S =    2 + 2² + 2³ + 2⁴ +......+2⁶⁰

2S =        2² +2³ +2⁴+..........+2⁶⁰ +2⁶¹

2S - S =                2⁶¹ - 2

S =                          2⁶¹ - 2

b, P = 3 + 3² + 3⁴ +..........+3²⁰²² 

  3P =        3² +3⁴ +...........+3²⁰²² +3²⁰²³

3P - P =   3²⁰²³ - 3

2P =           3²⁰²³ - 3

P = (3²⁰²³ - 3): 2

3^x + 4² = 19⁶ : ( 19³ . 19² ) - 3.1²⁰²²

<=> 3^x + `16 = 19⁶ : 19⁵ - 3

<=> 3^x + 16 = 19 - 3

<=> 3^x + 16 = 16

<=> 3^x = 0

=> Không tồn tại x

Xác suất bi vàng là

\(\dfrac{6}{4+6}=60\%\)

Xác suất bi vàng là:

    \(\dfrac{6}{4+6}=60\%\)

a) \(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{999.1000}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{999}-\dfrac{1}{1000}\)

\(A=1-\dfrac{1}{1000}=\dfrac{999}{1000}\)

b) \(B=\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+....+\dfrac{2}{999.1000}\)

\(B=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{999}-\dfrac{1}{1000}\right)\)

\(B=2.\dfrac{999}{1000}=\dfrac{999}{500}\)

c) \(C=\dfrac{3}{1.3}+\dfrac{3}{3.5}+\dfrac{3}{5.7}+....+\dfrac{3}{999.1001}\)

\(C=3\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{999}-\dfrac{1}{1001}\right)\)

\(C=3.\dfrac{1000}{1001}=\dfrac{3000}{1001}\)

a, \(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{999.1000}\)

    \(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{999}-\dfrac{1}{1000}\)

    \(A=1-\dfrac{1}{1000}=\dfrac{999}{1000}\)

b, \(B=\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{999.1000}\)

    \(B=2.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{999.1000}\right)\)

    \(B=2A=2.\dfrac{999}{1000}=\dfrac{999}{500}\)

c, \(C=\dfrac{3}{1.3}+\dfrac{3}{3.5}+\dfrac{3}{5.7}+...+\dfrac{3}{999.1001}\)

    \(C=\dfrac{3}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{999.1001}\right)\)

    \(C=\dfrac{3}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{999}-\dfrac{1}{1001}\right)\)

    \(C=\dfrac{3}{2}.\left(1-\dfrac{1}{1001}\right)=\dfrac{3}{2}.\dfrac{1000}{1001}=\dfrac{1500}{1001}\)

Theo quan hẹ trường độ như trên

=> 2³ móc đơn = 2⁴ móc kép

Móc đơn là a , móc kép là b

Ta có : 2³ a = 2⁴b 

\(\Leftrightarrow\dfrac{a}{b}=\dfrac{2^4}{2^3}\)

\(\Leftrightarrow\dfrac{a}{b}=2\)

\(\Leftrightarrow a=2b\)

Hay móc đơn = 2 móc kép 

\(\left(2a+3\right)^2=\left(2a\right)^2+2.2a.3+3^2=4a^2+12a+9\)

\(\dfrac{1}{3}+\dfrac{5}{6}+\dfrac{1}{12}+\dfrac{1}{24}+\dfrac{1}{4}\\ =\dfrac{8}{24}+\dfrac{20}{24}+\dfrac{2}{24}+\dfrac{1}{24}+\dfrac{6}{24}\\ =\dfrac{8+20+2+1+6}{24}\\ =\dfrac{37}{24}\)

\(\dfrac{1}{3}+\dfrac{5}{6}+\dfrac{1}{12}+\dfrac{1}{24}+\dfrac{1}{4}\)

\(=\dfrac{8+5\times4+2+1+6}{24}\)

\(=\dfrac{37}{24}\)

\(2⋮\left(x+1\right)=>x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\\ \begin{matrix}x+1&1&-1&2&-2\\x&0&-2&1&-3\end{matrix}=>x\in\left\{0;-2;1;-3\right\}\)

2 ⋮ (x+1) ⇔ (x+1) ϵ Ư(2) ⇔ (x+1) ϵ { -2; -1; 1; 2}

⇔ x ϵ {-3; -2; 0; 1}