Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=1+3^2+3^4+...+3^{2022}\)
\(3^2S=9S=3^2+3^4+3^6+...+3^{2024}\)
\(S=\dfrac{9S-S}{8}=\left(3^{2024}-1\right):8\)
d, không đáp án nào đúng
Lời giải:
$S=1+3^2+3^4+....+3^{2022}$
$9S=3^2S=3^2+3^4+3^6+...+3^{2024}$
$\Rightarrow 9S-S=3^{2024}-1$
$\Rightarrow S=\frac{3^{2024}-1}{8}$
Đáp án D.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu này cô làm rồi em nhá, em xem phần câu hỏi của tôi ý
![](https://rs.olm.vn/images/avt/0.png?1311)
P = 8.( 7 - 72 + 73 - 74 +...+ 72022)
Đặt B = 7 - 72 + 73 - 74+...+ 72022
7 \(\times\)B = 72 - 73 + 74-....- 72022 + 72023
7B + B = 7 + 72023
8B = ( 7 + 72023)
B = ( 7 + 72023): 8
P = 8 \(\times\) ( 7 + 72023) : 8
P = 7 + 72023
![](https://rs.olm.vn/images/avt/0.png?1311)
Q = \(\dfrac{1+x^4+x^8+...+x^{2020}}{1+x^2+...+x^{2022}}\)
Đặt A = 1 + \(x^4\) + \(x^8\) +...+ \(x^{2020}\)
Đặt B = 1 + \(x^2\) + ...+ \(x^{2022}\)
Thì Q = \(\dfrac{A}{B}\)
A = 1 + \(x^4\) + \(x^8\) + ...+ \(x^{2020}\)
A.\(x^4\) = \(x^4\) + \(x^8\) +....+ \(x^{2020}\) + \(x^{2024}\)
A.\(x^4\) - A = \(x^{2024}\) - 1
A = \(\dfrac{x^{2024}-1}{x^4-1}\)
B = 1 + \(x^2\) + \(x^4\) +...+ \(x^{2020}\) + \(x^{2022}\)
B.\(x^2\) = \(x^2\) + \(x^4\) +...+ \(x^{2020}\) + \(x^{2022}\) + \(x^{2024}\)
B\(x^2\) - B = \(x^{2024}\) - 1
B = \(\dfrac{x^{2024}-1}{x^2-1}\)
Q = \(\dfrac{\dfrac{x^{2024}-1}{x^4-1}}{\dfrac{x^{2024}-1}{x^2-1}}\)
Q = \(\dfrac{x^{2024}-1}{x^4-1}\) \(\times\)\(\dfrac{x^2-1}{x^{2024}-1}\)
Q = \(\dfrac{1}{x^2+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(x+1\right)\left(x^2-x+1\right)+x-\left(x-1\right)\left(x^2+x+1\right)+2021\)
\(=x^3+1+x-x^3+1+2021\)
\(=x+2023\)
=-2022+2023
=1
\(A=\left(x+1\right)\left(x^2-x+1\right)+x-\left(x-1\right)\left(x^2+x+1\right)+2021=x^3+1+x-x^2+1+2021=x+2023=-2022+2023=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\dfrac{1}{x}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+4}+...+\dfrac{1}{x+2020}-\dfrac{1}{x+2022}\)
\(=\dfrac{x+2022-x}{x\left(x+2022\right)}=\dfrac{2022}{x\left(x+2022\right)}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
tính riêng:
\(\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}\)
=\(\left(\frac{100}{99}-1\right)+\left(\frac{100}{98}-1\right)+\left(\frac{100}{97}-1\right)+...+\left(\frac{100}{2}-1\right)+99\)
=\(100.\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+...+\frac{1}{2}\right)+99-98\)
=\(100.\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+...+\frac{1}{2}\right)\)
vậy \(\left(\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}\right):\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)=100\)
chúc bạn học tốt ^^