lầy dạ??

a: \(\Leftrightarrow-\dfrac{23}{5}\cdot\dfrac{50}{23}< x< \dfrac{-13}{5}:\dfrac{21}{15}=\dfrac{-13}{5}\cdot\dfrac{5}{7}=\dfrac{-13}{7}\)

=>-10<x<-13/7

hay \(x\in\left\{-9;-8;-7;-6;-5;-4;-3;-2\right\}\)

b: \(\Leftrightarrow-\dfrac{13}{3}\cdot\dfrac{1}{3}< x< \dfrac{-2}{3}\cdot\dfrac{4-3-9}{12}\)

\(\Leftrightarrow-\dfrac{13}{9}< x< \dfrac{4}{9}\)

mà x là số nguyên

nên \(x\in\left\{-1;0\right\}\)

Từ gt ta có:
\(\dfrac{13}{3}.\left(-\dfrac{1}{3}\right)\le x\le\dfrac{2}{3}.\left(-\dfrac{11}{12}\right)\)
\(\Leftrightarrow\dfrac{-13}{9}\le x\le-\dfrac{11}{18}\)
\(\Leftrightarrow\dfrac{-26}{18}\le x\le-\dfrac{11}{18}\)
Suy ra \(26\ge x\ge11\)
Vậy \(11\le x\le26\) ( x thuộc Z ) là các giá trị cần tìm

\(4\dfrac{1}{3}.\left(\dfrac{1}{6}-\dfrac{1}{2}\right)\le x\le\dfrac{2}{3}.\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)\)

\(\dfrac{13}{3}.\dfrac{-1}{3}\le x\le\dfrac{2}{3}.\dfrac{-11}{12}\)

\(\dfrac{-13}{9}\)\(\le x\le\)\(\dfrac{-11}{18}\)

\(\dfrac{-26}{18}\)\(\le x\le\dfrac{-11}{18}\)

\(\Rightarrow x\in\left\{\dfrac{-12}{18};\dfrac{-13}{18};\dfrac{-14}{18};\dfrac{-15}{18};...;\dfrac{-24}{18};\dfrac{-25}{18}\right\}\)Tick hộ mình nha bạn

\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2018^2}< \dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{2018.2019}\)

=\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2018}-\dfrac{1}{2019}\)

\(=\dfrac{1}{2}-\dfrac{1}{2019}< 1\)

Vậy A < 1.

2: \(M=\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots-\frac{30}{3^{30}}\)

=>3M=\(1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots-\frac{30}{3^{29}}\)

=>3M+M=\(1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots-\frac{30}{3^{29}}+\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots-\frac{30}{3^{30}}\)

=>4M=\(1-\frac13+\frac{1}{3^2}-\cdots-\frac{1}{3^{29}}-\frac{30}{3^{30}}\)

Đặt \(A=-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{29}}\)

=>3A=-1+\(\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{28}}\)

=>3A+A=\(-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{28}}-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{29}}\)

=>4A=\(-1-\frac{1}{3^{29}}=\frac{-3^{29}-1}{3^{29}}\)

=>\(A=\frac{-3^{29}-1}{4\cdot3^{29}}\)

Ta có: \(4M=1-\frac13+\frac{1}{3^2}-\cdots-\frac{1}{3^{29}}-\frac{30}{3^{30}}\)

\(=1+\frac{-3^{29}-1}{4\cdot3^{29}}-\frac{30}{3^{30}}=1+\frac{-3^{30}-3-120}{4\cdot3^{30}}=1-\frac14-\frac{123}{4\cdot3^{30}}=\frac34-\frac{123}{4\cdot3^{30}}\)

=>4M<3/4

=>M<3/16