A=\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{26}{7}-\frac{13}{11}+\frac{13}{23}}+\frac{\frac{3}{5}+\frac{3}{13}-0,9}{\frac{7}{91}+0,2-\frac{3}{10}}\)
\(A=\frac{5.31-\frac{5.2}{7}-\frac{5}{11}+\frac{5}{23}}{13.31-\frac{13.2}{7}-\frac{13}{11}+\frac{13}{23}}+\frac{\frac{3}{5}+\frac{3}{13}-\frac{9}{10}}{\frac{1}{13}+\frac{1}{5}-\frac{3}{10}}\)
\(A=\frac{5.31-\frac{5.2}{7}-\frac{5}{11}+\frac{5}{23}}{13.31-\frac{13.2}{7}-\frac{13}{11}+\frac{13}{23}}+\frac{\frac{3}{5}+\frac{3}{13}-\frac{9}{10}}{\frac{1}{5}+\frac{1}{13}-\frac{3}{10}}\)
\(A=\frac{5}{13}+\frac{1}{3}=\frac{44}{13}\)
Bạn tham khảo nhé
Ta có :
\(A=\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{26}{7}-\frac{13}{11}+\frac{13}{23}}+\frac{\frac{3}{5}+\frac{3}{13}-0,9}{\frac{7}{91}+0,2-\frac{3}{10}}\)
\(A=\frac{5.31-5.\frac{2}{7}-5.\frac{1}{11}+5.\frac{1}{23}}{13.31-13.\frac{2}{7}-13.\frac{1}{11}+13.\frac{1}{23}}+\frac{3.\frac{1}{5}+3.\frac{1}{13}-3.\frac{3}{10}}{\frac{1}{13}+\frac{1}{5}-\frac{3}{10}}\)
\(A=\frac{5\left(31-\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}{13\left(31-\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}+\frac{3\left(\frac{1}{5}+\frac{1}{13}-\frac{3}{10}\right)}{\frac{1}{5}+\frac{1}{13}-\frac{3}{10}}\)
\(A=\frac{5}{13}+\frac{3}{1}=\frac{5}{13}+\frac{39}{13}=\frac{44}{13}\)
Vậy \(A=\frac{44}{13}\)
a) Ta có: BCNN(15,10) = 30 nên ta chọn mẫu số chung là 30
\(\frac{11}{15}+\frac{9}{10}=\frac{22}{30}+\frac{27}{30}=\frac{49}{30}\)
b) Ta có: BCNN(6,9,12) = 36 nên ta chọn mẫu số chung là 36
\(\frac{5}{6} + \frac{7}{9} + \frac{{11}}{{12}} = \frac{{30}}{{36}} + \frac{{28}}{{36}} + \frac{{33}}{{36}} = \frac{{91}}{{36}}\)
c) Ta có: BCNN(24,21) = 168 nên ta chọn mẫu số chung là 168
\(\frac{7}{{24}} - \frac{2}{{21}} = \frac{{49}}{{168}} - \frac{{16}}{{168}} = \frac{{33}}{{168}}=\frac{11}{56}\)
d) Ta có: BCNN(36,24) = 72 nên ta chọn mẫu số chung là 72
\(\frac{{11}}{{36}} - \frac{7}{{24}} = \frac{{22}}{{72}} - \frac{{21}}{{72}} = \frac{1}{{72}}\)
Ta có :
\(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}:\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
\(=\)\(\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}:\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\)
\(=\)\(\frac{2}{7}:\frac{2}{7}\)
\(=\)\(\frac{2}{7}.\frac{7}{2}\)
\(=\)\(1\)
Chúc bạn học tốt ~
\(=\frac{2-2+2}{7-7+7}:\frac{\frac{2}{6}-\frac{2}{8}+\frac{2}{10}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
\(=\frac{2}{7}:\frac{2-2+2}{7-7+7}\)
\(=\frac{2}{7}:\frac{2}{7}\)
\(=\frac{2}{7}.\frac{7}{2}\)
\(=\frac{2.7}{7.2}\)
\(=\frac{1.1}{1.1}\)
\(=\frac{1}{1}\)
\(=1\)
a) A=\(\frac{178}{179}+\frac{179}{180}+\frac{183}{181}\)
ta có :
\(A=\left(1-\frac{1}{179}\right)+\left(1-\frac{1}{180}\right)+\left(1+\frac{2}{181}\right)\)
\(\Rightarrow A=\left(1+1+1\right)-\left(\frac{1}{179}-\frac{1}{180}+\frac{2}{181}\right)\)
\(\Rightarrow A=3-\left(\frac{1}{179}-\frac{1}{180}+\frac{2}{181}\right)< 3\)
Vậy \(A< 3\)
a. Ta có :
\(\frac{178}{179}< 1\left(\frac{1}{179}\right)\)
\(\frac{179}{180}< 1\left(\frac{1}{180}\right)\)
\(\frac{183}{181}>1\left(\frac{3}{181}\right)\left(1\right)\)
Mà \(\frac{3}{181}>\frac{1}{179}+\frac{1}{180}\left(=\frac{359}{32220}< \frac{3}{181}\right)\left(2\right)\)
Từ \(\left(1\right)\&\left(2\right)\Rightarrow\frac{178}{179}+\frac{179}{180}+\frac{183}{181}< 1+1+1\)
Vậy \(A< 3\)