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C= 101 . 35 . 10101.23 / 35.10101 . 23. 101 = 1
Do D = 3535/3534 > 1 => C<D
\(\frac{121212+12121212-1212}{363636+36363636-3636}\)
\(=\frac{12.10101+12.1010101-12.101}{36.10101+36.1010101-36.101}\)
\(=\frac{12\left(10101+1010101-101\right)}{36\left(10101+1010101-101\right)}\)
\(=\frac{12}{12.3}\)
\(=\frac{1}{3}\)
Ta có: \(\frac{121212+12121212-1212}{363636+36363636-3636}\)
=\(\frac{12\times\left(10101+1010101-101\right)}{36\times\left(10101+1010101-101\right)}\)
=\(\frac{12}{36}\)
=\(\frac{1}{3}\)
= \(\frac{12}{15}\) +\(\frac{12}{35}\)+\(\frac{12}{63}\)+\(\frac{12}{99}\)
= 12 x (\(\frac{1}{15}\)+\(\frac{1}{35}\)+\(\frac{1}{63}\)+\(\frac{1}{99}\))
= 12 x ( \(\frac{1}{3x5}\)+\(\frac{1}{5x7}\)+\(\frac{1}{7x9}\)+\(\frac{1}{9x11}\))
= 12 x \(\frac{1}{2}\) x ( \(\frac{1}{3}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{9}\)+\(\frac{1}{9}\)-\(\frac{1}{11}\))
= 6 x ( \(\frac{1}{3}\) - \(\frac{1}{11}\))
= 6 x \(\frac{8}{33}\)
= \(\frac{48}{33}\)=\(\frac{16}{11}\)
Nhớ tk nha
\(M=\frac{7}{4}\times\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}+\frac{3333}{5656}\right)\)
\(M=\frac{7}{4}\times\left(\frac{11}{4}+\frac{33}{20}+\frac{11}{10}+\frac{11}{14}+\frac{33}{56}\right)\)
\(M=\frac{7}{4}\times\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}+\frac{33}{56}\right)\)
\(M=\frac{7}{4}\times\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}+\frac{33}{7.8}\right)\)
\(M=\frac{7}{4}\times\left[33\cdot\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\right]\)
\(M=\frac{7}{4}\times\left[33\times\left(\frac{1}{3}-\frac{1}{8}\right)\right]\)
\(M=\frac{7}{4}\times\left(33\times\frac{5}{24}\right)=\frac{7}{4}\times\frac{55}{8}=\frac{385}{32}\)
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
Gợi ý: rút gọn cho 101 rồi đặt 5 ra ngoài làm thừa số chung thì sẽ tìm ra kết quả là \(\frac{25}{24}\)
\(\frac{5454}{5757}-\frac{171717}{191919}=\frac{18.303}{19.303}-\frac{17.10101}{19.10101}=\frac{18}{19}-\frac{17}{19}=\frac{1}{19}\)
Ta có :
\(\frac{5454}{5757}=\frac{54.101}{57.101}=\frac{54}{57}\); \(\frac{171717}{191919}=\frac{17.10101}{19.10101}=\frac{17}{19}\)
Vậy \(\frac{54}{57}-\frac{17}{19}=\frac{1}{19}\)
\(\frac{3636}{3535}+\frac{5454}{4242}\)
\(=\frac{36}{35}+\frac{54}{42}\)
= \(\frac{36}{35}+\frac{9}{7}\)
= \(\frac{36}{35}+\frac{45}{35}\)
= \(\frac{81}{35}\)
81/35