A = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\)+ \(\dfrac{1}{3.2}\)+ ....+ \(\dfrac{1}{50.51}\)

A = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)+ \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{2}\)+...+ \(\dfrac{1}{50}\) - \(\dfrac{1}{51}\)

A = \(\dfrac{1}{1}\) - \(\dfrac{1}{51}\)

A = \(\dfrac{50}{51}\)

Kp = 50/51

a)

`1/1-1/2`

`=2/2-1/2`

`=1/2`

b)

`1/(1*2)+1/(2*3)`

`=1/1-1/2+1/2-1/3`

`=1/1-1/3`

`=3/3-1/3`

`=2/3`

c)

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{1}-\dfrac{1}{100}\\ =\dfrac{99}{100}\)

d) 

\(\dfrac{3}{1\cdot2}+\dfrac{3}{2\cdot3}+...+\dfrac{3}{99\cdot100}\) đề phải như thế này chứ nhỉ?

\(=\dfrac{1\cdot3}{1\cdot2}+\dfrac{1\cdot3}{2\cdot3}+...+\dfrac{1\cdot3}{99\cdot100}\\ =3\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\\ =3\cdot\dfrac{99}{100}\\ =\dfrac{297}{100}\)

`@` `\text {Ans}`

`\downarrow`

`3^3 * x^2 - 2^4 * x^2 = 8^2 * 5 - 4^2 * 3^2`

`=> x^2 . (3^3 - 2^4) = 2^6 . 5 - 2^4 . 3^2`

`=> x^2 . 11 = 2^4 . (2^2 . 5 - 3^2)`

`=> x^2 . 11 = 2^4 . 11`

`=> x^2 . 11 - 2^4 . 11 = 0`

`=> 11 . (x^2 - 16) = 0`

`=> x^2 - 16 = 0`

`=> x^2 = 16`

`=> x^2 = (+-4)^2`

`=> x = `\(\pm4\)

Vậy, `x \in`\(\left\{4;-4\right\}\)

_____

\(\left[\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{3}\right)^3\right]x+3^2\cdot2^2=4^2\cdot3\)

`=>`\(\left(\dfrac{1}{4}-\dfrac{1}{27}\right)x+\left(3\cdot2\right)^2=48\)

`=>`\(\dfrac{23}{108}\cdot x+6^2=48\)

`=>`\(\dfrac{23}{108}x=48-6^2\)

`=>`\(\dfrac{23}{108}x=48-36\)

`=>`\(\dfrac{23}{108}x=12\)

`=>`\(x=\dfrac{1296}{23}\)

Vậy, `x = `\(\dfrac{1296}{23}\)

\(3^3.x^2-2^4.x^2=8^2.5-4^3.3^2\)

\(\Leftrightarrow x^2\left(27-16\right)=2^6.5-2^6.9\)

\(\Leftrightarrow11x^2=2^6.\left(5-9\right)=-4.2^6=-2^8\)

\(\Leftrightarrow x^2=-\dfrac{2^6}{11}< 0\)

\(\Rightarrow x\in\varnothing\)

\(\left[\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{3}\right)^3\right]x+3^2.2^2=4^2.3\)

\(\Leftrightarrow\left(\dfrac{1}{4}-\dfrac{1}{27}\right)x+36=48\)

\(\Leftrightarrow\dfrac{23}{108}x=12\Leftrightarrow x=\dfrac{12.108}{23}=\dfrac{1296}{23}\)

a) Đặt A = 1.2 + 2.3 + ........ + (n-1)n

3A = 1.2.3 + 2.3.(4-1) + .... + (n-1)n[(n+1)-(n-2)]

3A = 1.2.3 + 2.3.4 - 1.2.3 + .... + (n-1)n(n+1) - (n-2)(n-1)n

3A = (1.2.3 - 1.2..3) + ... + (n-1)n(n+1)

A = \(\frac{\left(n-1\right)n\left(n+1\right)}{3}\)

b) Đặt B = 1² + 2² + ..... + n²

B = 1(2 - 1) + 2(3 - 1) + ..... + n[(n + 1) - 1]

B = 1.2 + 2.3 + .......... + n(n + 1) - (1+2+3+....+n)

B = A -  \(\frac{n\left(n+1\right)}{2}\)

5: \(u_n=5n\left(n\in N\right)\)

4: Ba số hạng đầu tiên là 1/2;1/3;1/4

3: Ba số hạng đầu tiên là 6;20;72

2C

1B

\(\dfrac{1}{2}\cdot\dfrac{3}{4}=\dfrac{3}{8}\\ \left(\dfrac{1}{2}\right)^3\cdot\left(\dfrac{3}{4}\right)^3=\left(\dfrac{1}{2}\cdot\dfrac{3}{4}\right)^3=\left(\dfrac{3}{8}\right)^3\\ \dfrac{3}{8}< \left(\dfrac{3}{8}\right)^3\Rightarrow\dfrac{1}{2}\cdot\dfrac{3}{4}< \left(\dfrac{1}{2}\right)^3\cdot\left(\dfrac{3}{4}\right)^3\)

`1)(x-3/4)xx2/3+1/3=4/3`

`(x-3/4)xx2/3=4/3-1/3=1`

`x-3/4=1:2/3=3/2`

`x=3/2+3/4=9/4`

Vậy `x=9/4`

`2)3/4:(x-1)+1/2=2`

`3/4:(x-1)=2-1/2=3/2`

`x-1=3/4:3/2`

`x-1=1/2`

`x=1+1/2=3/2`

Vậy `x=3/2`