199* 200+198 : 200 * 402 -404 = 39 793,98

c) - 34 + 54 - ( 157 + 43 ) = - 180

d ) 1+ 99 -(2 + 98) + 3 + 97 - (4 + 96) + .....+ (- 100 ) = - 100

a) 154 + 46 - ( 12 + 88 ) = 100

b) 2011 + 1989 - ( 596 + 404 ) = 3000

SAi rồi ! phải là 2666600 Mới đúng

Muốn biết thì bấm vào Đúng 0

404 < 440        200 + 5 < 250

765 > 756       440 - 40 > 399

899 < 900       500 + 50 + 5 = 555

\(H=0,25\times\left(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{19\times20}\right)\)

\(=0,25\times\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)\)

\(=0,25\times\left(1-\dfrac{1}{20}\right)=0,25\times\dfrac{19}{20}=\dfrac{19}{80}\)

\(H=\dfrac{0.25}{1\cdot2}+\dfrac{0.25}{2\cdot3}+...+\dfrac{0.25}{199\cdot200}\)

\(=\dfrac{1}{4}\cdot\dfrac{199}{200}=\dfrac{199}{800}\)

Lời giải:
Gọi tổng trên là $A$

$A=2(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{199.200})$

$=2(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{200-199}{199.200})$

$=2(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{199}-\frac{1}{200})$

$=2(\frac{1}{2}-\frac{1}{200})=1-\frac{1}{100}=\frac{99}{100}$

A = \(\dfrac{2}{2.3}\) + \(\dfrac{2}{3.4}\) + \(\dfrac{2}{4.5}\) + ... + \(\dfrac{2}{199.200}\)

A = 2. (\(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + ... + \(\dfrac{1}{199.200}\))

A = 2.(\(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + ... + \(\dfrac{1}{199}\) - \(\dfrac{1}{200}\))

A = 2.(\(\dfrac{1}{2}\) - \(\dfrac{1}{200}\))

A = 2. \(\dfrac{99}{200}\)

A = \(\dfrac{99}{100}\)