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\(K=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}...\frac{-9999}{10000}=\left(-1\right)^{99}.\frac{1.3.2.4...99.101}{2.2.3.3.4.4...100.100}=-\frac{1.2...99}{2.3...100}.\frac{3.4...101}{2.3...100}=-\frac{1}{100}.\frac{101}{2}=-\frac{101}{200}< -\frac{100}{200}=-\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) -Thay \(x=a\) vào K ta được:
\(K=\dfrac{16}{\left(a^2+2\right)+4}\)
-Thay \(x=-a\) vào K ta được:
\(K=\dfrac{16}{\left(\left(-a\right)^2+2\right)+4}=\dfrac{16}{\left(a^2+2\right)+4}\)
-Vậy tại x=a và x=-a (a∈R) thì 2 giá trị của K bằng nhau.
b) -Không có GTNN, chỉ có GTLN:
\(K=\dfrac{16}{\left(x^2+2\right)^2+4}\le\dfrac{16}{2^2+4}=2\)
\(K_{max}=2\Leftrightarrow x=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đổi: 675km = 67 500 000cm
Trên bản đồ tỉ lệ 1:2 500 000 quãng đường dài là:
67 500 000 : 2 500 000 = 27 (cm)
Đáp số: 27 cm
Xin lỗi nha
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(A=\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{2022}{2^{2022}}+\dfrac{2023}{2^{2023}}\)
\(2A=1+\dfrac{2}{2}+\dfrac{3}{2^2}+...+\dfrac{2022}{2^{2021}}+\dfrac{2023}{2^{2022}}\)
\(2A-A=\left(1+\dfrac{2}{2}+\dfrac{3}{2^2}+...+\dfrac{2022}{2^{2021}}+\dfrac{2023}{2^{2022}}\right)-\left(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{2022}{2^{2022}}+\dfrac{2023}{2^{2023}}\right)\)\(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\) - \(\dfrac{2023}{2^{2023}}\)
Đặt B = \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\)
2B = \(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\)
2B - B = \(\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\right)\)B = 2 - \(\dfrac{1}{2^{2022}}\)
Suy ra A = 2 - \(\dfrac{1}{2^{2022}}\) - \(\dfrac{2023}{2^{2023}}\) < 2
Vậy A < 2
\(A=\dfrac{1}{2}+\dfrac{2}{2^{2}}+\dfrac{3}{2^{3}}+...+\dfrac{2022}{2^{2022}}+\dfrac{2023}{2^{2023}}\)
\(2A=1+\dfrac22+\dfrac3{2^2}\ +\,.\!.\!.+\ \dfrac{2022}{2^{2021}}+\dfrac{2023}{2^{2022}}\\2A-A=\left(1+\dfrac22+\dfrac3{2^2}\ +\,.\!.\!.+\ \dfrac{2022}{2^{2021}}+\dfrac{2023}{2^{2022}}\right)-\left(\dfrac12+\dfrac2{2^2}+\dfrac3{2^3}\ +\,.\!.\!.+\ \dfrac{2022}{2^{2022}}+\dfrac{2023}{2^{2023}}\right)\\A=1+\dfrac12+\dfrac1{2^3}\ +\,.\!.\!.+\ \dfrac1{2^{2021}}+\dfrac1{2^{2022}}-\dfrac{2023}{2^{2023}}\\2\left(A+\dfrac{2023}{2^{2023}}\right)=2+1+\dfrac12+\dfrac1{2^2}\ +\,.\!.\!.+\ \dfrac1{2^{2020}}+\dfrac1{2^{2021}}\\A+\dfrac{2023}{2^{2023}}=2-\dfrac1{2^{2022}}\\A=2-\dfrac1{2^{2022}}+\dfrac{2023}{2^{2023}}<2\)
Đề bài hình như thiếu đó bạn ơi, bạn có thể viết đầy đủ được không!!!