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Gọi tổng là A
⇒ A = \(\dfrac{1}{28}+\dfrac{1}{70}+\dfrac{1}{130}+\dfrac{1}{208}+...+\dfrac{1}{3190}\)
⇒ 3A = \(3\left(\dfrac{1}{28}+\dfrac{1}{70}+\dfrac{1}{130}+\dfrac{1}{208}+...+\dfrac{1}{3190}\right)\)
⇒ 3A = \(\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+\dfrac{3}{13.16}+...+\dfrac{3}{55.58}\)
⇒ 3A = \(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}+...+\dfrac{1}{55}-\dfrac{1}{58}\)
⇒ 3A = \(\dfrac{1}{4}-\dfrac{1}{58}\) = \(\dfrac{29}{116}-\dfrac{2}{116}\) = \(\dfrac{27}{116}\)
⇒ A = \(\dfrac{27}{116}\): 3 = \(\dfrac{27}{116}\).\(\dfrac{1}{3}\) = \(\dfrac{9}{116}\)
a) 19 60 < 20 60 = 30 90 < 31 90
b) 15 23 > 14 23 = 70 115 > 70 117
\(A=\frac{7}{3\times13}+\frac{7}{13\times23}+...+\frac{7}{53\times63}\)
\(A=\frac{7}{10}.\left[\left(\frac{1}{3}-\frac{1}{13}\right)+\left(\frac{1}{13}-\frac{1}{23}\right)+....+\left(\frac{1}{53}-\frac{1}{63}\right)\right]\)
\(A=\frac{7}{10}.\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+....+\frac{1}{53}-\frac{1}{63}\right)\)
\(A=\frac{7}{10}.\left(\frac{1}{3}-\frac{1}{63}\right)\)
\(A=\frac{7}{10}.\frac{20}{63}\)
\(A=\frac{2}{9}\)
\(2040-\left[10\left(4^3-56\right):2^3+2^3\right].2022^0\)
\(=2040-\left[10.8:2^3+2^3\right].1\)
\(=2040-\left(10+2^3\right)\)
\(=2040-18=2022\)
2040−[10(43−56):23+23].202202040−[10(43−56):23+23].20220
=2040−[10.8:23+23].1=2040−[10.8:23+23].1
=2040−(10+23)=2040−(10+23)
=2040−18=2022
A=7*(1/3*13+1/13*23+1/23*33+1/33*43+1/43*53+1/53*63)
A=7/10(1/3-1/13+1/13-1/23+1/23-1/33+1/33-1/43+1/43-1/53+1/53-1/63)
A=7/10*(1/3-1/63)
A=7/10*20/63
A=2/9
125 . 5x + 5x+1 = 3250
125 . 5x + 5x . 5 = 3250
5x . ( 125 + 5 ) = 3250
5x . 130 = 3250
5x = 3250 : 130
5x = 25
5x = 52
=> x = 2
\(\left(3^{250}:3^{190}\right)-2^3-4^3\)
\(=3^{60}-8-64=3^{60}-72\)
P/s : Không ra được số nào gọn hơn hả bạn
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