trong các số tự nhiên nhỏ hơn 20 ,những số không phải phần tử của A

bài tập 6 ý ạ

bài 6 ạ

a/

\(VT=\dfrac{\left(x+4\right)-\left(x+2\right)}{\left(x+2\right)\left(x+4\right)}+\dfrac{\left(x+8\right)-\left(x+4\right)}{\left(x+4\right)\left(x+8\right)}+\dfrac{\left(x+14\right)-\left(x+8\right)}{\left(x+8\right)\left(x+14\right)}=\)

\(=\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+8}+\dfrac{1}{x+8}-\dfrac{1}{x+14}=\)

\(=\dfrac{1}{x+2}-\dfrac{1}{x+14}=\dfrac{12}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow\dfrac{12}{\left(x+2\right)\left(x+14\right)}=\dfrac{x}{\left(x+2\right)\left(x+14\right)}\left(x\ne-2;x\ne-14\right)\)

\(\Rightarrow x=12\)

\(\dfrac{x}{2023}+\dfrac{x+1}{2022}+...+\dfrac{x+2022}{1}+2023=0\)

\(\dfrac{1}{2023}x+\dfrac{1}{2022}x+\dfrac{1}{2022}\cdot1+...+\dfrac{1}{1}x+\dfrac{1}{1}\cdot2022+2023=0\)

\(x\left(\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}\right)+\left(\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023\right)=0\)

\(x\left(\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}\right)=\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023\)

\(x=\dfrac{\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023}{\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}}\)

\(x=\dfrac{\dfrac{1}{2022}+\dfrac{2022}{2022}+\dfrac{2}{2021}+\dfrac{2021}{2021}+...+\dfrac{2022}{1}+\dfrac{1}{1}}{\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}}\)

\(x=\dfrac{\dfrac{2023}{2022}+\dfrac{2023}{2021}+...+\dfrac{2023}{1}}{\dfrac{1}{2022}+\dfrac{1}{2021}+...+\dfrac{1}{1}}=2023\)

Vậy x = 2023

\(A=\dfrac{3}{5.6}+\dfrac{3}{6.7}+...+\dfrac{3}{91.92}\)

\(\Rightarrow A=3\left(\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{91.92}\right)\)

\(\Rightarrow A=3\left(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{91}-\dfrac{1}{92}\right)\)

\(\Rightarrow A=3\left(\dfrac{1}{5}-\dfrac{1}{92}\right)\)

\(\Rightarrow A=3.\dfrac{87}{460}=\dfrac{261}{460}\)

ai cứu với ạ

ko 

có chết đâu mà phải cứu

\(2018\equiv-1\left(mod2019\right)\)

\(\Rightarrow2018^{2019}\equiv-1^{2019}=-1\) (mod 2019)

\(\Rightarrow2018^{2019}\equiv-1\) (mod 2019)

\(\Rightarrow2018^{2018}+1⋮2019\)

mọi người giúp em nhanh với

ko giải đâu

đùa thôi =)

cúi với ạ mong là nhanh nhất có thể ạ

có thấy gì đâu mà cứu