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\(a,\left(\dfrac{7}{20}+\dfrac{11}{15}-\dfrac{15}{12}\right):\left(\dfrac{11}{20}-\dfrac{26}{45}\right).\)
\(=\left(\dfrac{21}{60}+\dfrac{44}{60}-\dfrac{75}{60}\right):\left(\dfrac{99}{180}-\dfrac{104}{180}\right).\)
\(=\left(\dfrac{65}{60}-\dfrac{75}{60}\right):\left(-\dfrac{5}{180}\right).\)
\(=-\dfrac{10}{60}:\left(-\dfrac{5}{180}\right).\)
\(=-\dfrac{1}{6}:\left(-\dfrac{1}{36}\right).\)
\(=-\dfrac{1}{6}.\left(-36\right).\)
\(=\dfrac{-1.\left(-36\right)}{6}=\dfrac{36}{6}=6.\)
Vậy......
\(b,\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}:\dfrac{15-\dfrac{15}{11}+\dfrac{15}{121}}{16-\dfrac{16}{11}+\dfrac{16}{121}}.\)
\(=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}:\dfrac{15\left(1-\dfrac{1}{11}+\dfrac{1}{121}\right)}{16\left(1-\dfrac{1}{11}+\dfrac{1}{121}\right)}.\)
\(=\dfrac{5}{8}:\dfrac{15}{16}.\)
\(=\dfrac{5}{8}.\dfrac{16}{15}=\dfrac{5.16}{8.15}=\dfrac{1.2}{1.3}=\dfrac{2}{3}.\)
Vậy......
c, (làm tương tự câu b).
~ Học tốt!!! ~
b) \(\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}=\dfrac{5}{8}\)
Vì không có thời gian nên mình chỉ làm câu khó nhất thôi, tick mình nhé
\(a.\)
\(\dfrac{17}{8}:\left(\dfrac{27}{8}+\dfrac{11}{2}\right)\)
\(=\dfrac{17}{8}:\left(\dfrac{27+44}{8}\right)=\dfrac{17}{8}:\dfrac{71}{8}=\dfrac{17}{8}\cdot\dfrac{8}{71}=\dfrac{17}{71}\)
\(b.\)
\(\dfrac{28}{15}\cdot\dfrac{1}{4^2}\cdot3+\left(\dfrac{8}{15}-\dfrac{69}{60}\cdot\dfrac{5}{23}\right):\dfrac{51}{54}\)
\(=\dfrac{28}{15}\cdot\dfrac{1}{4^2}\cdot3+\left(\dfrac{8}{15}-\dfrac{1}{4}\right):\dfrac{51}{54}\)
\(=\dfrac{28}{15}\cdot\dfrac{1}{4^2}\cdot3+\left(\dfrac{8\cdot4-15}{60}\right):\dfrac{51}{54}\)
\(=\dfrac{28}{15}\cdot\dfrac{1}{4^2}\cdot3+\dfrac{17}{60}:\dfrac{51}{54}\)
\(=\dfrac{28}{15}\cdot\dfrac{1}{16}\cdot3+\dfrac{17}{60}\cdot\dfrac{54}{51}\)
\(=\dfrac{7}{20}+\dfrac{3}{10}\)
\(=\dfrac{7+3\cdot2}{20}=\dfrac{13}{20}\)
a) \(\dfrac{-5}{9}+\dfrac{8}{15}+\dfrac{-2}{11}+\dfrac{4}{-9}+\dfrac{7}{15}\)
=\(\left(\dfrac{-5}{9}+\dfrac{-4}{9}\right)+\left(\dfrac{8}{15}+\dfrac{7}{15}\right)+\dfrac{-2}{11}\)
=\(\left(-1\right)+1+\dfrac{-2}{11}\)
=\(\dfrac{-2}{11}\)
b) \(\left(\dfrac{-5}{12}+\dfrac{6}{11}\right)+\left(\dfrac{7}{17}+\dfrac{5}{11}+\dfrac{5}{12}\right)\)
=\(\dfrac{-5}{12}+\dfrac{6}{11}+\dfrac{7}{17}+\dfrac{5}{11}+\dfrac{5}{12}\)
=\(\left(\dfrac{-5}{12}+\dfrac{5}{12}\right)+\left(\dfrac{6}{11}+\dfrac{5}{11}\right)+\dfrac{7}{17}\)
=\(0+0+\dfrac{7}{17}\)
=\(\dfrac{7}{17}\)
c) A= \(49\dfrac{8}{23}-\left(5\dfrac{7}{32}+14\dfrac{8}{23}\right)\)
A=\(49\dfrac{8}{23}-5\dfrac{7}{32}-14\dfrac{8}{23}\)
A=\(\left(49\dfrac{8}{23}-14\dfrac{8}{23}\right)-5\dfrac{7}{32}\)
A=\(35-5\dfrac{7}{32}\)
A=\(35-\dfrac{167}{32}=\dfrac{953}{32}\)
d) C=\(\dfrac{-3}{7}.\dfrac{5}{9}+\dfrac{4}{9}.\dfrac{-3}{7}+2\dfrac{3}{7}\)
C=\(\dfrac{-3}{7}.\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{17}{7}\)
C=\(\dfrac{-3}{7}.1+\dfrac{17}{7}\)
C=\(\dfrac{-3}{7}+\dfrac{17}{7}=2\)
a, `(-5)/9+8/15+(-2)/11+4/(-9)+7/15`
`=-5/9+8/15-2/11-4/9+7/15`
`=(-5/9-4/9)+(8/15+7/15)-2/11`
`=-9/9+15/15-2/11`
`=-1+1-2/11`
`=-2/11`
b, `((-5)/12+6/11)+(7/17+5/11+5/12)`
`=-5/12+6/11+7/17+5/11+5/12`
`=(-5/12+5/12)+(6/11+5/11)+7/17`
`=0+11/11+7/17`
`=1+7/17`
`=17/17+7/17`
`=24/17`
c, `A=49 8/23 - (5 7/32 + 14 8/23)`
`A=49 8/23 - 5 7/32 - 14 8/23`
`A=(49 8/23 - 14 8/23)-5 7/32`
`A=35 - 167/32`
`A=953/32`
d, `C=(-3)/7.5/9+4/9.(-3)/7+2 3/7`
`C=-3/7 . 5/9-4/9 . 3/7+17/7`
`C=-3/7.(5/9+4/9)+17/7`
`C=-3/7 . 1+17/7`
`C=2`
A -\(\dfrac{24}{25}\)
B -\(\dfrac{5}{21}\)
C -\(\dfrac{24}{47}\)
D -\(\dfrac{19}{42}\)
tick cho mk
1. \(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}=\dfrac{2}{4}=\dfrac{1}{2}\)
2. \(B=\dfrac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\dfrac{1}{5}\)
3.\(C=\dfrac{2^2.3^2.\text{4^2.5^2}.5^2}{1.2^2.3^2.4^2.5.6^2}=\dfrac{125}{36}\)
4.D=\(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right).\dfrac{4}{9}.\dfrac{1}{16}=\dfrac{19}{30}.\dfrac{1}{36}=\dfrac{19}{1080}\)
\(a.\)
\(\dfrac{5}{16}-\dfrac{5}{24}=\dfrac{5\cdot3-5\cdot2}{48}=\dfrac{15-10}{48}=\dfrac{5}{48}\)
\(b.\)
\(\dfrac{2}{11}+\left(\dfrac{-5}{11}-\dfrac{9}{11}\right)=\dfrac{2-5-9}{11}=-\dfrac{12}{11}\)
\(c.\)
\(\dfrac{1}{10}-\left(\dfrac{5}{12}-\dfrac{1}{15}\right)=\dfrac{1}{10}-\dfrac{5}{12}+\dfrac{1}{15}=\dfrac{6-5\cdot5+4}{60}=-\dfrac{15}{60}=-\dfrac{1}{4}\)
a) \(\dfrac{5}{16}-\dfrac{5}{24}=\dfrac{15}{48}-\dfrac{10}{48}=\dfrac{15-10}{48}=\dfrac{5}{48}\)
b)\(\dfrac{2}{11}+\left(\dfrac{-5}{11}-\dfrac{9}{11}\right)=\dfrac{2}{11}-\dfrac{5}{11}-\dfrac{9}{11}=\dfrac{2-5-9}{11}=\dfrac{-12}{11}\)
c)\(\dfrac{1}{10}-\left(\dfrac{5}{12}-\dfrac{1}{15}\right)=\dfrac{1}{10}-\dfrac{7}{20}=\dfrac{2}{20}-\dfrac{7}{20}=\dfrac{-5}{20}=\dfrac{-1}{4}\)
a)\(\dfrac{-10}{11}.\dfrac{8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
=\(\dfrac{10}{11}.\dfrac{-8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
=\(\dfrac{10}{11}(\dfrac{-8}{9}+\dfrac{7}{18})\)
=\(\dfrac{10}{11}.\dfrac{-1}{2}\)
=\(\dfrac{-5}{11}\)
b;
B = \(\dfrac{3}{14}\) : \(\dfrac{1}{28}\) - \(\dfrac{13}{21}\): \(\dfrac{1}{28}\) + \(\dfrac{29}{42}\) : \(\dfrac{1}{28}\) - 8
B = (\(\dfrac{3}{14}\) - \(\dfrac{13}{21}\) + \(\dfrac{29}{42}\)) - 8
B = (\(\dfrac{9}{42}\) - \(\dfrac{26}{42}\) + \(\dfrac{29}{42}\)) - 8
B = (\(\dfrac{-17}{42}\) + \(\dfrac{29}{42}\)) - 8
B = \(\dfrac{2}{7}\) - 8
B = \(\dfrac{2}{7}-\dfrac{56}{7}\)
B = - \(\dfrac{54}{7}\)
a: \(\dfrac{\left(\dfrac{1}{2}-0,75\right)\cdot\left(0,2-\dfrac{2}{5}\right)}{\dfrac{5}{9}-1\dfrac{1}{12}}\)
\(=\dfrac{\left(0,5-0,75\right)\left(0,2-0,4\right)}{\dfrac{5}{9}-\dfrac{13}{12}}=\dfrac{-0,25\cdot\left(-0,2\right)}{\dfrac{20}{36}-\dfrac{39}{36}}\)
\(=\dfrac{0,05}{-\dfrac{19}{36}}=0,05\cdot\dfrac{-36}{19}=\dfrac{-1,8}{19}=\dfrac{-9}{95}\)
b: Sửa đề: \(\dfrac{5-\dfrac{5}{3}-\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}-\dfrac{8}{9}-\dfrac{8}{27}}:\dfrac{15-\dfrac{15}{11}+\dfrac{15}{121}}{16-\dfrac{16}{11}+\dfrac{16}{121}}\)
\(=\dfrac{5\left(1-\dfrac{1}{3}-\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}-\dfrac{1}{9}-\dfrac{1}{27}\right)}:\dfrac{15\left(1-\dfrac{1}{11}+\dfrac{1}{121}\right)}{16\left(1-\dfrac{1}{11}+\dfrac{1}{121}\right)}\)
\(=\dfrac{5}{8}:\dfrac{15}{16}=\dfrac{5}{8}\cdot\dfrac{16}{15}=\dfrac{80}{120}=\dfrac{2}{3}\)