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1) \(5^{199}< 5^{200}=25^{100}\)
\(3^{300}=27^{100}>25^{100}\)
\(\Rightarrow3^{300}>5^{199}\)
\(\Rightarrow\dfrac{1}{3^{300}}< \dfrac{1}{5^{199}}\)
2) a) \(107^{50}=\left(107^2\right)^{25}=11449^{25}\)
\(73^{75}=\left(73^3\right)^{25}=389017^{25}>11449^{25}\)
\(\Rightarrow107^{50}< 73^{75}\)
b) \(54^4< 5^{12}< 21^{12}\Rightarrow54^4< 21^{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2019\cdot2020}+\dfrac{1}{2020\cdot2021}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2019}-\dfrac{1}{2020}+\dfrac{1}{2020}-\dfrac{1}{2021}\)
\(=\dfrac{1}{1}-\dfrac{1}{2021}=\dfrac{2021}{2021}-\dfrac{1}{2021}\)
\(=\dfrac{2020}{2021}\)
mà \(\dfrac{2020}{2021}< \dfrac{2021}{2021}=1\)
nên A<1
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Có : 10A = 10.(10^11-1)/10^12-1 = 10^12-10/10^12-1
Vì : 0 < 10^12-10 < 10^12-1 => 10A < 1 (1)
10B = 10.(10^10+1)/10^11+1 = 10^11+10/10^11+1
Vì : 10^11+10 > 10^11+1 > 0 => 10B > 1 (2)
Từ (1) và (2) => 10A < 10B
=> A < B
Tk mk nha
\(A=\frac{10^{11}-1}{10^{12}-1}\)
\(B=\frac{10^{10}+1}{10^{11}+1}\)
Mà \(\frac{10^{11}-1}{10^{12}-1}< 1\); \(\frac{10^{10}+1}{10^{11}+1}< 1\)
\(\Rightarrow\)\(A,B< 1\)
Ta có:
\(10^{11}-1>10^{10}+1\); \(10^{12}-1>10^{11}+1\)
\(\Rightarrow A>B\)
Vậy A > B
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)...\left(\frac{1}{100}-1\right)=\frac{-3}{4}.\frac{-8}{9}...\frac{-99}{100}=-\frac{3.8...99}{4.9....100}\)
\(=-\frac{1.3.2.4...9.11}{2.2.3.3....10.10}=-\frac{\left(1.2...9\right).\left(3.4...11\right)}{\left(2.3...10\right).\left(2.3...10\right)}=-\frac{1.11}{10.2}=-\frac{11}{20}< -\frac{11}{21}\)
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B=1/50+1/51+1/52+...+1/99
Ta có: 1/50=1/50
1/51<1/50
1/52<1/50
..............
1/99<1/50
1/50+1/51+1/52+...+1/99<1/50+1/50+1/50+...+1/50(50 phân số 1/50)
B<1
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài này dễ mà bạn cũng hỏi =(((
\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)....\left(\frac{1}{400}-1\right)\)
\(\Leftrightarrow A=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}....\frac{-399}{400}\)
\(=\frac{1.\left(-3\right)}{2.2}.\frac{2.\left(-4\right)}{3.3}.\frac{3.\left(-5\right)}{4.4}....\frac{19.\left(-21\right)}{20.20}\)
\(=\frac{\left(1.2.3...19\right).\left(\left(-3\right).\left(-4\right).\left(-5\right)...\left(-21\right)\right)}{\left(2.3.4...20\right)\left(2.3.4...20\right)}=\frac{1}{20}.\frac{\left(-21\right)}{2}=\frac{-21}{40}\)
Dễ dàng nhận thấy \(\frac{21}{40}>\frac{1}{2}\Rightarrow\frac{-21}{40}< \frac{-1}{2}\)
Vậy \(A< -\frac{1}{2}\)
Ta thấy: \(\frac{13^{16}+1}{13^{17}+1}\)< 1
\(\Rightarrow\)\(\frac{13^{16}+1}{13^{17}+1}\)< \(\frac{13^{16}+1+12}{13^{17}+1+12}\)= \(\frac{13^{16}+13}{13^{17}+13}\)=\(\frac{13\left(13^{15}+1\right)}{13\left(13^{16}+1\right)}\)= \(\frac{13^{15}+1}{13^{16}+1}\)
Vậy \(\frac{13^{16}+1}{13^{17}+1}\)< \(\frac{13^{15}+1}{13^{16}+1}\)
Chúc bạn học tốt!
Đặt A = \(\frac{13^{15}+1}{13^{16}+1}\)
=> 13A = \(\frac{13.\left(13^{15}+1\right)}{13^{16}+1}=\frac{13^{16}+13}{13^{16}+1}=1+\frac{12}{13^{16}+1}\)\(\)
Đặt B = \(\frac{13^{16}+1}{13^{17}+1}\)
=> 13B = \(\frac{13.\left(13^{16}+1\right)}{13^{17}+1}=\frac{13^{17}+13}{13^{17}+1}=1+\frac{12}{13^{17}+1}\)
Ta có : \(\frac{12}{13^{16}+1}>\frac{12}{13^{17}+1}\Rightarrow1+\frac{12}{13^{16}+1}>1+\frac{12}{13^{17}+1}\Rightarrow13A>13B\Rightarrow A>B\)
Vậy \(\frac{13^{15}+1}{13^{16}+1}>\frac{13^{16}+1}{13^{17}+1}\)