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A=(1.2)(2.2)+(2.2)(3.2)+...+(50.2)(51.2)
A=1.2.4+2.3.4+...+50.51.4
A=4(1.2+2.3+...+50.51)
M= 1.2+2.3+...+50.51
3M=1.2.3+2.3.(4-1)+...+50.51.(52-49)
=1.2.3+2.3.4-1.2.3+...+50.51.52-49.50.51
= 50.51.52
=132600
=> M=44200
=> A=4M=176800
\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{98.100}\\ =\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{98}-\dfrac{1}{100}\\ =\dfrac{1}{2}-\dfrac{1}{100}\\ =\dfrac{49}{100}\)
\(B=2.4+4.6+6.8+...+98.100\)
\(B=2.\left(1.2\right)+2.\left(2.3\right)+2.\left(3.4\right)+...+2.\left(49.50\right)\)
\(B=2\left(1.2+2.3+3.4+....+49.50\right)\)
Đặt:
\(A=1.2+2.3+3.4+...+49.50\)
\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)
\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+49.50.51-48.49.50\)
\(3A=49.50.51\)
\(A=\dfrac{49.50.51}{3}=41650\)
\(B=2A=41650.2=83300\)
\(\frac{2.4+4.6+6.8+...+98.100}{1.2+2.3+3.4+...+49.50}=\frac{4.\left(1.2+2.3+3.4+...+49.50\right)}{1.2+2.3+3.4+...+49.50}=\frac{4}{1}=4\)
M = \(\dfrac{5}{2.4}\) + \(\dfrac{5}{4.6}\)+ \(\dfrac{5}{6.8}\)+ ...+ \(\dfrac{5}{96.98}\)+ \(\dfrac{5}{98.100}\)
M = \(\dfrac{5}{2}\).( \(\dfrac{2}{2.4}\) + \(\dfrac{2}{4.6}\)+ \(\dfrac{2}{6.8}\)+...+ \(\dfrac{2}{96.98}\)+ \(\dfrac{2}{98.100}\))
M = \(\dfrac{5}{2}\).( \(\dfrac{1}{2}-\dfrac{1}{4}\)+ \(\dfrac{1}{4}-\dfrac{1}{6}\)+ \(\dfrac{1}{6}\) - \(\dfrac{1}{8}\)+...+ \(\dfrac{1}{96}\)-\(\dfrac{1}{98}\)+ \(\dfrac{1}{98}\)-\(\dfrac{1}{100}\))
M = \(\dfrac{5}{2}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{100}\))
M = \(\dfrac{49}{40}\)
\(x\) \(\times\) M - 1 = \(\dfrac{20}{29}\)
\(x\) \(\times\) \(\dfrac{49}{40}\) = \(\dfrac{20}{29}\) + 1
\(x\) \(\times\) \(\dfrac{49}{40}\) = \(\dfrac{49}{29}\)
\(x\) = \(\dfrac{49}{29}\) : \(\dfrac{49}{40}\)
\(x\) = \(\dfrac{40}{29}\)
6A = 2 . 4 . 6 + 4.6.6 + 6.8.6 + ... + 98 . 100 . 6 = 2.4.6 + 4.6.(8-2) + ... + 98. 100 . (102 - 96)
= 2.4.6 +4.6.8 - 2.4.6 + .... + 98.100 . 102 - 96.98.100
= 98. 100 . 102
= 999600
Suy ra A = 166600
Vậy ______________________
=166600