b: Ta có: \(\left(2+4+6+...+100\right)\cdot\left(36\cdot333-108\cdot111\right)\)

\(=\left(2+4+6+...+100\right)\cdot36\cdot111\cdot\left(3-3\right)\)

=0

Đặt A=\(\dfrac{1}{11}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{13}\)+...\(\dfrac{1}{69}\)+\(\dfrac{1}{70}\)

         =(\(\dfrac{1}{11}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{13}\)+...\(\dfrac{1}{19}\)+\(\dfrac{1}{20}\))+(\(\dfrac{1}{21}\)+\(\dfrac{1}{22}\)+\(\dfrac{1}{23}\)+...+\(\dfrac{1}{29}\)+\(\dfrac{1}{30}\))+(\(\dfrac{1}{31}\)+\(\dfrac{1}{32}\)+\(\dfrac{1}{33}\)+...+\(\dfrac{1}{59}\)+\(\dfrac{1}{60}\))+...+\(\dfrac{1}{70}\)

Nhận xét:

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

\(\dfrac{1}{21}\)+\(\dfrac{1}{22}\)+\(\dfrac{1}{23}\)+...+\(\dfrac{1}{29}\)+\(\dfrac{1}{30}\)>\(\dfrac{1}{30}\)+\(\dfrac{1}{30}\)+...+\(\dfrac{1}{30}\)=\(\dfrac{10}{30}\)=\(\dfrac{1}{3}\)

\(\dfrac{1}{31}\)+\(\dfrac{1}{32}\)+\(\dfrac{1}{33}\)+...+\(\dfrac{1}{59}\)+\(\dfrac{1}{60}\)>\(\dfrac{1}{60}\)+\(\dfrac{1}{60}\)+...+\(\dfrac{1}{60}\)=\(\dfrac{30}{60}\)=\(\dfrac{1}{2}\)

=>A>\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{2}\)+\(\dfrac{1}{61}\)+...+\(\dfrac{1}{69}\)+\(\dfrac{1}{70}\)>\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{2}\)=\(\dfrac{4}{3}\)

=>A>\(\dfrac{4}{3}\)

Vậy: \(\dfrac{1}{11}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{13}\)+...+\(\dfrac{1}{69}\)+\(\dfrac{1}{70}\)>\(\dfrac{4}{3}\) (ĐPCM)

Thấy đúng cho 1 tick và 1 follow nha!

Chúc bạn học tốt!

`-1 5/27-(3x-7/9)^3=-24/27`

`-32/27-(3x-7/9)^3=-24/27`

`(3x-7/9)^3=-8/27`

`(3x-7/9)^3=(-2/3)^3`

`3x-7/9=-2/3`

`3x=1/9`

`x=1/27`

`=>(3x-7/9)^3=-32/27+24/27`

`(3x-7/9)^3=-8/27`

`(3x-7/9)^3=(-2/3)^3`

`=>3x-7/9=-2/3`

`=>3x=-2/3+7/9`

`=>3x=-1/9`

`=>x=-1/27`

\(=\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{50}+\dfrac{1}{51}-\dfrac{102}{51\cdot52}\)

\(=\dfrac{1}{2}+\dfrac{1}{51}-\dfrac{102}{51\cdot52}\)

\(=\dfrac{1}{2}+\dfrac{52-102}{51\cdot52}=\dfrac{1}{2}+\dfrac{-50}{51\cdot52}=\dfrac{319}{663}\)

Bài 2:

a: \(x-\dfrac{1}{2}=\dfrac{7}{13}\cdot\dfrac{13}{28}\)

=>\(x-\dfrac{1}{2}=\dfrac{7}{28}=\dfrac{1}{4}\)

=>\(x=\dfrac{1}{4}+\dfrac{1}{2}=\dfrac{3}{4}\)

b: \(\dfrac{x}{15}=\dfrac{-3}{11}\cdot\dfrac{77}{36}\)

=>\(\dfrac{x}{15}=\dfrac{-3}{36}\cdot\dfrac{77}{11}=7\cdot\dfrac{-1}{12}=-\dfrac{7}{12}\)

=>\(x=-\dfrac{7}{12}\cdot15=-\dfrac{105}{12}=-\dfrac{35}{4}\)

c: \(x:\dfrac{15}{11}=\dfrac{-3}{12}:8\)

=>\(x:\dfrac{15}{11}=-\dfrac{1}{4}:8=-\dfrac{1}{32}\)

=>\(x=-\dfrac{1}{32}\cdot\dfrac{15}{11}=\dfrac{-15}{352}\)

Bài 1:

a: \(\dfrac{-12}{25}\cdot\dfrac{10}{9}=\dfrac{-12}{9}\cdot\dfrac{10}{25}=\dfrac{-4}{3}\cdot\dfrac{2}{5}=\dfrac{-8}{15}\)

b: \(\dfrac{10}{21}-\dfrac{3}{8}\cdot\dfrac{4}{5}\)

\(=\dfrac{10}{21}-\dfrac{12}{40}\)

\(=\dfrac{10}{21}-\dfrac{3}{10}=\dfrac{100-63}{210}=\dfrac{37}{210}\)

c: \(\dfrac{28}{11}:\dfrac{21}{22}\cdot9=\dfrac{28}{11}\cdot\dfrac{22}{21}\cdot9\)

\(=\dfrac{28}{21}\cdot\dfrac{22}{11}\cdot9=\dfrac{4}{3}\cdot2\cdot9=\dfrac{4}{3}\cdot18=24\)

d: \(-\dfrac{10}{21}\cdot\left[\dfrac{9}{15}+\left(\dfrac{3}{5}\right)^2\right]\)

\(=\dfrac{-10}{21}\cdot\left[\dfrac{3}{5}+\dfrac{9}{25}\right]\)

\(=\dfrac{-10}{21}\cdot\dfrac{15+9}{25}\)

\(=\dfrac{-10}{25}\cdot\dfrac{24}{21}=\dfrac{-2}{5}\cdot\dfrac{8}{7}=\dfrac{-16}{35}\)

e: \(\left(\dfrac{2}{3}-\dfrac{1}{2}-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}-\dfrac{1}{7}\right)\)

\(=\left(\dfrac{1}{3}-\dfrac{1}{2}\right)\cdot\dfrac{28-7-4}{28}\)

\(=\dfrac{-1}{6}\cdot\dfrac{17}{28}=\dfrac{-17}{168}\)

f: \(\left(\dfrac{15}{21}:\dfrac{5}{7}\right):\left(\dfrac{6}{5}:2\right)\)

\(=\left(\dfrac{5}{7}\cdot\dfrac{7}{5}\right):\left(\dfrac{6}{5\cdot2}\right)\)

\(=1:\dfrac{6}{10}=\dfrac{10}{6}=\dfrac{5}{3}\)

a) \(1+2+3+...+48=\dfrac{\left(48+1\right)\left(\dfrac{48-1}{1}+1\right)}{2}=1176\)

b) \(2+4+6+...+212=\dfrac{\left(212+2\right)\left(\dfrac{212-2}{2}+1\right)}{2}=11342\)

a: =>4y+15/16=1

=>4y=1/16

=>y=1/64

b: =>10y+1/2+1/4+...+1/1024=1

=>10y+1023/1024=1

=>10y=1/1024

=>y=1/10240