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\(\Rightarrow A=\frac{\frac{1}{2}-\frac{1}{5}+\frac{1}{7}}{\frac{3}{8}-\frac{3}{5}+\frac{3}{7}}+\frac{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}{\frac{3}{4}+\frac{1}{2}+\frac{3}{10}}\)
\(\Rightarrow A=\frac{\frac{1}{2}-\frac{1}{5}+\frac{1}{7}}{3.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{7}\right)}+\frac{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}{\frac{3}{2}.\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{5}\right)}\)
\(\Rightarrow A=\frac{1}{3}+\frac{2}{3}\)
\(\Rightarrow A=1\)
a) \(\frac{\left(17\frac{2}{9}-15\frac{2}{15}\right):5\frac{2}{9}}{\left(18+3,75\right):0,25}.25\%=\frac{\left(17+\frac{2}{9}-15-\frac{2}{15}\right):\frac{47}{9}}{\left(18+3,75\right).4}.\frac{1}{4}\)
\(=\frac{\left(2+\frac{4}{45}\right).\frac{9}{47}}{\left(18+3,75\right).16}=\frac{\frac{94}{45}.\frac{9}{47}}{288+60}=\frac{\frac{2}{5}}{348}=\frac{2}{5}.\frac{1}{348}=\frac{174}{5}=34,8\)
b) \(\frac{\frac{5}{3}-\frac{5}{7}+\frac{5}{9}}{\frac{10}{3}-\frac{10}{7}+\frac{10}{9}}=\frac{5\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{9}\right)}{10\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{9}\right)}=\frac{1}{2}\)
b) \(\frac{\frac{5}{3}-\frac{5}{7}+\frac{5}{9}}{\frac{10}{3}-\frac{10}{7}+\frac{10}{9}}=\frac{5.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{9}\right)}{10.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{9}\right)}=\frac{5}{10}=\frac{1}{2}\)
\(\begin{array}{l}B = \left( {\frac{{ - 3}}{{13}}} \right) + \frac{{16}}{{23}} + \left( {\frac{{ - 10}}{{13}}} \right) + \frac{5}{{11}} + \frac{7}{{23}}\\ = \left[ {\left( {\frac{{ - 3}}{{13}}} \right) + \left( {\frac{{ - 10}}{{13}}} \right)} \right] + \left[ {\frac{{16}}{{23}} + \frac{7}{{23}}} \right] + \frac{5}{{11}}\\ = - 1 + 1 + \frac{5}{{11}}\\ = \frac{5}{{11}}\end{array}\)
\(\frac{\left(\frac{53}{4}-\frac{59}{27}-\frac{65}{6}\right).\frac{5751}{25}+\frac{187}{4}}{\left(\frac{10}{7}+\frac{10}{3}\right):\left(\frac{37}{3}-\frac{100}{7}\right)}=\frac{\left(\frac{4293}{324}-\frac{708}{324}-\frac{3510}{324}\right).\frac{5751}{25}+\frac{187}{4}}{\left(\frac{30}{21}+\frac{70}{21}\right):\left(\frac{259}{21}-\frac{300}{21}\right)}=\frac{\frac{25}{108}.\frac{5751}{25}+\frac{187}{4}}{\frac{100}{21}:\left(-\frac{41}{21}\right)}\)=\(\frac{\frac{213}{4}+\frac{187}{4}}{-\frac{100}{41}}=100:\left(-\frac{100}{4}\right)=-4\)
\(30+\frac{14}{5}:\left(\frac{24}{150}-\frac{270}{150}-\frac{25}{150}\right)=30+\frac{14}{5}:\left(-\frac{271}{150}\right)=30+\left(-\frac{420}{271}\right)=\frac{7710}{271}\)
đặt \(A=\frac{7}{10}+\frac{7}{10^2}+\frac{7}{10^3}+\frac{7}{10^4}\)
\(A=7.\left(\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+\frac{1}{10^4}\right)\)
Lại đặt \(B=\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+\frac{1}{10^4}\)
\(10B=1+\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}\)
\(10B-B=\left(1+\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}\right)-\left(\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+\frac{1}{10^4}\right)\)
\(9B=1-\frac{1}{10^4}\)
\(\Rightarrow B=\frac{1-\frac{1}{10^4}}{9}\)
\(\Rightarrow A=7.\frac{1-\frac{1}{10^4}}{9}=\frac{7.\left(1-\frac{1}{10^4}\right)}{9}\)
Nhưng có vô hạn số hạng thì sao bạn