\(\frac{1}{4}+\frac{1}{3}:3.x=-5\)

\(\frac{1}{9}.x=-5-\frac{1}{4}\)

\(\frac{1}{9}.x=-\frac{21}{4}\)

\(x=-\frac{21}{4}:\frac{1}{9}\)

\(x=-\frac{189}{4}\)

\(\frac{1}{4}+\frac{1}{3}:3.x=-5\)

\(\frac{1}{4}+\frac{1}{9}x=-5\)

\(\frac{1}{9}x=-5-\frac{1}{4}\)

\(\frac{1}{9}x=\frac{-20}{4}-\frac{1}{4}\)

\(\frac{1}{9}x=\frac{-21}{4}\)

\(x=\frac{-21}{4}:\frac{1}{9}\)

\(x=\frac{-21}{4}.9\)

\(x=\frac{-189}{4}\)

          A = 1/3 + 2/3² + 3/3³ + ... + 100/3¹⁰⁰ 

=>    3A = 1 + 2/3 + 3/3² + ... + 100/3⁹⁹ 

     -    A =       1/3 + 2/3² + ... + 99/3⁹⁹ + 100/ 3¹⁰⁰ 

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=>    2A = 1 + 1/3 + 1/3² + ... + 1/3⁹⁹ + 100/3¹⁰⁰ 

=>    6A = 3 + 1 + 1/3 + ... + 1/3⁹⁸ + 100/3⁹⁹ 

     -  2A = 1 + 1/3 + 1/3² + ... + 1/3⁹⁸ + 1/3⁹⁹ +100/3¹⁰⁰ 

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    4A = 3 - 99/3⁹⁹ - 100/3¹⁰⁰  <  3

=>    4A  <  3

=>    A  <  3/4

\(\frac{x}{3}+76=\frac{x}{1000}+600\)

\(\Rightarrow\frac{x}{3}-\frac{x}{1000}=600-76\)

\(x.\left(\frac{1}{3}-\frac{1}{1000}\right)=524\)

\(x.\frac{997}{3000}=524\)

\(x=524:\frac{997}{3000}\)

mk ko bít tính nhanh 

chuyện vế bạn ơi

Bạn có sách công dân 6 không nếu có rỡ ra và tìm bài đó 

quan trong là mk ko bik viết chỗ nào đó bn ới

Bn tham khảo nhé : 

2 / 2 . 3 + 2 /3 . 4 + 2 / 4 .5  + ... + 2 / 2017 . 2018

= 2 . ( 1/2 . 3 + 1/3 . 4 + 1/4 . 5 + ... + 1/ 2017 . 2018 

= 2 . ( 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/2017 - 1/2018 ) 

= 2 . ( 1/2 - 1/2018)

= 2 . 1008/2018

= 2016/2018

= 1008/1009

\(2\times(\frac{1}{2\times3}\times\frac{1}{3\times4}\times...\times\frac{1}{2017\times2018}))\)

\(2\times(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018})\)

\(2\times(\frac{1}{2}-\frac{1}{2018})\)

\(2\times\frac{504}{1009}=\frac{1008}{1009}\)

\(\frac{5}{2}.x+\frac{2}{3}=\frac{9}{2}.x\)

\(\frac{5}{2}.x-\frac{9}{2}.x=\frac{-2}{3}\)

\(x.\left(\frac{5}{2}-\frac{9}{2}\right)=\frac{-2}{3}\)

\(x.\left(-2\right)=\frac{-2}{3}\)

\(x=\frac{-2}{3}:\left(-2\right)\)

\(x=\frac{1}{3}\)

Chúc bn học tốt !!

\(\frac{5}{2}x+\frac{2}{3}=\frac{9}{2}x\)

\(\frac{9}{2}x-\frac{5}{2}x=\frac{2}{3}\)

\(x(\frac{9}{2}-\frac{5}{2})=\frac{2}{3}\)

\(x\times2=\frac{2}{3}\)

\(x=\frac{2}{3}\div2\)

\(x=\frac{1}{3}\)

vậy x=\(\frac{1}{3}\)

A=3.(\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\) )

A=3.(\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}\)\(+...+\frac{1}{97}-\frac{1}{100}\))

A=3.\(\left(1-\frac{1}{100}\right)\)=\(\frac{297}{100}\)

\(A=3\left(\frac{3}{1.4}\right)+3\left(\frac{3}{4.7}\right)+3\left(\frac{3}{7.10}\right)+...+3\left(\frac{3}{97.100}\right)\)

\(=3\left(1-\frac{1}{4}\right)+3\left(\frac{1}{4}-\frac{1}{7}\right)+3\left(\frac{1}{7}-\frac{1}{10}\right)+...+3\left(\frac{1}{97}-\frac{1}{100}\right)\)

\(=3-\frac{3}{4}+\frac{3}{4}-\frac{3}{7}+\frac{3}{7}-\frac{3}{10}+...+\frac{3}{97}-\frac{3}{100}\)

\(=3-\frac{3}{100}\)

\(=\frac{297}{100}\)

mình cũng không hiểu sao ra 9/10 nữa

ta có: \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}< \frac{1}{101}+\frac{1}{101}+...+\frac{1}{101}=\frac{100}{101}\)

mà \(\frac{100}{101}< 1\)

\(\Rightarrow\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}< 1\left(đpcm\right)\)

Chúc bn học tốt !!!

CT:>1