3/15x18+3/18x21+3/21x24...3/87x90
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 58b;
D = \(\dfrac{3^9.2^{10}}{6^8}\): (\(\dfrac{5^5.2^4}{10^4}.\dfrac{2^6.3^4}{6^4}\))
D = \(\dfrac{\left(3.2\right)^9.2}{6^8}\):(\(\dfrac{\left(5.2\right)^4.5}{10^4}\).\(\dfrac{\left(2.3\right)^4.2^2}{6^4}\))
D = \(\dfrac{6^9.2}{6^8}\): (\(\dfrac{10^4.5}{10^4}\).\(\dfrac{6^4.4}{6^4}\))
D = 6.2 : (5.4)
D = 12:20
D = \(\dfrac{3}{5}\)
Bài 55b;
B = \(\dfrac{18}{2.5}\) + \(\dfrac{18}{5.8}\) + ... + \(\dfrac{18}{203.206}\)
B = 6.(\(\dfrac{3}{2.5}\) + \(\dfrac{3}{5.8}\) + ... + \(\dfrac{3}{203.206}\))
B = 6.(\(\dfrac{1}{2}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{8}\) + ... + \(\dfrac{1}{203}\) - \(\dfrac{1}{206}\))
B = 6.(\(\dfrac{1}{2}\) - \(\dfrac{1}{206}\))
B = 6.\(\dfrac{51}{103}\)
B = \(\dfrac{306}{103}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Giải:
Gọi số học sinh trường đó là: \(x\) (học sinh); 25 < \(x\) \(\in\) N*
Theo bài ra ta có: \(\left\{{}\begin{matrix}192-12⋮x\\116-8⋮x\end{matrix}\right.\)
\(\left\{{}\begin{matrix}180⋮x\\108⋮x\end{matrix}\right.\)
\(x\) \(\in\) ƯC(180; 108)
180 = 22.32.5; 108 = 22.33
ƯCLN(180; 108) = 22.32 = 36
\(x\) \(\in\) Ư(36) = {1; 2; 3; 4; 6; 9; 11; 18; 36}
vì 25 < \(x\) \(\in\) N* nên \(x=36\)
Kết luận: Số học sinh nhận được quà là 36 học sinh.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a^2-b^2=-2022\) là số chẵn \(\Rightarrow a^2;b^2\) cùng tính chẵn lẻ
\(\Rightarrow a;b\) cùng tính chẵn lẻ
\(\Rightarrow a-b\) và \(a+b\) đều là số chẵn
Khi đó: \(a^2-b^2=\left(a-b\right)\left(a+b\right)\) là tích của 2 số chẵn nên luôn chia hết cho 4
Mà \(-2022\) không chia hết cho 4
\(\Rightarrow\) Không tồn tại a;b nguyên thỏa mãn \(a^2-b^2=-2022\)
![](https://rs.olm.vn/images/avt/0.png?1311)
d.
\(\dfrac{2^3.5^4.49}{35.2^2.175}=\dfrac{2^3.5^4.7^2}{5.7.2^2.7.5^2}=\dfrac{2^3.5^4.7^2}{2^2.5^3.7^2}=2.5=10\)
\(\dfrac{12.2^a}{3^2.8^b}=\dfrac{3.2^2.2^a}{3^2.\left(2^3\right)^b}=\dfrac{3.2^{a+2}}{3^2.2^{3b}}=\dfrac{2^{a-3b+2}}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{2.4+2.4.8+4.8.16+8.16.32}{3.4+2.6.8+4.12.16+8.24.32}\)
\(=\dfrac{1.2.4+2.1.2.2.2.4+4.1.4.2.4.4+8.1.8.2.8.4}{3.4+2.1.2.3.2.4+4.1.4.3.4.4+8.1.8.3.8.4}\)
\(=\dfrac{1.2.4+2^3.1.2.4+4^3.1.2.4+8^3.1.2.4}{1.3.4+2^3.1.3.4+4^3.1.3.4+8^3.1.3.4}\)
\(=\dfrac{1.2.4.\left(1+2^3+4^3+8^3\right)}{1.3.4.\left(1+2^3+4^3+8^3\right)}\)
\(=\dfrac{2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
S= 3/2-3/3+3/3-3/4+3/4-3/5+....+3/199-3/200
S=3/2-3/200
S=300/200-3/200
S=297/200
\(S=\dfrac{3}{2.3}+\dfrac{3}{3.4}+\dfrac{3}{4.5}+...+\dfrac{3}{199.200}\\ \Rightarrow S=3.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+..+\dfrac{1}{199.200}\right)\\ \Rightarrow S=3.\left(\dfrac{1}{2}-\dfrac{1}{200}\right)\\ \Rightarrow S=3.\dfrac{99}{200}\\ \Rightarrow S=\dfrac{501}{200}.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=2003-\dfrac{1}{2.3}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{18.19}+\dfrac{1}{19.20}\right)\)
Đặt
\(B=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{19.20}=\)
\(=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{20-19}{19.20}=\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}=\)
\(=1-\dfrac{1}{20}=\dfrac{19}{20}\)
\(\Rightarrow A=2023-\dfrac{1}{1.2}.B=2023-\dfrac{1}{6}.\dfrac{19}{20}=\)
\(\dfrac{3}{15.18}+\dfrac{3}{18.21}+\dfrac{3}{21.24}+...+\dfrac{3}{87.90}\)
=\(\dfrac{1}{15}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{24}+...+\dfrac{1}{87}-\dfrac{1}{90}\)
\(=\dfrac{1}{15}-\dfrac{1}{90}\)
\(=\dfrac{5}{90}\)
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