\(\dfrac{6}{7}+\dfrac{8}{7}=\)
BT2: Tính nhanh
1) \(\dfrac{5}{6}-\dfrac{6}{7}+\dfrac{7}{8}-\dfrac{8}{9}+\dfrac{10}{9}-\dfrac{5}{6}+\dfrac{6}{7}-\dfrac{7}{8}+\dfrac{8}{9}\)
2) \(\dfrac{1}{13}+\dfrac{16}{7}+\dfrac{3}{105}-\dfrac{9}{7}-\dfrac{-12}{13}\)
1) \(\dfrac{5}{6}-\dfrac{6}{7}+\dfrac{7}{8}-\dfrac{8}{9}+\dfrac{10}{9}-\dfrac{5}{6}+\dfrac{6}{7}-\dfrac{7}{8}+\dfrac{8}{9}\)
\(=\left(\dfrac{5}{6}-\dfrac{5}{6}\right)-\left(\dfrac{6}{7}+\dfrac{6}{7}\right)+\left(\dfrac{7}{8}-\dfrac{7}{8}\right)-\left(\dfrac{8}{9}+\dfrac{8}{9}\right)+\dfrac{10}{9}\)
\(=0-0+0-0+\dfrac{10}{9}\)
\(=\dfrac{10}{9}\)
2) \(\dfrac{1}{13}+\dfrac{16}{7}+\dfrac{3}{105}-\dfrac{9}{7}-\dfrac{-12}{13}\)
\(=\left(\dfrac{1}{13}-\left(-\dfrac{12}{13}\right)\right)+\left(\dfrac{16}{7}-\dfrac{9}{7}\right)+\dfrac{3}{105}\)
\(=1+1+\dfrac{3}{105}\)
\(=\dfrac{213}{105}=\dfrac{71}{35}\)
a)\(\dfrac{7}{8}\)x\(\dfrac{3}{13}\)+\(\dfrac{4}{9}\)x\(\dfrac{4}{13}\)
b)\(\dfrac{6}{5}\)+\(\dfrac{7}{3}\)+\(\dfrac{8}{9}\)
c)23: \(\dfrac{5}{14}\)+\(\dfrac{6}{7}\)+\(\dfrac{4}{9}\)
d)\(4\dfrac{1}{4}\)+\(7\dfrac{3}{7}\)-\(2\dfrac{4}{17}\)
e)8-(9\(\dfrac{2}{11}\)+\(\dfrac{8}{33}\))
a, \(\dfrac{7}{8}\) \(\times\) \(\dfrac{3}{13}\) + \(\dfrac{4}{9}\) \(\times\) \(\dfrac{4}{13}\)
= \(\dfrac{1}{13}\) \(\times\)( \(\dfrac{21}{8}\) + \(\dfrac{16}{9}\))
= \(\dfrac{1}{13}\) \(\times\)( \(\dfrac{189}{72}\) + \(\dfrac{128}{72}\))
= \(\dfrac{1}{13}\) \(\times\) \(\dfrac{317}{73}\)
= \(\dfrac{317}{949}\)
b, \(\dfrac{6}{5}\) + \(\dfrac{7}{3}\) + \(\dfrac{8}{9}\)
= \(\dfrac{54}{45}\) + \(\dfrac{105}{45}\) + \(\dfrac{40}{45}\)
= \(\dfrac{199}{45}\)
c, 23 : \(\dfrac{5}{14}\) + \(\dfrac{6}{7}\) + \(\dfrac{4}{9}\)
= \(\dfrac{322}{5}\) + \(\dfrac{6}{7}\) + \(\dfrac{4}{9}\)
= \(\dfrac{20286}{315}\) + \(\dfrac{270}{315}\) + \(\dfrac{140}{315}\)
= \(\dfrac{20696}{315}\)
d, 4\(\dfrac{1}{4}\) + 7\(\dfrac{3}{7}\) - 2\(\dfrac{4}{17}\)
= 4 + \(\dfrac{1}{4}\) + 7 + \(\dfrac{3}{7}\) - 2 - \(\dfrac{4}{17}\)
= (4+7-2) + (\(\dfrac{1}{4}\) + \(\dfrac{3}{7}\) - \(\dfrac{4}{17}\))
= 9 + \(\dfrac{119}{476}\) + \(\dfrac{204}{476}\) - \(\dfrac{112}{476}\)
= 9\(\dfrac{211}{476}\) = \(\dfrac{4495}{476}\)
e, 8 - (9\(\dfrac{2}{11}\) + \(\dfrac{8}{33}\))
= 8 - 9 - \(\dfrac{2}{11}\) - \(\dfrac{8}{33}\)
= -1 - \(\dfrac{2}{11}\) - \(\dfrac{8}{33}\)
= \(\dfrac{-33}{33}\) - \(\dfrac{-6}{33}\) - \(\dfrac{8}{33}\)
= - \(\dfrac{47}{33}\)
\(\dfrac{3}{7}\) + 4 = ?
\(\dfrac{5}{9}\) x 3 = ?
\(\dfrac{7}{8}\) - \(\dfrac{2}{9}\)= ?
5 - \(\dfrac{3}{4}\)= ?
\(\dfrac{5}{7}\) : 6 = ?
\(\dfrac{5}{6}\) x \(\dfrac{2}{7}\)= ?
3 + \(\dfrac{3}{4}\)= ?
\(\dfrac{3}{5}x\dfrac{4}{8}\) = ?
5 : \(\dfrac{3}{8}\)= ?
Mong mn trả lời
\(\dfrac{3}{7}+\dfrac{4}{1}=\dfrac{3}{7}+\dfrac{28}{7}=\dfrac{31}{7}\)
\(\dfrac{5}{9}\times3=\dfrac{5}{9}\times\dfrac{3}{1}=\dfrac{15}{9}=\dfrac{5}{3}\)
\(\dfrac{7}{8}-\dfrac{2}{9}=\dfrac{63}{72}-\dfrac{16}{72}=\dfrac{47}{72}\)
\(\dfrac{5}{1}-\dfrac{3}{4}=\dfrac{20}{4}-\dfrac{3}{4}=\dfrac{17}{4}\)
\(\dfrac{5}{7}:\dfrac{6}{1}=\dfrac{5}{7}\times\dfrac{1}{6}=\dfrac{5}{42}\)
\(\dfrac{5}{6}\times\dfrac{2}{7}=\dfrac{5\times2}{6\times7}=\dfrac{10}{42}=\dfrac{5}{21}\)
\(\dfrac{3}{1}+\dfrac{3}{4}=\dfrac{12}{4}+\dfrac{3}{4}=\dfrac{15}{4}\)
\(\dfrac{3}{5}\times\dfrac{4}{8}=\dfrac{3}{5}\times\dfrac{1}{2}=\dfrac{3\times1}{5\times2}=\dfrac{3}{10}\)
\(\dfrac{5}{1}:\dfrac{3}{8}=\dfrac{5}{1}\times\dfrac{8}{3}=\dfrac{40}{3}\)
a) \(\dfrac{3}{7}+4=\dfrac{3}{7}+\dfrac{4}{1}=\dfrac{3}{7}+\dfrac{28}{7}=\dfrac{31}{7}\)
b) \(\dfrac{5}{9}\times3=\dfrac{5\times3}{9}=\dfrac{15}{9}=\dfrac{5}{3}\)
c) \(\dfrac{7}{8}-\dfrac{2}{9}=\dfrac{63}{72}-\dfrac{16}{72}=\dfrac{47}{72}\)
d) \(5-\dfrac{3}{4}=\dfrac{5}{1}-\dfrac{3}{4}=\dfrac{20}{4}-\dfrac{3}{4}=\dfrac{17}{4}\)
e) \(\dfrac{5}{7}:6=\dfrac{5}{7}:\dfrac{6}{1}=\dfrac{5}{7}\times\dfrac{1}{6}=\dfrac{5}{42}\)
f) \(\dfrac{5}{6}\times\dfrac{2}{7}=\dfrac{5\times2}{6\times7}=\dfrac{10}{42}=\dfrac{5}{21}\)
g) \(3+\dfrac{3}{4}=\dfrac{3}{1}+\dfrac{3}{4}=\dfrac{12}{4}+\dfrac{3}{4}=\dfrac{15}{4}\)
h) \(\dfrac{3}{5}\times\dfrac{4}{8}=\dfrac{3\times4}{5\times8}=\dfrac{12}{40}=\dfrac{3}{10}\)
i) \(5:\dfrac{3}{8}=\dfrac{5}{1}:\dfrac{3}{8}=\dfrac{5}{1}\times\dfrac{8}{3}=\dfrac{40}{3}\)
e) \(\dfrac{9}{17}.\dfrac{3}{7}+\dfrac{9}{17}\div\dfrac{7}{4}\)
f) \(\dfrac{6}{7}.\dfrac{8}{13}+\dfrac{6}{13}.\dfrac{9}{7}-\dfrac{4}{13}.\dfrac{6}{7}\)
e).
= \(\dfrac{9}{17}.\dfrac{3}{7}+\dfrac{9}{17}.\dfrac{4}{7}\)
= \(\dfrac{9}{17}.\left(\dfrac{3}{7}+\dfrac{4}{7}\right)=\dfrac{9}{17}.1=\dfrac{9}{17}\)
f) .
cj nghĩ đề sai đk=)? mạnh dạn đoán=)
\(\dfrac{5}{7}...\dfrac{8}{7}\) ; \(^{\dfrac{6}{7}...\dfrac{8}{7}}\)
so sánh
chung mẫu nên so sánh tử
5<8
nên \(\dfrac{5}{7}< \dfrac{8}{7}\)
chung mẫu nên so sánh tử
6 < 8
nên \(\dfrac{6}{7}< \dfrac{8}{7}\)
1/Thực hiện phép tính:
𝒂) ( \(\dfrac{7}{5}\) − \(\dfrac{8}{7}\) ) + \(3\dfrac{2}{5}\) ∶ 𝟎,𝟔
b) \(\dfrac{11}{6}\) + \(\dfrac{7}{6}\) ∶ \(\dfrac{-5}{4}\)
c) \(\dfrac{7}{4}\) − \(\dfrac{5}{4}\) . \(\dfrac{8}{15}\)
2/ Tìm x biết
b) 𝒙 + 𝟐,𝟓 = \(\dfrac{-8}{3}\)
b) 𝟐𝒙 − 𝟐𝟎% = \(\dfrac{9}{5}\)
𝒄) ( \(\dfrac{5}{4}\) )2 + 𝒙 = \(2\dfrac{3}{4}\)
\(1\)/
\(a\)) \(=\left(\dfrac{7}{5}-\dfrac{8}{7}\right)+\dfrac{17}{5}:0,6\)
\(=\dfrac{9}{35}+\dfrac{17}{3}\)
\(=\dfrac{622}{105}\)
\(b\)) \(=\dfrac{11}{6}+\dfrac{-14}{15}\)
\(=\dfrac{9}{10}\)
\(c\)/ \(=\dfrac{7}{4}-\dfrac{2}{3}\)
\(=\dfrac{13}{12}\)
\(\dfrac{3}{4}.\dfrac{7}{9}.\dfrac{1}{9}.\dfrac{7}{4}\)
\(\dfrac{6}{7}.\dfrac{8}{13}-\dfrac{6}{9}.\dfrac{9}{7}+\dfrac{5}{13}.\dfrac{6}{7}\)
2.11.\(\dfrac{3}{4}.\dfrac{9}{121}\)
\(\dfrac{3}{4}\cdot\dfrac{7}{9}\cdot\dfrac{1}{9}\cdot\dfrac{7}{4}\)
\(=\dfrac{3\cdot7\cdot1\cdot7}{4\cdot9\cdot9\cdot4}=\dfrac{3\cdot7\cdot1\cdot7}{4\cdot3\cdot3\cdot9\cdot4}\)
\(=\dfrac{7\cdot1\cdot7}{4\cdot3\cdot9\cdot4}=\dfrac{49}{432}\)
\(\dfrac{6}{7}\cdot\dfrac{8}{13}-\dfrac{6}{9}\cdot\dfrac{9}{7}+\dfrac{5}{13}\cdot\dfrac{6}{7}\)
\(=\dfrac{6}{7}\cdot\dfrac{8}{13}-\dfrac{6}{7}+\dfrac{5}{13}\cdot\dfrac{6}{7}\)
\(=\dfrac{6}{7}\left(\dfrac{8}{13}-1+\dfrac{5}{13}\right)\)
\(=\dfrac{6}{7}\cdot0\)
\(=0\)
\(2\cdot11\cdot\dfrac{3}{4}\cdot\dfrac{9}{121}\)
\(=\dfrac{2\cdot11\cdot3\cdot9}{4\cdot121}=\dfrac{2\cdot11\cdot3\cdot9}{2\cdot2\cdot11\cdot11}\)
\(=\dfrac{3\cdot9}{2\cdot11}=\dfrac{27}{22}\)
a. \(\dfrac{3.7.1.7}{4.3.3.9.4}=\dfrac{49}{432}\)
b. \(\dfrac{6}{7}.\left(\dfrac{8}{13}+\dfrac{5}{13}-1\right)=\dfrac{6}{7}.0=0\)
\(\dfrac{2.11.3.9}{2.2.11.11}=\dfrac{27}{22}\)
Tính:
a)\(\dfrac{6}{{13}}.\dfrac{8}{7}.\dfrac{{ - 26}}{3}.\dfrac{{ - 7}}{8}\)
b) \(\dfrac{6}{5}.\dfrac{3}{{13}} - \dfrac{6}{5}.\dfrac{{16}}{{13}}\)
a) \(\dfrac{6}{{13}}.\dfrac{8}{{7}}.\dfrac{{ - 26}}{3}.\dfrac{{ - 7}}{8}\)
\(\begin{array}{l} = \left( {\dfrac{6}{{13}}.\dfrac{{ - 26}}{3}} \right).\left( {\dfrac{8}{7}.\dfrac{{ - 7}}{8}} \right)\\ = \dfrac{{6.\left( { - 26} \right)}}{{13.3}}.\dfrac{{8.\left( { - 7} \right)}}{{7.8}}\\= (- 4).\left( { - 1} \right) = 4\end{array}\)
b) \(\dfrac{6}{5}.\dfrac{3}{{13}} - \dfrac{6}{5}.\dfrac{{16}}{{13}}\)
\(\begin{array}{l} = \dfrac{6}{5}.\left( {\dfrac{3}{{13}} - \dfrac{{16}}{{13}}} \right)\\ = \dfrac{6}{5}.\dfrac{{3 - 16}}{{13}}\\ = \dfrac{6}{5}.\dfrac{{-13}}{{13}}\\= \dfrac{6}{5}.\left( { - 1} \right)\\ = \dfrac{{ - 6}}{5}\end{array}\)
Phân số nào nhỏ nhất trong các phân số sau: \(\dfrac{6}{7}\);\(\dfrac{8}{7}\);\(\dfrac{12}{14}\);\(\dfrac{3}{4}\)
A. \(\dfrac{6}{7}\) B. \(\dfrac{8}{7}\) C. \(\dfrac{12}{14}\) D. \(\dfrac{3}{4}\)
Tính:
a) \(\dfrac{-7}{8}\) . \(\dfrac{3}{5}\) - \(\dfrac{2}{5}\) . \(\dfrac{7}{8}\) + \(3\dfrac{7}{8}\)
b) -1,6 : ( 1 + \(\dfrac{2}{3}\) )
c) \(\dfrac{6}{7}\) + \(\dfrac{5}{8}\) : 5 - \(\dfrac{3}{16}\) . ( -2 )\(^2\)
d) \(\dfrac{1^2}{1.2}\) . \(\dfrac{2^2}{2.3}\) . \(\dfrac{3^2}{3.4}\) . \(\dfrac{4^2}{4.5}\)
a: \(=\dfrac{-7}{8}\left(\dfrac{3}{5}+\dfrac{2}{5}\right)+3+\dfrac{7}{8}=\dfrac{-7}{8}+\dfrac{7}{8}+3=3\)
b: \(=-\dfrac{8}{5}:\dfrac{5}{3}=-\dfrac{24}{25}\)
c: \(=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{6}{8}=\dfrac{6}{7}-\dfrac{5}{8}=\dfrac{48}{56}-\dfrac{35}{56}=\dfrac{13}{56}\)