999 - 888 - 111 + 111 - 111 + 111 - 111

= 111 - 111 + 111 -111 + 111 - 111

= 0 + 111 - 111 + 111 - 111

= 111 - 111 + 111 - 111

= 0 + 111 - 111

= 111 - 111

= 0

Đáp số: 0

k cho minh nha

noichung ai k minh thi minh k cho

$A=\dfrac12-\dfrac{2}{2^2}+\dfrac{3}{2^3}-\dfrac{4}{2^4}+\cdots+\dfrac{99}{2^{99}}-\dfrac{100}{2^{100}}$

Nhóm từng 2 số:

$A=\left(\dfrac12-\dfrac{2}{2^2}\right)+\left(\dfrac3{2^3}-\dfrac4{2^4}\right)+\cdots+\left(\dfrac{99}{2^{99}}-\dfrac{100}{2^{100}}\right)$

$=0+\dfrac18+\dfrac{2}{32}+\dfrac3{128}+\cdots+\dfrac{49}{2^{99}}$

$=\sum_{k=1}^{50}\dfrac{k-1}{2^{2k-1}}$

Ta có: $\dfrac{k-1}{2^{2k-1}}=\dfrac{2(k-1)}{4^k}$

Mà: $\sum_{k=1}^{\infty}\dfrac{k-1}{4^k}=\dfrac{1}{9}$

Nên: $A<2\cdot\dfrac19$ $=\dfrac29$

Vậy: $A<\dfrac29$

$4=1\cdot4,\quad28=4\cdot7,\quad70=7\cdot10,\ldots$

tức là: $E=\dfrac3{1\cdot4}+\dfrac3{4\cdot7}+\dfrac3{7\cdot10}+\cdots+\dfrac3{n(n+3)}$

Với $n=1,4,7,\ldots$.

Ta có: $\dfrac3{n(n+3)}=\dfrac1n-\dfrac1{n+3}$

Do đó: $E=\left(1-\dfrac14\right)+\left(\dfrac14-\dfrac17\right)+\left(\dfrac17-\dfrac1{10}\right)+\cdots+\left(\dfrac1n-\dfrac1{n+3}\right)$

$=1-\dfrac1{n+3}$

Vì: $\dfrac1{n+3}>0$ nên: $1-\dfrac1{n+3}<1$

A = 3/1 + 3/1+2 + 3/1+2+3 + 3/1+2+3+4 + ...+3/1+2+..+100

A = 3/1 + 3/3 + 3/6 + 3/10 +..+3/5050

A = 2/2 .( 3/1 + 3/3 + 3/6 + 3/10 +...+ 3/5050)

A = 6/2 + 6/6 + 6/12 + 6/20 +..+6/10100)

A = 6 .(1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 +.. +1/100.101)

A = 6. (1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ...+1/100 - 1/101)

A = 6 (1 - 1/101)

A = 6 . 100/101

A = 600/101

ai làm hộ mình đi mình k mà

a)

= 8/22 + 5/22

= 11/22

= 1/2

b)

= 24/8 - 3/8

= 21/8

c)

= 45/25

= 9/5

d)

= 3/7 x 1/4

= 3/28

:)

a) \(\dfrac{4}{11}+\dfrac{5}{22}=\dfrac{8}{22}+\dfrac{5}{22}=\dfrac{13}{22}\)

b) \(3-\dfrac{3}{8}=\dfrac{3}{1}-\dfrac{3}{8}=\dfrac{24}{8}-\dfrac{3}{8}=\dfrac{21}{8}\)

c) \(\dfrac{9}{25}\times5=\dfrac{45}{25}=\dfrac{9}{5}\)

d) \(\dfrac{3}{7}\div4=\dfrac{3}{28}\)

A=9954