Ta có:\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+..+\frac{2}{210}=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+..+\frac{2}{14.15}\)

\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+..+\frac{1}{14}-\frac{1}{15}\right)\)

\(=2.\left(\frac{1}{2}-\frac{1}{15}\right)\)

\(=1-\frac{2}{15}=\frac{15-2}{15}=\frac{13}{15}\)

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A = \(\dfrac{1}{2}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) + \(\dfrac{1}{30}\)+...+\(\dfrac{1}{182}\)+ 210

 A = \(\dfrac{1}{1\times2}\) + \(\dfrac{1}{2\times3}\)+\(\dfrac{1}{3\times4}\)+ \(\dfrac{1}{4\times5}\)+...+ \(\dfrac{1}{13\times14}\)+ 210

A = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)+ \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)+ \(\dfrac{1}{3}\)- \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)+...+\(\dfrac{1}{13}\) - \(\dfrac{1}{14}\) + 210

A = 1 - \(\dfrac{1}{14}\) + 210

A = 211 - \(\dfrac{1}{14}\)

A = \(\dfrac{2953}{14}\)

b: \(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)

\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+3^8\cdot2^{10}\cdot5}\)

\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{11}\cdot3^9}\)

\(=\dfrac{1}{2}\cdot\dfrac{-2}{3}=\dfrac{-1}{3}\)

Sửa đề câu a) phải là (-2020/2021)⁰

Xin lỗi mn

\(a,=\dfrac{3}{4}+\dfrac{15}{6}\cdot\dfrac{1}{5}-\dfrac{1}{36}\cdot9-\dfrac{2020}{2021}=\dfrac{3}{4}+\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{2020}{2021}=1-\dfrac{2020}{2021}=\dfrac{1}{2021}\\ b,=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2\cdot5}=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^9\cdot3^8\left(2+5\right)}=\dfrac{2\left(-2\right)}{3}=-\dfrac{4}{3}\)

+) \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)

\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)

\(\Rightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)\)

\(\Rightarrow A=1-\dfrac{1}{2^{10}}=\dfrac{2^{10}-1}{2^{10}}\)

Vậy \(A=\dfrac{2^{10}-1}{2^{10}}\)

+) \(F=\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+...+\dfrac{1}{190}\)

\(\Rightarrow\dfrac{1}{2}F=\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+...+\dfrac{1}{380}\)

\(=\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{19.20}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{20}\)

\(=\dfrac{1}{5}-\dfrac{1}{20}=\dfrac{3}{20}\Rightarrow F=\dfrac{3}{20}:\dfrac{1}{2}=\dfrac{3}{10}\)

Vậy \(F=\dfrac{3}{10}\)

+) \(G=\dfrac{12}{84}+\dfrac{12}{210}+\dfrac{12}{390}+...+\dfrac{12}{2100}\)

\(=\dfrac{4}{28}+\dfrac{4}{70}+\dfrac{4}{130}+...+\dfrac{4}{700}=\dfrac{4}{4.7}+\dfrac{4}{7.10}+...+\dfrac{4}{25.28}\)

\(=\dfrac{4}{3}.\left(\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{25.28}\right)\)

\(=\dfrac{4}{3}.\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)

\(=\dfrac{4}{3}.\left(\dfrac{1}{4}-\dfrac{1}{28}\right)=\dfrac{4}{3}.\dfrac{3}{14}=\dfrac{2}{7}\)

Vậy \(G=\dfrac{2}{7}\)

\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)

\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)

\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)\)

\(A=1-\dfrac{1}{2^{10}}=\dfrac{1024-1}{1024}=\dfrac{1023}{1024}\)

\(F=\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+...+\dfrac{1}{190}\)

\(=\dfrac{2}{30}+\dfrac{2}{42}+\dfrac{2}{56}+...+\dfrac{2}{380}\)

\(=\dfrac{2}{5.6}+\dfrac{2}{6.7}+\dfrac{2}{7.8}+...+\dfrac{2}{19.20}\)

\(=2\left(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{19.20}\right)\)

\(=2\left(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)\)

\(=2\left(\dfrac{1}{5}-\dfrac{1}{20}\right)=2.\dfrac{3}{20}=\dfrac{3}{10}\)

\(G=\dfrac{12}{84}+\dfrac{12}{210}+\dfrac{12}{390}+...+\dfrac{12}{2100}\)

\(=\dfrac{4}{28}+\dfrac{4}{70}+\dfrac{4}{130}+...+\dfrac{4}{700}\)

\(=\dfrac{4}{4.7}+\dfrac{4}{7.10}+\dfrac{4}{10.13}+...+\dfrac{4}{25.28}\)

\(=\dfrac{4}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)

\(=\dfrac{4}{3}\left(\dfrac{1}{4}-\dfrac{1}{28}\right)\)

\(=\dfrac{4}{3}.\dfrac{3}{14}=\dfrac{2}{7}\)

a) 39                    b) 20                        c) 29                       d) ...

Dãy số cách đều: 2;4;6;8...; b

Có số số hạng là: \(\frac{\left(b-2\right)}{2}+1=\frac{b}{2}\)

Có tổng là: \(\frac{1}{2}\cdot\left(2+b\right)\cdot\frac{b}{2}=210\Leftrightarrow b\left(b+2\right)=840=28\cdot30=28\cdot\left(28+2\right)\)

Do đó, b = 28

a) (125 - x) + 25 = 98

125 - x = 98 - 25

125 - x = 73

x = 125 - 73

x = 52

b) 100 : (x + 20) = 68 : 34

100 : (x + 20) = 2

x + 20 = 100 : 2

x + 20 = 50

x = 50 - 20

x = 30

c) (3x + 12) . 5 = 25 . 4 - 10

(3x + 12) . 5 = 100 - 10

(3x + 12) . 5 = 90

3x + 12 = 90 : 5

3x + 12 = 18

3x = 18 - 12

3x = 6

x = 6 : 2

x = 3

d) 210 : (2x - 3) - 20 = 10

210 : (2x - 3) = 10 + 20

210 : (2x - 3) = 30

2x - 3 = 210 : 30

2x - 3 = 70

2x = 70 + 3

2x = 73

x = 73/2

a, ( 125 - x ) + 25 = 98

⇒ 125 - x = 73

⇒ x = 52.

Vậy..

b, 100 : ( x + 20 ) = 68 : 34

⇒ 100 : ( x + 20 ) = 2

⇒ x + 20 = 50

⇒ x = 30

Vậy...

c, ( 3 . x + 12 ) . 5 = 25 . 4 - 10

⇒ ( 3 . x + 12 ) . 5 = 90

⇒ 3 .x + 12 = 18

⇒ 3x = 6

⇒ x = 2.

Vậy...

d, 210 : ( 2 .x - 3 ) - 20 = 10

⇒ ( 2 .x - 3 ) - 20 = 21

⇒ 2x - 3 = 41

⇒ 2x = 44

⇒ x = 22.

Vậy..

Cho mk chỉnh lại câu cuối:

 d, 210 : ( 2 .x - 3 ) - 20 = 10

⇒ 210 : ( 2 .x - 3 ) = 30

⇒ 2 . x - 3 = 7

⇒ 2x = 10

⇒ x = 5.

Vậy..