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\(\frac{1010+1111+1212+1313+1414+1515+1616+1717}{2020+2121+2222+2323+2424+2525+2626+2727}\)
\(=\frac{101.10+101.11+...+101.17}{101.20+101.21+...+101.27}\)
\(=\frac{101.\left(10+11+...+17\right)}{101.\left(20+21+...+27\right)}\)
\(=\frac{108}{188}\)
\(=\frac{27}{47}\)
\(2>\left(\frac{1}{6}+\frac{2}{15}+\frac{3}{40}+\frac{4}{96}\right)\cdot5.y>\frac{5}{6}\)
\(\Rightarrow2>\left(\frac{1}{6}+\frac{2}{15}+\frac{3}{40}+\frac{1}{24}\right):5.y>\frac{5}{6}\)
\(\Rightarrow2>\left(\frac{20}{120}+\frac{16}{120}+\frac{9}{120}+\frac{5}{120}\right):5.y>\frac{5}{6}\)
\(\Rightarrow2>\frac{5}{12}:5.y>\frac{5}{6}\)
\(\Rightarrow2>\frac{1}{12}.y>\frac{5}{6}\)
Đặt :\(\frac{1}{12}.y=2\Rightarrow y=2:\frac{1}{12}=24\)
\(\frac{1}{12}.y=\frac{5}{6}\Rightarrow y=\frac{5}{6}:\frac{1}{12}=10\)
\(\Rightarrow24>y>10\)
\(\Rightarrow y\in\left\{11;12;...;23\right\}\)
A)1212/1313<1313/1414.B)1717/2525<515151/727272.23/28,7/9,5/7,11/18
Ta có :
\(\frac{1313}{9191}\)= \(\frac{1}{7}\)= \(\frac{11}{77}\)
\(\frac{1111}{7373}\)= \(\frac{11}{73}\)
Vì \(\frac{11}{77}\)< \(\frac{11}{73}\)( 77>73)
=> \(\frac{1313}{9191}\)> \(\frac{1111}{7373}\)
\(\frac{1313}{9191}=\frac{1}{7}=\frac{1.11}{7.11}=\frac{11}{77}\)
\(\frac{1111}{7373}=\frac{11}{73}\)
\(\Rightarrow\frac{11}{77}< \frac{11}{73}\Rightarrow\frac{1313}{9191}>\frac{1111}{7373}\)
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{15}\right)\)
\(=\frac{1}{2}.\frac{14}{15}\)
\(=\frac{7}{15}\)
a) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{15}\right)=\frac{1}{2}.\frac{14}{15}\)\(=\frac{7}{15}\)
b)\(\frac{1414+1515+...+1919}{2020+2121+...+2525}\)
\(\Rightarrow\frac{101\left(14+15+16+17+18+19\right)}{101\left(20+21+22+23+24+25\right)}\)
\(=\frac{14+15+16+17+18+19}{20+21+22+23+24+25}\)
\(=\frac{\left(19+14\right).6:2}{\left(25+20\right).6:2}=\frac{19+14}{25+20}=\frac{33}{45}=\frac{11}{15}\)
a,Ta có phân số chung gian 123/343. mà:123/341>123/343(so sánh mẫu số khi tử bằng nhau)vaf123/343>103/343.
Qua 2 so sánh trên có thể chứng minh:123/341>103/343.
B,Ta có :1-105/107=2/107 và 1-107/109=2/109.
Mà:2/107>2/109.Vậy 105/107<107/109.(So sánh phần bù)
\(\dfrac{121212}{131313}=\dfrac{121212:10101}{131313:10101}=\dfrac{12}{13}=1-\dfrac{1}{13}\)
\(\dfrac{1313}{1414}=\dfrac{1313:101}{1414:101}=\dfrac{13}{14}=1-\dfrac{1}{14}\)
Vì \(\dfrac{1}{13}>\dfrac{1}{14}\) nên \(\dfrac{12}{13}< \dfrac{13}{14}\) hay \(\dfrac{1212}{1313}< \dfrac{1313}{1414}\)
b) \(\dfrac{1717}{2525}=\dfrac{1717:101}{2525:101}=\dfrac{17}{25}=\dfrac{51}{75}\)
\(\dfrac{515151}{727272}=\dfrac{515151:10101}{727272:10101}=\dfrac{51}{72}\)
Vì \(\dfrac{51}{75}< \dfrac{51}{72}\) nên \(\dfrac{17}{25}< \dfrac{51}{72}\) hay \(\dfrac{1717}{2525}< \dfrac{515151}{727272}\)
a) \(\dfrac{1212}{1313}=\dfrac{101x12}{101x13}=\dfrac{12}{13}< \dfrac{12+1}{13+1}=\dfrac{13}{14}\)
\(\dfrac{1313}{1414}=\dfrac{101x13}{101x14}=\dfrac{13}{14}\)
Vậy \(\dfrac{1212}{1313}< \dfrac{1313}{1414}\)
Làm tương tự câu b