giúp mk vs

\(a=1+2+2^2+...+2^{2021}\)

\(\Rightarrow2a=2+2^2+2^3+...+2^{2022}\)

\(\Rightarrow2a-a=2+2^2+2^3+...+2^{2022}-1-2-2^2-...-2^{2021}\)

\(\Rightarrow a=2^{2022}-1\)

\(\Rightarrow a=2^{2022}-1=b\)

\(a=1+2+2^2+2^3+...+2^{2021}\)

\(2a=2+2^2+2^3+2^4...+2^{2021}+2^{2022}\)

\(2a-a=\)\(\left(2+2^2+2^3+2^4...+2^{2021}+2^{2022}\right)-\left(1+2+2^2+2^3+...+2^{2021}\right)\)

\(a=2^{2022}-1\)

⇒ a=b

Giải:

Ta có:

A=\(\dfrac{10^{2019}-1}{10^{2020}+1}\) 

10A=\(\dfrac{10^{2020}-10}{10^{2020}+1}\) 

10A=\(\dfrac{10^{2020}+1-11}{10^{2020}+1}\) 

10A=\(1+\dfrac{-11}{10^{2020}+1}\) 

Tương tự:

B=\(\dfrac{10^{2020}-1}{20^{2021}+1}\) 

10B=\(1+\dfrac{-11}{10^{2021}+1}\) 

Vì \(\dfrac{-11}{10^{2020}+1}< \dfrac{-11}{10^{2021}+1}\) nên 10A<10B

⇒A<B

Chúc bạn học tốt!

Ta có:

\(A=1+2+2^2+2^3+...+2^{2021}+2^{2022}\)

\(\Rightarrow2A=2\left(1+2+2^2+...+2^{2022}\right)\)

\(\Rightarrow2A=2+2^3+2^4+...+2^{2023}\)

\(\Rightarrow2A-A=\left(2+2^2+2^3+...+2^{2023}\right)-\left(1+2+2^2+...+2^{2022}\right)\)

\(\Rightarrow A=2^{2023}-1\)

Ta thấy: \(2^{2023}-1=2^{2023}-1\)

Vậy: \(A=B\)

Chứng minh cái gì hả bạn?

Bài 1 :

\(M=\dfrac{30-2^{20}}{2^{18}}=\dfrac{2.15-2^{20}}{2^{18}}=\dfrac{15}{2^{17}}-2^2=\dfrac{15}{2^{17}}-4< 0\left(\dfrac{15}{2^{17}}< 1\right)\)

\(N=\dfrac{3^5}{1^{2021}+2^3}=\dfrac{3^5}{9}=\dfrac{3^5}{3^2}=3^3=27\)

\(\Rightarrow M< N\)

Bài 3 :

a) \(t^2+5t-8\) khi \(t=2\)

\(=5^2+2.5-8\)

\(=25+10-8\)

\(=27\)

b) \(\left(a+b\right)^2-\left(b-a\right)^3+2021\left(1\right)\)

\(\left\{{}\begin{matrix}a=5\\b=a+1=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=11\\b-a=1\end{matrix}\right.\)

\(\left(1\right)=11^2-1^3+2021=121-1+2021=2141\)

c) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)

\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)

\(\left(1\right)=1^3=1\)

Bài 2 :

a) \(...=2^3\left(19-14\right)+1=8.5+1=41\)

b) \(...=100-\left[60:\left(5^2-15\right)\right]=100-\left[60:10\right]=100-6=94\)

c) \(...=160:\left[17+\left(9.5-\left(14+2^3\right)\right)\right]=160:\left[17+\left(45-22\right)\right]=160:\left[17+23\right]=160:40=4\)

d) \(...=798+100\left[16-2\left(25-22\right)\right]=798+100\left[16-2.3\right]=798+100.10=798+1000=1798\)

\(A=2+2^2+2^3+...+2^{2021}\\ \Leftrightarrow2A=2^2+2^3+2^4+...+2^{2022}\\ \Leftrightarrow2A-A=\left(2^2+2^3+2^4+...+2^{2022}\right)-\left(2+2^2+2^3+...+2^{2021}\right)\\ \Leftrightarrow A=2^{2022}-2\\ 2^{2022}-2< 2^{2022}\Rightarrow A< B\)

A = 2 + 2 2 + 2 3 + . . . + 2 2021 ⇔ 2 A = 2 2 + 2 3 + 2 4 + . . . + 2 2022 ⇔ 2 A − A = ( 2 2 + 2 3 + 2 4 + . . . + 2 2022 ) − ( 2 + 2 2 + 2 3 + . . . + 2 2021 ) ⇔ A = 2 2022 − 2 2 2022 − 2 < 2 2022 ⇒ A < B

Ta có \(b-a=9.10^{2019}-\dfrac{9}{10^{2021}}>0\Rightarrow b>a\).