A = 6771 x (8771 - 1) = 6771 x 8771 - 6771

B = (6771 - 1) x 8771 = 6771 x 8771 - 8771

Vì 6771 < 8771 nên 6771 x 8771 - 6771 > 6771 x 8771 - 8771 => A > B

bài làm

A = 6771 x (8771 - 1)

= 6771 x 8771 - 6771

B = (6771 - 1) x 8771

= 6771 x 8771 - 8771

Do 6771 < 8771 nên

6771 x 8771 - 6771 > 6771 x 8771 - 8771

=> A > B

=> Đpcm

hok tốt

\(a+692\cdot3-51=1965\)

\(a+2076-51=1965\)

\(a+2076=2016\)

\(a=-60\)

\(\frac{b}{5}\cdot8+6769=6771\)

\(\frac{8b}{5}=2\)

\(8b=10\)

\(b=\frac{5}{4}\)

\(c-61582+55=31073\)

\(c-61582=31018\)

\(c=92600\)

\(x-\left(6954\cdot65-6854+6754-1528\right)=102548\)

\(x-450382=102548\)

\(x=552930\)

\(\frac{1100}{y}\cdot101=2525\)

\(\frac{1100}{y}=25\)

\(y=44\)

\(1987536-58\cdot z=1978082\)

\(58z=9454\)

\(z=163\)

a: 43/52>26/52=1/2=60/120

b: 17/68=1/4<1/3=35/105<35/103

c: \(\dfrac{2018\cdot2019-1}{2018\cdot2019}=1-\dfrac{1}{2018\cdot2019}\)

\(\dfrac{2019\cdot2020-1}{2019\cdot2020}=1-\dfrac{1}{2019\cdot2020}\)

2018*2019<2019*2020

=>-1/2018*2019<-1/2019*2020

=>\(\dfrac{2018\cdot2019-1}{2018\cdot2019}< \dfrac{2019\cdot2020-1}{2019\cdot2020}\)

ko hiểu

\(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+199\right)+\left(x+200\right)=200\)

\(\Leftrightarrow200x+\left(1+2+3+..+200\right)=200\)

\(\Leftrightarrow200x+\dfrac{\left(200+1\right).200}{2}=200\)

\(\Leftrightarrow200x+20100=200\)

\(\Leftrightarrow200x=-19900\)

\(\Leftrightarrow x=-99,5\)

a) Ta có: \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+199\right)+\left(x+200\right)=200\)

\(\Leftrightarrow x+x+1+x+2+...+x+199+x+200=200\)

\(\Leftrightarrow201x+20100=200\)

\(\Leftrightarrow201x=-19900\)

hay \(x=\dfrac{-19900}{201}\)

Vậy: \(x=\dfrac{-19900}{201}\)

\(B=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)...\left(1-\frac{1}{81}\right)\left(1-\frac{1}{100}\right)\)

\(B=\frac{3}{4}\cdot\frac{8}{9}\cdot...\cdot\frac{80}{81}\cdot\frac{99}{100}\)

\(B=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot...\cdot\frac{8.10}{9.9}\cdot\frac{9.11}{10.10}\)

\(B=\frac{\left(1\cdot2\cdot...\cdot8\cdot9\right).\left(3\cdot4\cdot...\cdot10\cdot11\right)}{\left(2\cdot3\cdot..\cdot9\cdot10\right).\left(2\cdot3\cdot...\cdot9\cdot10\right)}\)

\(B=\frac{1\cdot2\cdot...\cdot8\cdot9}{2\cdot3\cdot...\cdot9\cdot10}\cdot\frac{3\cdot4\cdot...\cdot10\cdot11}{2\cdot3\cdot...\cdot9\cdot10}\)

\(B=\frac{1}{10}\cdot\frac{11}{2}=\frac{11}{20}\)

Vì 20 < 21 nên 11/20 > 11/21

Vậy ..... 

bạn vào link này nè:https://olm.vn/hoi-dap/question/980572.html

\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)\)

\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{18}{19}\cdot\frac{19}{20}\)

\(A=\frac{1}{20}\)

Vì 20 < 21 nên 1/20 > 1/21 

Vậy ............

\(\frac{-216}{-217}=\frac{216}{217}>\frac{-15}{16}\)

\(\frac{-216}{-217}=\frac{216}{217}>0\left(1\right)\)

\(\frac{-15}{16}< 0\left(2\right)\)

\(\text{Từ (1) và (2) }\Rightarrow\frac{-216}{-217}>\frac{-15}{16}\)

a) Có vẻ đề o đúng lắm . Theo mình o phải là 11/11 mà 1/11

Ta có \(\frac{1}{11}>\frac{1}{12}>\frac{1}{13}>...>\frac{1}{19}>\frac{1}{20}\)

\(\Rightarrow\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{19}+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{10}{20}=\frac{1}{2}\)

hay \(S>\frac{1}{2}\)

b)Ta có 1998 x 1999 + 3997=(2000-2) x 1999 +3997 = 2000 x 1999 - 2 x 1999 +3997 = 1999 x 2000 -3998 +3997 =1999 x 2000 -1

< 1999 x 2000 +2 

=> 1999 x 2000 +2 / 1998 x 1999 +3997 > 1 hay M>1

Thanks you . Mình sẽ kết bạn với cậu nhé

Ta có : x< y hay a/m <b/m \(\Rightarrow\)a<b

So sánh X, Y ,Z ta chuyển chúng cùng mẫu : 2

mà a<b

Suy ra : a+a<b+a

Hay 2a < a+b

Suy ra x<z (1)

Mà a<b 

Suy ra a+b<b+b

Hay a+b< 2b

Suy ra Z<y (2)

Từ (1) và (2) kết luận x < z<y

Tích nha Bạn