So sánh:A=\(\frac{13^{40}+2}{10^{40}-1}\)và B=\(\frac{13^{39}+2}{13^{39}-1}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a\)\(3^{39}\)và \(11^{21}\)
\(3^{39}< 11^{21}\)vì
\(3^{39}< 3^{40}=3^{39}< \left(3^2\right)^{20}=9^{20}< 11^{21}\)
\(\Rightarrow3^{39}< 11^{29}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, 339 < 340 = (32)20=920<1121
b, chép đúng đề bài ko pn
c, 1340< 1640= (24)40= 2160<2162
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
Ta có:
\(\left(\frac{1}{10}\right)^{15}=\left(\frac{1}{5}\right)^{3.5}=\left(\frac{1}{125}\right)^5\)
\(\left(\frac{3}{10}\right)^{20}=\left(\frac{3}{10}\right)^{4.5}=\left(\frac{81}{10000}\right)^5\)
Lại có:
\(\frac{1}{125}=\frac{80}{10000}< \frac{81}{10000}\Rightarrow\left(\frac{1}{125}\right)^5< \left(\frac{81}{10000}\right)^5\)
\(\Rightarrow\left(\frac{1}{10}\right)^{15}< \left(\frac{3}{10}\right)^{20}\)
Bài 2:
Ta có:
\(A=\frac{13^{15}+1}{13^{16}+1}\Rightarrow13A=\frac{13^{16}+13}{13^{16}+1}=1+\frac{12}{13^{16}+1}\)
\(B=\frac{13^{16}+1}{13^{17}+1}\Rightarrow13B=\frac{13^{17}+13}{13^{17}+1}=1+\frac{12}{13^{17}+1}\)
Mà \(\frac{12}{13^{16}+1}>\frac{12}{13^{17}+1}\)
\(\Rightarrow1+\frac{12}{13^{16}+1}>1+\frac{12}{13^{17}+1}\)
\(\Rightarrow13A>13B\Rightarrow A>B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\frac{x}{y}+\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{-\frac{2}{12}-\frac{10}{24}+\frac{14}{39}}\)
\(B=\frac{x}{y}+\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{-\left(\frac{2}{12}+\frac{10}{24}-\frac{14}{39}\right)}\)
\(B=\frac{x}{y}+\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{-\frac{2}{3}\left(\frac{1}{4}+\frac{5}{8}-\frac{7}{13}\right)}\)
\(B=\frac{x}{y}+\frac{1}{-\frac{2}{3}}\)
\(B=\frac{x}{y}-\frac{3}{2}\)
Thế x = 0, 5 = 1/2 ; y = 3 ta được :
\(B=\frac{\frac{1}{2}}{3}-\frac{3}{2}=\frac{1}{6}-\frac{9}{6}=-\frac{8}{6}=-\frac{4}{3}\)
Ta có:\(B=\frac{x}{y}+\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{\frac{-2}{12}-\frac{10}{24}+\frac{14}{39}}\)
\(B=\frac{x}{y}+\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{-\left(\frac{2}{12}+\frac{10}{24}-\frac{14}{39}\right)}\)
\(B=\frac{x}{y}+\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{-\frac{2}{3}\left(\frac{1}{4}+\frac{5}{8}-\frac{7}{13}\right)}\)
\(B=\frac{x}{y}+\frac{1}{-\frac{2}{3}}\)(Do\(\frac{1}{4}+\frac{5}{8}-\frac{7}{13}\ne0\))
\(B=\frac{x}{y}-\frac{3}{2}\)
Thay x = 0,5; y = 3 vào B ta được:
\(B=\frac{0,5}{3}-\frac{3}{2}\)
\(B=\frac{1}{6}-\frac{3}{2}\)
\(B=\frac{1}{6}-\frac{9}{6}\)
\(B=-\frac{4}{3}\)
Vậy\(B=-\frac{4}{3}\)tại x = 0,5; y = 3
Linz
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{5}{9}:\frac{13}{7}+\frac{5}{9}:\frac{13}{9}-1\frac{2}{3}\\ =\frac{5}{9}\cdot\frac{7}{13}+\frac{5}{9}\cdot\frac{9}{13}-\frac{5}{3}\\ =\frac{5}{9}\left(\frac{7}{13}+\frac{9}{13}\right)-\frac{5}{3}\\ =\frac{5}{9}\cdot\frac{16}{13}-\frac{5}{3}\\ =\frac{80}{117}-\frac{5}{3}\\ =\frac{80}{117}-\frac{195}{117}=\frac{-115}{117}\)
b) \(\left(15-6\frac{13}{18}\right):11\frac{1}{27}-2\frac{1}{8}:1\frac{11}{40}\\ =\left(\frac{270}{18}-\frac{121}{18}\right):\frac{298}{27}-\frac{17}{8}:\frac{51}{40}\\ =\frac{149}{18}\cdot\frac{27}{298}-\frac{17}{8}\cdot\frac{40}{51}\\ =\frac{3}{4}-\frac{5}{3}\\ =\frac{9}{12}-\frac{20}{12}=\frac{-11}{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\frac{\frac{5}{11.2}+\frac{3}{13}-\frac{1}{2}}{\frac{4}{13}-\frac{1}{11}+\frac{3}{2}}=\frac{5}{\frac{2}{4}}=\frac{5}{\frac{1}{2}}\)