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}

Theo tính chất của 2 đường thẳng song song : để Ax // By 

<=> 2 góc trong cùng phía bù nhau 

Hay \(\widehat{B\text{Ax}}+\widehat{ABy}=180^o\)

\(\Leftrightarrow a+3a=180^o\)

\(\Leftrightarrow4a=180^o\Leftrightarrow a=45^o\)

\(2⋮x=>x\inƯ\left(2\right)\\ Ư\left(2\right)=\left\{\pm1;\pm2\right\}\\ =>x\in\left\{\pm1;\pm2\right\}\)

theo bài ra ta có   với a,b ϵ N*               

     a =  40b + 18

a : 6 = ( 40b + 18) : 6  = (36b + 4b + 18) : 6

nếu b ⋮ 3 thì a ⋮6 nếu b \(⋮̸\) 3 thì a \(⋮̸\) 6

a : 7 = (40b + 18) : 7 = ( 35b + 14 + 5b + 4): 7

nếu 5b + 4 ⋮ 7 thì a ⋮ 7 , nếu 5b + 4 \(⋮̸\)7 thì a \(⋮̸\) 7              

\(A=\left(x+2\right)^2-\left(x+2\right)\left(x-2\right)+\left(x-4\right)^2\\ =x^2+4x+4-\left(x^2-4\right)+x^2-8x+16=x^2-4x+24\\ \cdot x=-2=>A=\left(-2\right)^2-4.\left(-2\right)+24=36\\ \cdot x=0=>A=0^2-4.0+24=24\\ \cdot x=2=>A=2^2-4.2+24=20\\ A=\left(x-2\right)^2+20>0\left(DPCM\right)\)