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\(\dfrac{m}{n}=\dfrac{7}{1\cdot4}+\dfrac{7}{4\cdot7}+...+\dfrac{7}{37\cdot40}\)
\(=\dfrac{7}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{37\cdot40}\right)\)
\(=\dfrac{7}{3}\left(1-\dfrac{1}{40}\right)\)
\(=\dfrac{7}{3}\cdot\dfrac{39}{40}=\dfrac{91}{40}\)
\(\Leftrightarrow\left(m,n\right)=\left(91;40\right)\)
Suy ra: S=91+40=131
a, 21.52.17 = 2.25.17 = 50.17 = 850
b, 22 + 23 + 24 = 4 + 8 + 16 = 28
c, 25.3 + 24:8 + 50: 52
= 32.3 + 16:8 + 50:25
=96 + 2 + 2
= 100
d, 112 - 102 - 32
= 121 - 100 - 9
= 21 - 9
= 12
e, 13 + 23 + 33 + 43 + 53
= ( 1+ 2+3+4+5)2
= 152
= 225
3/3. (3/1.4 + 3/4.7 + 3/7.10 + .... + 3/67.70)=3/3. (1-1/4+1/4-1/7+1/7-1/10+ .... + 1/67-1/70)= 3/3. (1-1/70)=3/3. 69/70 = 207/210=69/70
\(a,\dfrac{\left(-11\right).3+7.\left(-11\right)}{\left(-15\right).22}=\dfrac{\left(-11\right).\left(3+7\right)}{\left(-15\right).22}=\dfrac{-11.10}{-15.22}=\dfrac{-11.2.5}{-11.3.5+2.11}=\dfrac{-1}{-3}=\dfrac{1}{3}\\ \dfrac{25.11}{15.37-15.48}\\ =\dfrac{5.5.11}{15.\left(37-48\right)}=\dfrac{5.5.11}{15.\left(-11\right)}=\dfrac{5.5.11}{3.5.\left(-11\right)}=\dfrac{5}{-3}=-\dfrac{5}{3}\\ \dfrac{-2.3.5^2}{3^2.5^3}=\dfrac{-2}{3.5}=-\dfrac{2}{15}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\)
\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)(ĐPCM)
a: \(=\dfrac{6\left(12-7\right)}{60}=\dfrac{6\cdot5}{6\cdot10}=\dfrac{5}{10}=\dfrac{1}{2}\)
b: \(=\dfrac{35\cdot\left(18-1\right)}{34\cdot7}=5\cdot\dfrac{1}{2}=\dfrac{5}{2}\)
c: \(=\dfrac{42\left(1-11\right)}{21\cdot\left(-15\right)}=2\cdot\dfrac{-10}{-15}=2\cdot\dfrac{2}{3}=\dfrac{4}{3}\)
d: \(=\dfrac{2^5\cdot3^2}{2^5\cdot3^3}=\dfrac{1}{3}\)
e: \(=\dfrac{-36\cdot14}{27\left(14+7\right)}=\dfrac{-36}{27}\cdot\dfrac{14}{21}=\dfrac{-4}{3}\cdot\dfrac{2}{3}=-\dfrac{8}{9}\)
\(A=\dfrac{7}{1.9}+\dfrac{7}{9.17}+\dfrac{7}{17.25}+...+\dfrac{7}{81.89}\)
\(\dfrac{8}{7}A=\dfrac{8}{1.9}+\dfrac{8}{9.17}+\dfrac{8}{17.25}+...+\dfrac{8}{81.89}\)
\(\dfrac{8}{7}A=1-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{25}+...+\dfrac{1}{81}-\dfrac{1}{89}\)
\(\dfrac{8}{7}A=1-\dfrac{1}{89}=\dfrac{88}{89}\Rightarrow A=\dfrac{88}{89}:\dfrac{8}{7}=\dfrac{77}{89}\)
\(B=\dfrac{5^2}{1.4}+\dfrac{3^2}{4.7}+\dfrac{3^2}{7.10}+...+\dfrac{3^2}{37.40}\)
\(B=\dfrac{25}{1.4}+\dfrac{9}{4.7}+\dfrac{9}{7.10}+...+\dfrac{9}{37.40}\)
\(\dfrac{1}{3}B=\dfrac{25}{12}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{37.40}\)
\(\dfrac{1}{3}B=\dfrac{25}{12}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{37}-\dfrac{1}{40}\)
\(\dfrac{1}{3}B=\dfrac{25}{12}+\dfrac{1}{4}-\dfrac{1}{40}=\dfrac{277}{120}\Rightarrow B=\dfrac{277}{120}:\dfrac{1}{3}=\dfrac{277}{40}\)
\(A=\dfrac{7}{1.9}+\dfrac{7}{9.17}+\dfrac{7}{17.25}+...+\dfrac{7}{81.89}\)
\(=7\left(\dfrac{8}{1.9}+\dfrac{8}{9.17}+\dfrac{8}{17.25}+...+\dfrac{8}{81.89}\right)\)
\(=7\left(1-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{25}+\dfrac{1}{25}+...+\dfrac{1}{81}-\dfrac{1}{89}\right)\)
\(=7.\left(1-\dfrac{1}{89}\right)=7.\dfrac{88}{89}=\dfrac{616}{89}\)