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\(M=\frac{7}{4}\times\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}+\frac{3333}{5656}\right)\)
\(M=\frac{7}{4}\times\left(\frac{11}{4}+\frac{33}{20}+\frac{11}{10}+\frac{11}{14}+\frac{33}{56}\right)\)
\(M=\frac{7}{4}\times\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}+\frac{33}{56}\right)\)
\(M=\frac{7}{4}\times\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}+\frac{33}{7.8}\right)\)
\(M=\frac{7}{4}\times\left[33\cdot\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\right]\)
\(M=\frac{7}{4}\times\left[33\times\left(\frac{1}{3}-\frac{1}{8}\right)\right]\)
\(M=\frac{7}{4}\times\left(33\times\frac{5}{24}\right)=\frac{7}{4}\times\frac{55}{8}=\frac{385}{32}\)
A = \(\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{4242}+\frac{3333}{3030}\right)\)
A = \(\frac{7}{4}.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
A = \(\frac{7}{4}.33.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
A = \(\frac{7}{4}.33.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
A = \(\frac{7}{4}.33.\left(\frac{1}{3}-\frac{1}{7}\right)=\frac{7}{4}.33.\frac{4}{21}\)
=> A = \(\frac{1}{3}.33\)
=> A = 11
\(\frac{63.303}{505.36}=\frac{9.7.101.3}{101.5.9.4}=\frac{21}{20}\)
\(\frac{270.450+135.505.2}{2+4+6+...+16+18}\)
\(=\frac{270.450+135.2.505}{2.\left(1+2+3+...+8+9\right)}\)
\(=\frac{270.450+270.505}{2.\left(1+9\right).9:2}\)
\(=\frac{270.\left(450+505\right)}{10.9}=\frac{270.955}{90}=3.955=2865\)
Gợi ý: rút gọn cho 101 rồi đặt 5 ra ngoài làm thừa số chung thì sẽ tìm ra kết quả là \(\frac{25}{24}\)
=25/24 đấy