A = ( 6 - \(\dfrac{2}{3}\) + \(\dfrac{1}{2}\)) - ( 5 + \(\dfrac{5}{3}\) - \(\dfrac{3}{2}\)) - (3 - \(\dfrac{7}{3}\) + \(\dfrac{5}{2}\))
a, \(\dfrac{1}{3}-\dfrac{1}{4}:\dfrac{2}{5}\)
b, \(\dfrac{6}{7}-\left(\dfrac{5}{6}+\dfrac{1}{3}\right)-\left(\dfrac{2}{3}+\dfrac{1}{7}\right)\)
c, \(\dfrac{-5}{9}.\dfrac{2}{5}+4\dfrac{5}{9}+\dfrac{5}{9}.\dfrac{-3}{5}\)
d, \(3\dfrac{1}{2}-\left(5\dfrac{4}{7}-1\dfrac{1}{2}\right):0,75\)
a) `1/3 - 1/4 : 2/5 = 1/3 - 5/8 = -7/24`
b) `6/7-(5/6+1/3)-(2/3+1/7) = 6/7-5/6-1/3-2/3-1/7`
`=(6/7-1/7)-(1/3+2/3)-5/6`
`=5/7-1-5/6`
`=-47/42`
c) `-5/9 . 2/5 + 4 5/9 + 5/9 . (-3/5)`
`= -5/9 . 2/5 + 4 + 5/9 + (-5/9) . 3/5`
`=-5/9 . (2/5 + 3/5-1) + 4`
`=-5/9 . 0 +4`
`=4`
d) 3 1/2 - (5 4/7 - 1 1/2) : 0,75`
`=7/2 - (39/7 - 3/2) : 3/4`
`= 7/2 - 57/14 : 3/4`
`=7/2 - 38/7`
`=-27/14`
Tính A= \(\dfrac{1}{2}\)- \(\dfrac{2}{3}\)+\(\dfrac{3}{4}\)-\(\dfrac{4}{5}\)+\(\dfrac{5}{6}\)-\(\dfrac{6}{7}\)-\(\dfrac{6}{5}\)+\(\dfrac{4}{5}\)-\(\dfrac{3}{4}\)+\(\dfrac{2}{3}\)-\(\dfrac{1}{2}\)
\(\dfrac{1}{2}-\dfrac{2}{3}+\dfrac{3}{4}-\dfrac{4}{5}+\dfrac{5}{6}-\dfrac{6}{7}-\dfrac{6}{5}+\dfrac{4}{5}-\dfrac{3}{4}+\dfrac{2}{3}-\dfrac{1}{2}\)
\(=\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(-\dfrac{2}{3}+\dfrac{2}{3}\right)+\left(\dfrac{3}{4}-\dfrac{3}{4}\right)+\left(-\dfrac{4}{5}+\dfrac{4}{5}\right)+\left(\dfrac{5}{6}-\dfrac{6}{7}-\dfrac{6}{5}\right)\)
\(=0+0+0+0-\dfrac{257}{210}\)
\(=\dfrac{257}{210}\)
Có ai biết câu này không, làm giúp mình với
a,\(\dfrac{2}{3}\)x\(\dfrac{5}{2}\):\(\dfrac{9}{5}\)
b,\(\dfrac{1}{3}\)x\(\dfrac{1}{4}\)+\(\dfrac{5}{6}\)
c,\(\dfrac{1}{2}\)-\(\dfrac{7}{8}\):\(\dfrac{7}{4}\)
d,\(\dfrac{6}{5}\)-\(\dfrac{4}{5}\)x\(\dfrac{3}{2}\)
\(\dfrac{3}{5}+\dfrac{a}{b}=5\) \(\dfrac{a}{b}-\dfrac{4}{7}=\dfrac{5}{6}\) \(\dfrac{2}{3}\) x \(\dfrac{a}{b}=\dfrac{3}{5}\)
\(\dfrac{a}{b}:\dfrac{2}{7}=3+\dfrac{2}{3}\) \(\dfrac{7}{5}-\dfrac{a}{b}=\dfrac{2}{5}:2\)
\(\dfrac{3}{5}:\dfrac{a}{b}=\dfrac{2}{7}\) ÉT O ÉT
a)\(\dfrac{a}{b}=5-\dfrac{3}{5}=\dfrac{25}{5}-\dfrac{3}{5}=\dfrac{22}{5}\)
b)\(\dfrac{a}{b}=\dfrac{5}{6}+\dfrac{4}{7}=\dfrac{35}{42}+\dfrac{24}{42}=\dfrac{59}{42}\)
c)\(\dfrac{a}{b}=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{3}{5}\times\dfrac{3}{2}=\dfrac{9}{10}\)
d)\(\dfrac{a}{b}=3\times\dfrac{2}{7}=\dfrac{6}{7}\)
e)\(\dfrac{a}{b}=\dfrac{7}{5}-\left(\dfrac{2}{5}\times\dfrac{1}{2}\right)=\dfrac{7}{5}-\dfrac{1}{5}=\dfrac{6}{5}\)
a)\(0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}-\dfrac{1}{6}-\dfrac{4}{35}\)
b)\(\left(3-\dfrac{1}{4}+\dfrac{2}{3}\right)-\left(5+\dfrac{1}{3}-\dfrac{6}{5}\right)-\left(-6-\dfrac{7}{4}+\dfrac{3}{2}\right)\)
c)\(\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)
\(a,0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}-\dfrac{1}{6}-\dfrac{4}{35}\\ =\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{5}+\dfrac{5}{7}-\dfrac{1}{6}-\dfrac{4}{35}\\ =\dfrac{5}{6}+\dfrac{39}{35}-\dfrac{1}{6}-\dfrac{4}{35}\\ =\left(\dfrac{5}{6}-\dfrac{1}{6}\right)+\left(\dfrac{39}{35}-\dfrac{4}{35}\right)\\ =\dfrac{2}{3}+1\\ =\dfrac{4}{3}.\)
\(b,\left(3-\dfrac{1}{4}+\dfrac{2}{3}\right)-\left(5+\dfrac{1}{3}-\dfrac{6}{5}\right)-\left(-6-\dfrac{7}{4}+\dfrac{3}{2}\right)\\ =3-\dfrac{1}{4}+\dfrac{2}{3}-5-\dfrac{1}{3}+\dfrac{6}{5}+6+\dfrac{7}{4}-\dfrac{3}{2}\\ =\left(3-5+6\right)+\left(-\dfrac{1}{4}+\dfrac{7}{4}\right)+\left(\dfrac{2}{3}-\dfrac{1}{3}\right)+\left(\dfrac{6}{5}+\dfrac{7}{4}\right)\\ =4-\dfrac{3}{2}+\dfrac{1}{3}+\dfrac{59}{20}\\ =\dfrac{5}{2}+\dfrac{1}{3}+\dfrac{59}{20}\\ =\dfrac{17}{6}+\dfrac{59}{20}\\ =\dfrac{347}{60}.\)
\(c,\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\\ =\dfrac{1}{3}+\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\\ =\left(\dfrac{1}{3}-\dfrac{2}{9}\right)+\left(\dfrac{3}{4}-\dfrac{1}{36}\right)+\left(\dfrac{3}{5}+\dfrac{1}{15}\right)+\dfrac{1}{64}\\ =\dfrac{1}{9}+\dfrac{13}{18}+\dfrac{2}{3}+\dfrac{1}{64}\\ =\dfrac{3}{2}+\dfrac{1}{64}\\ =\dfrac{65}{64}.\)
Đ, S?
a) \(\dfrac{1}{2}-\dfrac{1}{8}=\dfrac{3}{8}\) b) \(\dfrac{7}{10}-\dfrac{1}{5}=\dfrac{6}{5}\)
c) \(\dfrac{5}{4}+\dfrac{5}{12}=\dfrac{5}{16}\) d) \(\dfrac{3}{6}+\dfrac{2}{3}=\dfrac{7}{6}\)
a) Đ
b) S
\(\dfrac{7}{10}-\dfrac{1}{5} \\ =\dfrac{7}{10}-\dfrac{2}{10}\\ =\dfrac{5}{10}=\dfrac{1}{2}\)
c) S
\(\dfrac{5}{4}+\dfrac{5}{12}\\ =\dfrac{15}{12}+\dfrac{5}{12}\\ =\dfrac{20}{12}=\dfrac{5}{3}\)
d) Đ
Tính giá trị biểu thức :
\(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\left(\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}\right)-\left(\dfrac{1}{5}+\dfrac{2}{5}+\dfrac{3}{5}+\dfrac{4}{5}\right)+\left(\dfrac{1}{6}+\dfrac{2}{6}+\dfrac{3}{6}+\dfrac{4}{6}+\dfrac{5}{6}\right)-\left(\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{3}{7}+\dfrac{4}{7}+\dfrac{5}{7}+\dfrac{6}{7}\right)+...+\left(100+...+\dfrac{99}{100}\right)\)
b) \(\dfrac{2}{5}\)+\(\dfrac{3}{5}\):(\(\dfrac{3}{5}\)+\(\dfrac{-2}{3}\))\(-3\dfrac{1}{2}\)
c) (\(4-\dfrac{5}{12}\)):3+\(\dfrac{7}{36}\)
d) (\(2+\dfrac{5}{6}\)):\(1\dfrac{1}{5}\)+\(\dfrac{-7}{12}\)
b: \(=\dfrac{2}{5}+\dfrac{3}{5}:\dfrac{9-10}{15}-\dfrac{7}{2}\)
\(=\dfrac{4-35}{10}+\dfrac{3}{5}\cdot\dfrac{15}{-1}\)
\(=\dfrac{-31}{10}-9=\dfrac{-31}{10}-\dfrac{90}{10}=-\dfrac{121}{10}\)
c: \(=\dfrac{48-5}{12}\cdot\dfrac{1}{3}+\dfrac{7}{36}=\dfrac{43}{36}+\dfrac{7}{36}=\dfrac{50}{36}=\dfrac{25}{18}\)
d: \(=\dfrac{17}{6}:\dfrac{6}{5}+\dfrac{-7}{12}\)
\(=\dfrac{85}{36}-\dfrac{7}{12}=\dfrac{85}{36}-\dfrac{21}{36}=\dfrac{64}{36}=\dfrac{16}{9}\)
Bài 7: Tính
a) \(4\dfrac{2}{5}x8\dfrac{3}{4}-2\dfrac{3}{4}=...............................\)
b) \(2\dfrac{2}{3}+1\dfrac{2}{5}-\dfrac{2}{15}\)=..............................................
c) \(3\dfrac{1}{3}-2\dfrac{2}{3}+1\dfrac{5}{6}=...............................\)
Bài 7: Tính
a) \(4\dfrac{2}{5}\times8\dfrac{3}{4}-2\dfrac{3}{4}\)
\(=\dfrac{22}{5}\times\dfrac{35}{4}-\dfrac{11}{4}\)
\(=\dfrac{77}{2}-\dfrac{11}{4}\)
\(=\dfrac{143}{4}\)
b) \(2\dfrac{2}{3}+1\dfrac{2}{5}-\dfrac{2}{15}\)
\(=\dfrac{8}{3}+\dfrac{7}{5}-\dfrac{2}{15}\)
\(=\dfrac{47}{15}-\dfrac{2}{15}\)
\(=\dfrac{45}{15}=3\)
c) \(3\dfrac{1}{3}-2\dfrac{2}{3}+1\dfrac{5}{6}\)
\(=\dfrac{10}{3}-\dfrac{8}{3}+\dfrac{11}{6}\)
\(=\dfrac{2}{3}+\dfrac{11}{6}\)
\(=\dfrac{15}{6}\)
1. Tính :
a, \(A=\dfrac{\dfrac{1}{3}-\dfrac{5}{2}}{\dfrac{3}{4}-\dfrac{1}{2}}.\dfrac{\dfrac{5}{6}+\dfrac{7}{3}}{1-\dfrac{5}{6}}.\dfrac{\dfrac{-2}{5}+1}{\dfrac{2}{5}-1}\).
b, \(B=\dfrac{\dfrac{1}{3}-\dfrac{4}{5}}{\dfrac{1}{3}+\dfrac{4}{5}}.\dfrac{\dfrac{3}{4}-\dfrac{5}{3}}{\dfrac{3}{4}+\dfrac{5}{3}}:\dfrac{\dfrac{4}{5}-1}{1-\dfrac{2}{3}}\).