độ dài đoạn CI là

CI=CE+EI=2+5=7(cm)`

Ta có : \(A\text{=}\dfrac{2023^{2023}}{2023^{2024}}\text{=}\dfrac{1}{2023}\)

và \(B\text{=}\dfrac{2023^{2022}}{2023^{2023}}\text{=}\dfrac{1}{2023}\)

\(\Rightarrow A\text{=}B\)

Ta có :

A=\(\dfrac{2023^{2023}}{2023^{2024}}\)=\(\dfrac{2023^{2022}.2023}{2023^{2023}.2023}\)=\(\dfrac{2023^{2022}}{2023^{2023}}\)

Mà B=\(\dfrac{2023^{2023}}{2023^{2024}}\)

Vậy A=B

\(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{117\cdot120}\)

\(=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{117}-\dfrac{1}{120}\\ =\dfrac{1}{2}-\dfrac{1}{120}\\ =\dfrac{59}{120}\)

Gọi ƯCLN của 6n+7 và 2n+1 là : a

\(\Rightarrow6n+7⋮a\) và \(2n+1⋮a\)

\(\Rightarrow3\left(2n+1\right)⋮a\)

\(\Rightarrow(6n+7-6n-3)⋮a\)

\(\Rightarrow4⋮a\)

\(\Rightarrow a\inƯ\left(4\right)\in\left(1;-1;2;-2;4;-4\right)\)

mà \(2n+1\) là số lẻ nên không có ước là : (2;-2;4;-4)

\(\Rightarrow a\in\left(1;-1\right)\)

\(\Rightarrow A\) tối giản

\(1.\dfrac{-7}{18}+\dfrac{-5}{12}-\dfrac{-13}{18}\text{=}\left(\dfrac{-7}{18}-\dfrac{-13}{18}\right)+\dfrac{-5}{12}\text{=}\dfrac{1}{3}+\dfrac{-5}{12}\text{=}\dfrac{-1}{12}\)

\(2.\dfrac{-13}{17}+\dfrac{-13}{21}+\dfrac{-4}{17}\text{=}\left(\dfrac{-13}{17}+\dfrac{-4}{17}\right)+\dfrac{-13}{21}\text{=}-1+\dfrac{-13}{21}\text{=}\dfrac{-34}{21}\)

\(3.\dfrac{-13}{10}-\dfrac{-4}{13}+\dfrac{-11}{10}\text{=}\dfrac{-12}{5}-\dfrac{-4}{13}\text{=}\dfrac{-136}{65}\)

\(4.\dfrac{13}{17}\times\left(\dfrac{-4}{5}+\dfrac{-3}{4}\right)\text{=}\dfrac{13}{17}\times\dfrac{-31}{20}\text{=}\dfrac{-403}{340}\)

\(5.\left(\dfrac{-5}{12}\times\dfrac{-9}{20}\right)\times\dfrac{-7}{17}\text{=}\dfrac{3}{16}\times\dfrac{-7}{17}\text{=}\dfrac{-21}{272}\)

\(6.\dfrac{11}{23}\times\left(\dfrac{5}{9}+\dfrac{17}{9}-\dfrac{13}{9}\right)\text{=}\dfrac{11}{23}\times1\text{=}\dfrac{11}{23}\)

\(\dfrac{x-4}{2022}+\dfrac{x-3}{2021}+\dfrac{x-2}{2020}+\dfrac{x-1}{2019}\text{=}-4\)

\(\dfrac{x-4}{2022}+\dfrac{x-3}{2021}+\dfrac{x-2}{2020}+\dfrac{x-1}{2019}+4\text{=}0\)

\(\left(\dfrac{x-4}{2022}+1\right)+\left(\dfrac{x-3}{2021}+1\right)+\left(\dfrac{x-2}{2020}+1\right)+\left(\dfrac{x-1}{2019}+1\right)\text{=}0\)

\(\dfrac{x-2018}{2022}+\dfrac{x-2018}{2021}+\dfrac{x-2018}{2020}+\dfrac{x-2018}{2019}\text{=}0\)

\(\left(x-2018\right)\left(\dfrac{1}{2022}+\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}\right)\text{=}0\)

\(Do:\) \(\dfrac{1}{2022}+\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}\ne0\)

\(x-2018\text{=}0\)

\(x\text{=}2018\)

\(Vậy...\)

\(\dfrac{-10}{15}=\dfrac{x}{9}=\dfrac{-8}{y}=\dfrac{z}{-21}\)

có: \(\dfrac{-10}{15}=\dfrac{x}{9}\\ =>15x=-90\\ =>x=-6\)

có

\(\dfrac{-6}{9}=\dfrac{-8}{y}\\ =>-6y=-72\\ =>y=12\)

có:

\(\dfrac{-8}{12}=\dfrac{z}{-21}\\ =>12z=168\\ =>z=14\)

Ta có : \(\overline{ab}+\overline{ba}\text{=}10a+b+10b+a\)

                       \(\text{=}\left(10a+a\right)+\left(10b+b\right)\)

                       \(\text{=}11a+11b\)

                        \(\text{=}11\left(a+b\right)\)

\(\Rightarrow\overline{ab}+\overline{ba}⋮11\)

Ta có :

ab=a.10+b.1

ba=b.10+a.1

⇒ab+ba=a.11+b.11

⇒ab+ba=11.(a+b)

Vì 11⋮11 ⇒ 11.(a+b)

⇒11.(a+b) ⋮ 11

⇒(ab+ba)⋮ 11

Vậy(ab+ba)⋮11

Cái này là abc=a.b.c 

hay \(\overline{abc}\) vậy bạn

Đề sai. Bạn coi lại