Tìm tập giá trị lớn nhất, giá trị nhỏ nhất của hàm số sau  y = 2 sin x   + 3

A.  m a x   y = 5 , m i n   y = 2

B.   m a x   y = 5 , m i n   y = 3

C.  m a x   y = 5 , m i n   y = 1

D.  m a x   y = 5 , m i n   y = 2 5

\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}+\dfrac{1}{3^{100}}\)

\(\Rightarrow3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}+\dfrac{1}{3^{99}}\)

\(\Rightarrow3A-A=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}+\dfrac{1}{3^{99}}+\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow2A=1-\dfrac{1}{3^{100}}\Leftrightarrow A=\dfrac{1-\dfrac{1}{3^{100}}}{2}\)

P/s: Chúc bạn thi tốt

Use the correct form of the word given in each sentence.

Do you surf the Internet for fun or ……………………….? (educate)

tấm vải thứ nhất là 81

tấm vải thứ hai là 90

tấm vải thứ ba là 63 . 100% luôn

\(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}=\dfrac{\left(2^2\right)^5\cdot\left(3^2\right)^4-2\cdot\left(2\cdot3\right)^9}{2^{10}\cdot3^8+\left(2\cdot3\right)^8\cdot4\cdot5}=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8}=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+1\right)}=\dfrac{1-3}{1+1}=-\dfrac{2}{2}=-1\)

\(\left(x-3\right)^5=\left(x-3\right)^7\)

\(\Rightarrow\left(x-3\right)^7-\left(x-3\right)^5=0\)

\(\Rightarrow\left(x-3\right)^5\left[\left(x-3\right)^2-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-3\right)^5=0\\\left(x-3\right)^2-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\\left(x-3\right)^2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\end{matrix}\right.\)

Vậy x = 2; x = 3; x = 4

Ta có:

7⁴ⁿ⁺¹ = ...7

=> 7⁴ⁿ = ...7:7 = ...1

=> 7²⁰¹⁶ = ...1

9^2k+1 = ...9

=> 9²⁰¹⁷ = ...9

===> 7²⁰¹⁶ + 9²⁰¹⁷ = ...1 + ...9 = ...10 = ...0 chia hết cho 10