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 = \(\dfrac{27}{116}\).\(\dfrac{1}{3}\) = \(\dfrac{9}{116}\)

\(\dfrac{3}{2}+\dfrac{3}{14}.\dfrac{7}{9}-0,75\)

\(=\dfrac{3}{2}+\dfrac{1}{6}-0,75\)

\(=\dfrac{5}{3}-0,75\)

= \(\dfrac{11}{12}\)

\(A=\dfrac{5^2.2^{19}.3^{11}+2^{14}.3^{10}.5^2}{2^{17}.3^{12}.5^4-3^{11}.2^{18}.5^3}\)
\(=\dfrac{5^2.2^{14}.3^{10}\left(2^5.3+1\right)}{2^{17}.3^{11}.5^3\left(3.5-2\right)}\)
\(=\dfrac{97}{2^3.3.5.13}\)
\(=\dfrac{97}{1560}\)
#Đang Bận Thở

Lời giải:

\(A=\frac{5^2.2^{14}.3^{10}(2^5.3+1)}{2^{17}.3^{11}.5^3(3.5-2)}\)

\(=\frac{5^2.2^{14}.3^{10}.97}{2^{17}.3^{11}.5^3.13}=\frac{97}{5.2^3.3.13}=\frac{97}{1560}\)

A = \(\dfrac{3^2}{1.4}+\dfrac{3^2}{4.7}+...+\dfrac{3^2}{196.199}\)

A = \(\dfrac{3.3}{1.4}+\dfrac{3.3}{4.7}+...+\dfrac{3.3}{196.199}\)

A = \(3.\dfrac{3}{1.4}+3.\dfrac{3}{4.7}+...+3.\dfrac{3}{196.199}\)

A = \(3\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{196.199}\right)\)

A = \(3\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{196}-\dfrac{1}{199}\right)\)

A = \(3\left(1-\dfrac{1}{199}\right)\) = \(3.\dfrac{198}{199}\) = \(\dfrac{594}{199}\)

Bạn chú ý đăng đúng môn học nhé!

đây là toán mà ,ko phải vật lý

Lời giải:
PT $\Leftrightarrow (\frac{x+1}{2022}+1)+(\frac{x+2}{2021}+1)+...+(\frac{x+23}{2000}+1)=0$

$\Leftrightarrow \frac{x+2023}{2022}+\frac{x+2023}{2021}+...+\frac{x+2023}{2000}=0$

$\Leftrightarrow (x+2023)(\frac{1}{2022}+\frac{1}{2021}+...+\frac{1}{2000})=0$
Dễ thấy tổng trong () luôn dương 

$\Rightarrow x+2023=0$

$\Leftrightarrow x=-2023$

Tổng số kẹo mà bạn có là : \(12:\dfrac{2}{5}=12\times\dfrac{5}{2}=30\left(cái\right)\)

Vậy bạn có 30 cái 

do những số đó bé hơn 1 nên cộng lại vẫn bé hơn 1

  A =  \(\dfrac{1}{3}\) +    \(\dfrac{1}{6}\) +  \(\dfrac{1}{10}\)  + \(\dfrac{1}{15}\) + ..+ \(\dfrac{1}{55}\)+ \(\dfrac{1}{66}\)

A  = 2  \(\times\) ( \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\)  + \(\dfrac{1}{20}\) + \(\dfrac{1}{30}\) +...+ \(\dfrac{1}{110}\) + \(\dfrac{1}{132}\))

A  = 2 \(\times\) ( \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) +  \(\dfrac{1}{4.5}\)+ \(\dfrac{1}{5.6}\) +...+ \(\dfrac{1}{10.11}\)+ \(\dfrac{1}{11.12}\))

A = 2 \(\times\) ( \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) +...+ \(\dfrac{1}{10}\) - \(\dfrac{1}{11}\)+ \(\dfrac{1}{11}\) - \(\dfrac{1}{12}\))

A = 2 \(\times\) ( \(\dfrac{1}{2}\) - \(\dfrac{1}{12}\))

A = 1 - \(\dfrac{1}{6}\) < 1

Vậy A = \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + \(\dfrac{1}{15}\) + ...+ \(\dfrac{1}{55}\)+ \(\dfrac{1}{66}\) < 1