Lời giải:

$=(-7+|13|)-(13-|-7|-25)-(25+|-10|-9)$

$=(-7+13)-(13-7-25)-(25+10-9)$

$=-7+13-13+7+25-25-10+9$

$=(-7+7)+(13-13)+(25-25)-10+9=-10+9=-1$

\(\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{4}\right)\left(1+\dfrac{1}{5}\right)+...+\left(1+\dfrac{1}{2021}\right)\)

= \(\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot\dfrac{6}{5}\cdot...\cdot\dfrac{2022}{2021}\)

= \(\dfrac{2022}{3}\)

Đề thiếu. Bạn coi lại.

A=20/1.21+20/2.22+...+20/80.100

=1-1/21+1/2-1/22+...+1/80-1/100

=(1+1/2+...+1/80)-(1/21+1/22+...+1/100)

80B=80/1.81+80/2.82+...+8/20.100

=1-1/81+1/2-1/82+...+1/20-1/100

=(1+1/2+...+1/20)-(1/81+1/82+...+1/100)

=(1+1/2+1/3+...+1/20+1/21+1/22+...+1/80)-(1/21+1/22+...1/80+1/81+1/82+...1/100)

=>20A=80B

=>A=4B

Ta có : +) A= 1/5 -1/7 +1/7 -1/12 +1/12 - 1/19 +1/19 - 1/28 +1/28 - 1/39 +1/30.40  ⇔ A=1/5 -1/39 +1/30.40

+) B= 2.(1/5.8 +1/8.11 +1/11.14 +1/14.17 + 1/17.20 ) 

⇔B=2. 1/3.(1/5 - 1/8 +1/8 - 1/11 +1/11- 1/14 +1/14 -1/17 +1/17 -1/20 )

⇔B=2/3.( 1/5-1/20 )  Ta luôn có :B luôn <1/5 - 1/20

Mà 1/5 -1/20 <1/5 -1/39 +1/30.40 =A 

⇒ A>B (dpcm) Tích mình với nha bn .

ok

\(\left(-\dfrac{1}{3}\right)^2\).18-3⁵:3³+(-5).4

= \(\dfrac{1}{9}\).18-3²+(-5).4

=2-9+(-20)

=-27

\(11.\left(3-x\right)=6.\left(5-x\right)\)

\(33-11x=30-6x\)

\(3=5x\)

\(x=\dfrac{3}{5}\)

11(3 - x) = (5 - x)6

33 - 11x = 30 - 6x

(33 - 11x) - (30 - 6x) = 0

33 - 11x - 30 + 6x = 0

33 - 30 - 11x + 6x = 0

3 - 5x = 0

5x = 0 + 3

5x = 3

x = 3: 5

x = 3/5

\(=\dfrac{5}{28}+\dfrac{5}{70}+\dfrac{5}{130}+...+\dfrac{5}{700}\)

\(=\dfrac{5}{4.7}+\dfrac{5}{7.10}+\dfrac{5}{10.13}+...+\dfrac{5}{25.28}\)

\(=\dfrac{5}{3}.\left(\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{3}{25.28}\right)\)

\(=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)

\(=\dfrac{5}{3}.\left(\dfrac{1}{4}-\dfrac{1}{28}\right)=\dfrac{5}{14}\)

5/14