\(\dfrac{45}{105}va\dfrac{84}{147}\)

b)

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3 tháng 8 2017

a,

\(\dfrac{13}{17}=1-\dfrac{4}{17}\\ \dfrac{25}{29}=1-\dfrac{4}{29}\\ \dfrac{4}{17}>\dfrac{4}{29}\Rightarrow1-\dfrac{4}{17}< 1-\dfrac{4}{29}\Leftrightarrow\dfrac{13}{17}< \dfrac{25}{29}\)

Vậy \(\dfrac{13}{17}< \dfrac{25}{29}\)

b,

\(\dfrac{59}{101}>\dfrac{56}{101}>\dfrac{56}{105}\\ \Rightarrow\dfrac{59}{101}>\dfrac{56}{105}\)

Vậy \(\dfrac{59}{101}>\dfrac{56}{105}\)

c,

\(\dfrac{14}{55}>\dfrac{14}{56}=\dfrac{1}{4}=\dfrac{20}{80}>\dfrac{20}{83}\)

Vậy \(\dfrac{14}{55}>\dfrac{20}{83}\)

3 tháng 8 2017

d,

\(\dfrac{13}{57}< \dfrac{13}{39}=\dfrac{1}{3}=\dfrac{29}{87}< \dfrac{29}{73}\)

Vậy \(\dfrac{13}{57}< \dfrac{29}{73}\)

e,

\(\dfrac{17}{21}=\dfrac{17\cdot101}{21\cdot101}=\dfrac{1717}{2121}\)

Vậy \(\dfrac{17}{21}=\dfrac{1717}{2121}\)

16 tháng 5 2018

F=(9.75.21\(\dfrac{3}{7}\)+\(\dfrac{39}{4}\).18\(\dfrac{4}{7}\)).\(\dfrac{15}{78}\)

=(\(\dfrac{39}{4}\).21\(\dfrac{3}{7}\)+\(\dfrac{39}{4}\).18\(\dfrac{4}{7}\)).\(\dfrac{15}{78}\)

=[\(\dfrac{39}{4}\).(21\(\dfrac{3}{7}\)+18\(\dfrac{4}{7}\))].\(\dfrac{15}{78}\)

=[\(\dfrac{39}{4}\).(21+18)+(\(\dfrac{3}{7}\)+\(\dfrac{4}{7}\))].\(\dfrac{15}{78}\)

=[\(\dfrac{39}{4}\).(39+1)].\(\dfrac{15}{78}\)

=(\(\dfrac{39}{4}\).40).\(\dfrac{15}{78}\)

=390.\(\dfrac{15}{78}\)=75

25 tháng 6 2017

\(B=71\dfrac{38}{45}-\left(43\dfrac{8}{45}-1\dfrac{17}{57}\right)\)

\(B=71\dfrac{38}{45}-43\dfrac{8}{45}-1\dfrac{17}{57}\)

\(B=28\dfrac{2}{3}-1\dfrac{17}{57}=27\dfrac{11}{57}\)

\(D=\left(19\dfrac{5}{8}:\dfrac{7}{12}-13\dfrac{1}{4}:\dfrac{7}{12}\right).\dfrac{4}{5}\)

\(D=\dfrac{12}{7}.\left(19\dfrac{5}{8}-13\dfrac{1}{4}\right).\dfrac{4}{5}\)

\(D=\dfrac{12}{7}.\dfrac{51}{8}.\dfrac{4}{5}=\dfrac{306}{35}\)

Câu còn lại làm tương tự!

Chúc bạn học tốt!!!

3 tháng 8 2017

3. a) Ta có : 13.29 = 377
25.17 = 425
=> \(\dfrac{13}{17}< \dfrac{25}{29}\)
b) Ta có : 59.105 > 56.101
=> \(\dfrac{59}{101}>\dfrac{56}{105}\)
c) Ta có : 14.83 = 1162
20.55 = 1100
=> \(\dfrac{14}{55}>\dfrac{20}{83}\)
d) Ta có : 13.73 = 949
29.57 = 1653
=> \(\dfrac{13}{57}< \dfrac{29}{73}\)
e) Ta có : \(\dfrac{1717}{2121}=\dfrac{17}{21}\)
=> \(\dfrac{17}{21}=\dfrac{1717}{2121}\)
@Đặng Vũ Hoài Anh

3 tháng 8 2017

4. Gọi các phân số cần tìm có dạng \(\dfrac{x}{3}\)
Ta có : \(\dfrac{-1}{2}< \dfrac{x}{3}< \dfrac{1}{2}\)

=> \(\dfrac{-3}{6}< \dfrac{2x}{6}< \dfrac{3}{6}\)

=> -3 < 2x < 3
=> 2x = -2; 0; 2
=> x = -1; 0; 1 (thỏa mãn)
@Đặng Vũ Hoài Anh

9: \(=\dfrac{47}{51}\cdot\dfrac{17}{94}-\dfrac{47}{51}\cdot\dfrac{53}{91}-\dfrac{53}{91}\cdot\dfrac{91}{53}+\dfrac{53}{91}\cdot\dfrac{47}{51}\)

\(=\dfrac{1}{6}-1=-\dfrac{5}{6}\)

10: \(=\dfrac{13}{19}\cdot\dfrac{19}{26}-\dfrac{13}{19}\cdot\dfrac{71}{43}+\dfrac{71}{43}\cdot\dfrac{13}{19}-\dfrac{71}{43}\cdot\dfrac{86}{71}\)

\(=\dfrac{1}{2}-2=-\dfrac{3}{2}\)

5 tháng 2 2023

bạn giải chi tiết đi

 

 

14 tháng 4 2018

a)

\(3\dfrac{14}{19}+\dfrac{13}{17}+\dfrac{35}{43}+6\dfrac{5}{19}+\dfrac{8}{43}\\ =\left(3\dfrac{14}{19}+6\dfrac{5}{19}\right)+\left(\dfrac{35}{43}+\dfrac{8}{43}\right)+\dfrac{13}{17}\\ =10+1+\dfrac{13}{17}\\ =11\dfrac{13}{17}\)

b)

\(\dfrac{-5}{7}\cdot\dfrac{2}{11}+\dfrac{-5}{7}\cdot\dfrac{9}{11}+1\dfrac{5}{7}\\ =\dfrac{-5}{7}\cdot\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+1\dfrac{5}{7}\\ =\dfrac{-5}{7}\cdot1+1\dfrac{5}{7}\\ =\dfrac{-5}{7}+1\dfrac{5}{7}\\ =1\)

14 tháng 4 2018

a) \(3\dfrac{14}{19}+\dfrac{13}{17}+\dfrac{35}{43}+6\dfrac{5}{19}+\dfrac{8}{43}\)

\(=\left(3\dfrac{14}{19}+6\dfrac{5}{19}\right)+\left(\dfrac{35}{43}+\dfrac{8}{43}\right)+\dfrac{13}{17}\)

\(=\left[\left(3+6\right)+\left(\dfrac{14}{19}+\dfrac{5}{19}\right)\right]+1+\dfrac{13}{17}\)

\(=\left[9+1\right]+1+\dfrac{13}{17}\)

\(=10+1+\dfrac{13}{17}\)

\(=11+\dfrac{13}{17}\)

\(=\dfrac{187}{17}+\dfrac{13}{17}\)

\(=\dfrac{200}{17}\)

b) \(\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}.\dfrac{9}{11}+1\dfrac{5}{7}\)

\(=\dfrac{-5}{7}.\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\)

\(=\dfrac{-5}{7}.1+\dfrac{12}{7}\)

\(=\dfrac{-5}{7}+\dfrac{12}{7}\)

\(=\dfrac{7}{7}\)

\(=1\)

c) \(11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)

= \(11\dfrac{3}{13}-2\dfrac{4}{7}-5\dfrac{3}{13}\)

\(=\left(11\dfrac{3}{13}-5\dfrac{3}{13}\right)-2\dfrac{4}{7}\)

\(=\left[\left(11-5\right)+\left(\dfrac{3}{13}-\dfrac{3}{13}\right)\right]-\dfrac{18}{7}\)

\(=\left[6+0\right]-\dfrac{18}{7}\)

\(=6-\dfrac{18}{7}\)

\(=\dfrac{42}{7}-\dfrac{18}{7}\)

\(=\dfrac{24}{7}\)

d) \(\dfrac{2}{7}.5\dfrac{1}{4}-\dfrac{2}{7}.3\dfrac{1}{4}\)

\(=\dfrac{2}{7}.\left(5\dfrac{1}{4}-3\dfrac{1}{4}\right)\)

\(=\dfrac{2}{7}.\left[\left(5-3\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)\right]\)

\(=\dfrac{2}{7}.\left[2+0\right]\)

\(=\dfrac{2}{7}.2\)

= \(\dfrac{4}{7}\)

10 tháng 11 2017

Câu 1:

a) \(\dfrac{-15}{17}\)\(\dfrac{-19}{21}\)

Ta có: \(\dfrac{-15}{17}=-1+\dfrac{2}{17}\); \(\dfrac{-19}{21}=-1+\dfrac{2}{21}\)

\(\dfrac{2}{17}>\dfrac{2}{21}\)

Do đó: \(\dfrac{-15}{17}>\dfrac{19}{-23}\)

b) \(\dfrac{-13}{19}\)\(\dfrac{19}{-23}\)

Ta có: \(\dfrac{19}{23}>\dfrac{19}{25}\); \(\dfrac{13}{19}=1-\dfrac{6}{19}\); \(\dfrac{19}{25}=1-\dfrac{6}{25}\)

\(\dfrac{6}{19}>\dfrac{6}{25}\) \(\Rightarrow\dfrac{13}{19}< \dfrac{19}{25}< \dfrac{19}{23}\)

\(\dfrac{13}{19}< \dfrac{19}{23}\Rightarrow\dfrac{-13}{19}>\dfrac{19}{-23}\)

c) \(\dfrac{-24}{35}\)\(\dfrac{-19}{30}\)

Ta có: \(\dfrac{-24}{35}=-1+\dfrac{19}{35}\);\(\dfrac{-19}{30}=-1+\dfrac{11}{30}\)

\(\dfrac{11}{35}< \dfrac{11}{30}\)

Do đó: \(\dfrac{-24}{35}< \dfrac{-19}{30}\)

d) \(\dfrac{-1941}{1931}\)\(\dfrac{-2011}{2001}\); \(\dfrac{-2011}{2001}=-1+\dfrac{10}{2001}\)

\(\dfrac{10}{1931}< \dfrac{10}{1001}\)

Do đó: \(\dfrac{-1941}{1931}< \dfrac{-2011}{2001}\)

Ta có: \(\dfrac{-1941}{1931}=-1+\dfrac{10}{1931}\)

10 tháng 11 2017

Sorry câu d mình viết ngược:

Làm lại:

d) \(\dfrac{-1941}{1931}\)\(\dfrac{-2011}{2001}\)

Ta có: \(\dfrac{-1941}{1931}=-1+\dfrac{10}{1931};\)

\(\dfrac{-2011}{2001}=-1+\dfrac{10}{2001}\)

\(\dfrac{10}{1931}< \dfrac{10}{1001}\)

Do đó: \(\dfrac{-1941}{1931}< \dfrac{-2011}{2001}\)

12 tháng 7 2017

Theo quy ước với mọi phân số lớn hơn 0 thì ta có:

\(\dfrac{a}{b}>0=>\dfrac{a}{b}< \dfrac{a+n}{b+n}\left(n\in N;n\ne0\right)\)

Áp dụng với bài trên ta => ĐPCM

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