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![](https://rs.olm.vn/images/avt/0.png?1311)
\(5^2.5^3=5^{2+3}=5^5\)
\(9^{10}:9^2=9^{10-2}=9^8\)
\(8^{10}.8^2=8^{10+2}=8^{12}\)
\(5^{2+3}=5^5\)
\(9^{10-2}=9^8\)
\(8^{10+2}=8^{12}\)
Oh , tích mik nha bà con cô bác
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\frac{-\frac{1}{9}+1-\frac{2}{10}+1-\frac{3}{11}+1-...-\frac{92}{100}+1}{\frac{1}{9}+\frac{1}{10}+...+\frac{1}{100}}\)
\(=\frac{\frac{8}{9}+\frac{8}{10}+\frac{8}{11}+...+\frac{8}{100}}{\frac{1}{9}+\frac{1}{10}+...+\frac{1}{100}}\)
\(=\frac{8\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\right)}{\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}}\)
= 8
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=9+1+2+3+4+5+6=31\)
\(=2^5\)
\(=0\)
Mong các pạn ủng hộ nhiều
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{9}{1}+\frac{8}{2}+...+\frac{1}{9}\)
=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{10-1}{1}+\frac{10-2}{2}+...+\frac{10-9}{9}\)
=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{10}{1}-1+...+\frac{10}{9}-1\)
=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right]x=10-9+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}\)= \(\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}+\frac{10}{10}\)
=>\(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right]x=10\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)\)
=> \(x=10\)
b) Tương tự câu a
![](https://rs.olm.vn/images/avt/0.png?1311)
Xét TH1 : ( S < 8/9 )
\(\frac{1}{2\cdot2}< \frac{1}{1\cdot2};\frac{1}{3\cdot3}< \frac{1}{2\cdot3};...;\frac{1}{9\cdot9}< \frac{1}{8\cdot9}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}\)
hay \(S< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}\)
\(S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}\)
\(S< 1-\frac{1}{9}=\frac{8}{9}\left(1\right)\)
TH2 : ( S > 2/5 )
\(\frac{1}{2\cdot2}>\frac{1}{2\cdot3};\frac{1}{3\cdot3}>\frac{1}{3\cdot4};...;\frac{1}{9\cdot9}>\frac{1}{9\cdot10}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}>\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)
hay \(S>\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)
\(S>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(S>\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\left(2\right)\)
Từ (1) và (2) => đpcm
Ko tk thì ko phải là ng` nx rồi :)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1 :
36/1212 = 3/101
13/1313 = 1/101
3/101 + 1/101 = 4/101
Vậy 36/1212 + 13/1313 = 4/101.
Bài 2 :
A = 5/13 + 1/2 + -5/9 + -3/6 + 4/-9
A = 5/13 + 1/2 + -5/9 + -1/2 + -4/9
A = (1/2 + -1/2) + (-5/9 + -4/9) + 5/13
A = 0 + (-1) + 5/13
A = (-1) + 5/13 = -13/13 + 5/13 = 8/13.
Chúc bạn học giỏi nhé.
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có : \(\frac{10^9+2}{10^9-1}=\frac{10^9}{10^9-3}\)
\(\Leftrightarrow\left(10^9+2\right)\left(10^9-3\right)=\left(10^9-1\right)10^9\)
\(\Leftrightarrow10^{18}-10^9.3+2.10^9-6=10^{18}-10^9\)
\(\Rightarrow10^{18}-10^9.3+2.10^9-6=10^{18}-\left(10^9.3-2.10^9+6\right)\)
\(=10^{18}-\left(10^9+6\right)\)
vì \(-10^9>-\left(10^9+6\right)\Rightarrow10^{18}-10^9>10^{18}-\left(10^9+6\right)\)
\(\Rightarrow A>B\)
Ta có: A=\(\frac{10^9+2}{10^9-1}=\frac{10^9-1+3}{10^9-1}=1+\frac{3}{10^9-1}\)
B=\(\frac{10^9}{10^9-3}=\frac{10^9-3+3}{10^9-3}=1+\frac{3}{10^9-3}\)
Mà \(\frac{3}{10^9-1}< \frac{3}{10^9-3}\Rightarrow1+\frac{3}{10^9-1}< 1+\frac{3}{10^9-3}\Rightarrow A< B\)
Vậy A<B
\(5000.100:10=500000:10=50000\)
\(\frac{9}{1}-\frac{2}{2}=\frac{18}{2}-\frac{2}{2}=\frac{16}{2}=\frac{8}{1}=8\)
\(9^2-2^2=81-4=77\)
\(nhae\)
=77
k đi
k lại