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![](https://rs.olm.vn/images/avt/0.png?1311)
Tử số=(1+1/3+1/5+1/7+...+1/97+1/99)x1/5 ={ ( 1+1/99) + ( 1/3 + 1/97 ) + ( 1/5 + 1/95) +.....+(1/49 + 1/51)} X 1/5 = (100/ 1 x 99 + 100/ 3 x 97 + 100/ 5 x 95 + ...+ 100/ 49 x 51)X 1/5 = ( 1/1x 99 + 1/ 3 x 97 + 1/ 5 x 95 +...+ 1/ 49 x 51) x 20 Mẫu số=2/1x99+2/3x97+2/5x95+...+2/49x51 = ( 1/1x 99 + 1/ 3 x 97 + 1/ 5 x 95 +...+ 1/ 49 x 51) x 2 Vậy phân số có giá trị = 20/2 = 10
![](https://rs.olm.vn/images/avt/0.png?1311)
Tử số=(1+1/3+1/5+1/7+...+1/97+1/99)x1/5
={ ( 1+1/99) + ( 1/3 + 1/97 ) + ( 1/5 + 1/95) +.....+(1/49 + 1/51)} X 1/5
= (100/ 1 x 99 + 100/ 3 x 97 + 100/ 5 x 95 + ...+ 100/ 49 x 51)X 1/5
= ( 1/1x 99 + 1/ 3 x 97 + 1/ 5 x 95 +...+ 1/ 49 x 51) x 20
Mẫu số=2/1x99+2/3x97+2/5x95+...+2/49x51
= ( 1/1x 99 + 1/ 3 x 97 + 1/ 5 x 95 +...+ 1/ 49 x 51) x 2
Vậy phân số có giá trị = 20/2 = 10
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{2}{3}+\dfrac{3}{4}< x< 1\dfrac{1}{3}+\dfrac{4}{5}\)
\(\dfrac{2\times4}{3\times4}+\dfrac{3\times3}{4\times3}< x< \dfrac{\left(1\times3+1\right)\times5}{3\times5}+\dfrac{4\times3}{5\times3}\)
\(\dfrac{8}{12}+\dfrac{9}{12}< x< \dfrac{20}{15}+\dfrac{12}{15}\\ \dfrac{17}{12}< x< \dfrac{32}{15}\)
Ước tính: \(\dfrac{17}{12}=1,4\) và \(\dfrac{32}{15}=2,1\). Vậy số tự nhiên x = 2 sẽ thõa mãn 1,4 < x < 2,1
b)
\(\dfrac{5}{6}-\dfrac{1}{4}< x< 2\dfrac{1}{3}-\dfrac{2}{5}\\ \dfrac{5\times4}{6\times4}-\dfrac{1\times6}{4\times6}< x< \dfrac{\left(2\times3+1\right)\times5}{3\times5}-\dfrac{2\times3}{5\times3}\\ \dfrac{20}{24}-\dfrac{6}{24}< x< \dfrac{35}{15}-\dfrac{6}{15}\\ \dfrac{14}{24}< x< \dfrac{29}{15}\)
Ước tính \(\dfrac{14}{24}=0,5\) và \(\dfrac{29}{15}=1,9\)
Vậy với x là số tự nhiên x = 1 sẽ thõa mãn 0,5 < x < 1,9
![](https://rs.olm.vn/images/avt/0.png?1311)
1 + 1/3 + 1/9 + 1/27 + 1/81
= 1 + (1/3 + 1/27) + (1/9 + 1/81)
= 1 + (9/27 + 1/27) + (9/81 + 1/81)
= 1 + 10/27 + 10/81
= 1 + 30/81 + 10/81
= 1 + 40/81
= 121/81
![](https://rs.olm.vn/images/avt/0.png?1311)
C = \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) + \(\dfrac{1}{64}\) + \(\dfrac{1}{128}\)
2\(\times\)C = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) + \(\dfrac{1}{64}\)
2 \(\times\) C - C = 1 - \(\dfrac{1}{128}\)
C = \(\dfrac{127}{128}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{49\cdot63}{14\cdot54}\) \(=\frac{7\cdot7\cdot7\cdot9}{7\cdot2\cdot6\cdot9}\) \(=\) \(\frac{7\cdot7}{2\cdot6}\) \(=\frac{49}{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
=>\(D=7\left(\dfrac{5}{42\cdot37}+\dfrac{1}{42\cdot43}+\dfrac{6}{43\cdot49}+\dfrac{10}{49\cdot59}\right)\)
\(=7\left(\dfrac{1}{37}-\dfrac{1}{42}+\dfrac{1}{42}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{49}+\dfrac{1}{49}-\dfrac{1}{59}\right)\)
=7(1/37-1/59)
=7*22/2183
\(E=5\left(\dfrac{8}{37\cdot45}+\dfrac{2}{45\cdot47}+\dfrac{3}{47\cdot50}+\dfrac{9}{50\cdot59}\right)\)
\(=5\left(\dfrac{1}{37}-\dfrac{1}{45}+\dfrac{1}{45}-\dfrac{1}{47}+...+\dfrac{1}{50}-\dfrac{1}{59}\right)\)
=5(1/37-1/59)
=>D/E=7/5
Ủa bạn ơi , mới học lớp 5 đã học cái này rồi á
Mình tưởng lớp 7 ,8 cơ😏😏😏