\(\dfrac{4}{9}-\dfrac{1}{6}+\dfrac{5}{9}+2=\left(\dfrac{4}{9}+\dfrac{5}{9}+2\right)-\dfrac{1}{6}\)

\(=3-\dfrac{1}{6}=\dfrac{17}{6}\)

17/6 nha

x=2022 nên x+1=2023

\(M\left(x\right)=x^{2023}-2023\left(x^{2022}-x^{2021}+x^{2020}-...+x^2-x\right)\)

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

\(=x^{2023}-x^{2023}-x^{2022}+x^{2022}+x^{2021}+...-x^3-x^2+x^2+x\)

=x

=2022

\(\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2018\times2019}\)

\(=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2018}-\dfrac{1}{2019}\)

\(=\left(\dfrac{2020}{2019}-\dfrac{1}{2019}\right)+\left(\dfrac{2019}{2018}-\dfrac{1}{2018}\right)\)

\(=1+1=2\)

\(A=\dfrac{1}{2^1}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{2023}{2^{2023}}\)

\(\Rightarrow2A=1+\dfrac{2}{2^1}+\dfrac{3}{2^2}+...+\dfrac{2023}{2^{2022}}\)

Trừ vế cho vế:

\(\Rightarrow2A-A=1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}-\dfrac{2023}{2^{2023}}\)

\(\Rightarrow A=1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}-\dfrac{2023}{2^{2023}}\)

\(\Rightarrow2A=2+1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2021}}-\dfrac{2023}{2^{2022}}\)

Trừ vế cho vế:

\(2A-A=2-\dfrac{2024}{2^{2022}}+\dfrac{2023}{2^{2023}}\)

\(\Rightarrow A=2-\dfrac{1}{2^{2022}}\left(2024-\dfrac{2023}{2}\right)\)

\(\Rightarrow A=2-\dfrac{2025}{2^{2023}}< 2\)

Vậy \(A< 2\)

chịu

có nhé 

a: Sửa đề: ΔAKB và ΔAKC

Xét ΔAKB và ΔAKC có

AK chung

\(\widehat{KAB}=\widehat{KAC}\)

AB=AC

Do đó: ΔAKB=ΔAKC

b: ΔAKB=ΔAKC

=>KB=KC

=>ΔKBC cân tại K

c: Ta có: ΔABC cân tại A

mà AD là đường phân giác

nên AD\(\perp\)BC

Câu 1: C

Câu 2: D

Câu 3: A

Câu 4: B

Câu 5; D

Câu 6: B

Câu 7: D

Câu 8: A

Câu 9: B

Câu 10: A

Câu 11: D

Câu 12: C

Câu 13: B

Câu 14: A

Câu 15: A

Câu 16: B

Câu 17: C

Câu 18: B

Câu 19: C