2\(\dfrac{43}{46}\) = 2,93478

A = \(\dfrac{1}{10}\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{21}\) + ... + \(\dfrac{1}{120}\)

A = \(\dfrac{2}{2}\).(\(\dfrac{1}{10}\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{21}\) + ... + \(\dfrac{1}{120}\))

A = \(2\).(\(\dfrac{1}{20}\) + \(\dfrac{1}{30}\) + \(\dfrac{1}{42}\)... + \(\dfrac{1}{240}\))

A = 2.(\(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\) + ... + \(\dfrac{1}{15.16}\))

A = 2.(\(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\) + ... + \(\dfrac{1}{15}\) - \(\dfrac{1}{16}\))

A = 2.(\(\dfrac{1}{4}\) - \(\dfrac{1}{16}\))

A = 2.\(\dfrac{3}{16}\)

A = \(\dfrac{3}{8}\) 

\(\dfrac{-3}{8}\cdot16\cdot\dfrac{8}{17}-0,375\cdot7\cdot\dfrac{9}{17}\)

\(=-\dfrac{3}{8}\cdot\dfrac{128}{17}-\dfrac{3}{8}\cdot\dfrac{63}{17}\)

\(=-\dfrac{3}{8}\left(\dfrac{128}{17}+\dfrac{63}{17}\right)=-\dfrac{3}{8}\cdot\dfrac{191}{17}=\dfrac{-573}{136}\)

- \(\dfrac{3}{8}\).16.\(\dfrac{8}{17}\) - 0,375.7\(\dfrac{9}{17}\)

Đề như này phải không em?

Bài 8:

\(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\)

\(=\dfrac{1}{3}-\dfrac{1}{99}=\dfrac{33}{99}-\dfrac{1}{99}=\dfrac{32}{99}\)

Bài 6:

a: 

b: I là trung điểm của MN

=>\(MI=\dfrac{MN}{2}=\dfrac{7}{2}=3,5\left(cm\right)\)

Nhanh giúp mình vơi

200 chia 50

a: SỐ tiền lãi anh Duy nhận được sau 1 năm là:

\(200\cdot10^6\cdot5,6\%=11200000\left(đồng\right)\)

Số tiền cả gốc lẫn lãi anh Duy nhận được là:

\(200000000+11200000=211200000\left(đồng\right)\)

b: Số tiền lãi năm thứ hai anh Duy nhận được là:

\(211200000\cdot8\%=16896000\left(đồng\right)\)

Tổng số tiền anh Duy nhận được là:

\(211200000+16896000=228096000\left(đồng\right)\)

Ta có:

\(2020+2021+2022< 2021+2022+2023\)

\(\Rightarrow\dfrac{2020+2021+2022}{2021+2022+2023}< 1\)

\(\Rightarrow Q< 1\)

Lại có: \(2020.2>2021.1\Rightarrow\dfrac{2020}{2021}>\dfrac{1}{2}\)

\(2021.2>2022.1\Rightarrow\dfrac{2021}{2022}>\dfrac{1}{2}\)

\(2022.2>2023.1\Rightarrow\dfrac{2022}{2023}>\dfrac{1}{2}\)

\(\Rightarrow\dfrac{2020}{2021}+\dfrac{2021}{2022}+\dfrac{2022}{2023}>\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{3}{2}\)

\(\Rightarrow P>\dfrac{3}{2}>1\)

\(\Rightarrow P>Q\)

^{\(\dfrac{2020}{2021}\) > \(\dfrac{2020}{2021+2022+2023}\)}

^{\(\dfrac{2021}{2022}\) > \(\dfrac{2021}{2021+2022+2023}\)}

^{\(\dfrac{2022}{2023}\) > \(\dfrac{2022}{2021+2022+2023}\)}

^{Cộng vế với vế ta có: P = \(\dfrac{2020}{2021}\) + \(\dfrac{2021}{2022}\) + \(\dfrac{2022}{2023}\) > \(\dfrac{2020+2021+2022}{2021+2022+2023}\) = Q}

\(-\dfrac{2}{7}+\dfrac{5}{12}\cdot\dfrac{18}{35}\)

\(=\dfrac{-2}{7}+\dfrac{18}{12}\cdot\dfrac{5}{35}\)

\(=-\dfrac{2}{7}+\dfrac{3}{2}\cdot\dfrac{1}{7}=\dfrac{-2}{7}+\dfrac{3}{14}=\dfrac{-4+3}{14}=-\dfrac{1}{14}\)

-2/7 + 5/12 . 18/35

= -2/7 + 1/2 . 3/7

= -2/7 + 3/14

= -4/14 + 3/14

= 1/14

\(\dfrac{1}{2}x+\dfrac{2}{3}x-1=-3\dfrac{1}{3}\)

=>\(x\left(\dfrac{1}{2}+\dfrac{2}{3}\right)=-\dfrac{10}{3}+1\)

=>\(x\cdot\dfrac{7}{6}=\dfrac{-7}{3}\)

=>\(x=-\dfrac{7}{3}:\dfrac{7}{6}=-\dfrac{7}{3}\cdot\dfrac{6}{7}=-2\)