\(\left(\frac{11}{12}:\frac{33}{16}\right).\frac{3}{5}\)

\(=\frac{11}{12}.\frac{16}{33}.\frac{3}{5}\)

\(=\frac{11.16.3}{12.33.5}\)

\(=\frac{11.4.4.3}{4.3.3.11.5}\)

\(=\frac{4}{15}\)

\(\left(\frac{11}{12}:\frac{33}{16}\right).\frac{3}{5}\)
\(\left(\frac{11}{12}.\frac{16}{33}\right).\frac{3}{5}\)
\(\frac{11}{12}.\frac{16}{33}.\frac{3}{5}=\frac{11.16.3}{12.33.5}=\frac{16}{4.3.5}=\frac{4}{3.5}=\frac{4}{15}\)

~Chúc bạn học giỏi~
                                                                           ~TMT_Nhók

Cách 1:
\(A=\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)
  \(=\left(\frac{36}{6}-\frac{4}{6}+\frac{3}{6}\right)-\left(\frac{30}{6}+\frac{10}{6}-\frac{9}{6}\right)-\left(\frac{18}{6}-\frac{14}{6}+\frac{15}{6}\right)\)
  \(=\frac{35}{6}-\frac{31}{6}-\frac{19}{6}\)
  \(=-\frac{15}{6}\)
  \(=-\frac{5}{2}\)

Cách 2:
\(A=\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)
  \(=6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)
  \(=\left(6-5-3\right)+\left(-\frac{2}{3}-\frac{5}{3}+\frac{7}{3}\right)+\left(\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\right)\)
  \(=-2+0-\frac{1}{2}\)
  \(=-\frac{4}{2}-\frac{1}{2}\)
  \(=-\frac{5}{2}\)

C1
23+12)−(5+53−32)−(3−73+52)=(6.66−2.26+36)−(5.66+5.26−3.36)−(3.66−7.26+5.36)=36−4+36−30+10−96−18−14+156=356−316−196=35−31−196=−156=−52=−212.

23+12)−(5+53−32)−(3−73+52)=6−23+12−5−53+32−3+73−52=(6−5−3)+(−23−53+73)+(12+32−52)=−2+−2−5+73+1+3−52=−2+0−12=−52=−212." role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:16.94px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">A=(6−23+12)−(5+53−32)−(3−73+52)=6−23+12−5−53+32−3+73−52=(6−5−3)+(−23−53+73)+(12+32−52)=−2+−2−5+73+1+3−52=−2+0−12=−52=−212

Bài này đúng ko

Em hãy tìm cách " nối" các số ở những chiếc là bằng dấu các phép tính cộng, trừ, nhân, chia và dấu ngoặc để được một biểu thức có giá trị đúng bằng số ở bông hoa?

Giải

Có nhiều cách nối, chẳng hạn:

    4.(-25) + 10 : (-2)

= -100 + (-5)

= -105

\(\frac{1}{2}\) (-100) - 5,6 : 8

= -50 - 0,7

= -50 + (-0,7)

= -50,7 

(14,78-a)/(2,87+a)=4/1

14,78+2,87=17,65

Tổng số phần bằng nhau là 4+1=5

Mỗi phần có giá trị bằng 17,65/5=3,53

=>2,87+a=3,53

=>a=0,66.

Ừ, bài trong SGK trang 13

\(\left(-2\right).\frac{-38}{21}.\frac{-7}{4}.\frac{-3}{8}\)

\(=\left(-2.\frac{-7}{4}.\frac{-3}{8}\right).\frac{-38}{21}\)

\(=\left(\frac{7}{2}.\frac{-3}{8}\right).\frac{-38}{21}\)

\(=\frac{-21}{16}.\frac{-38}{21}\)

\(=\frac{-38}{-16}=\frac{19}{8}\)

\(\left(-2\right).\frac{-38}{21}.\frac{-7}{4}.\left(\frac{-3}{8}\right)\)

\(=\left(-2.\frac{-7}{2}.\frac{-3}{8}\right).\frac{-38}{21}\)

\(=\left(7.\frac{-3}{8}\right).\frac{-38}{21}\)

\(=\frac{-21}{8}.\frac{-38}{21}\)

\(=\frac{-38}{-8}=\frac{19}{4}\)

\(A=49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)

\(A=49\frac{8}{23}-5\frac{7}{32}+14\frac{8}{23}\)

\(A= \left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}\)

\(A=\left[\left(49-14\right)-\left(\frac{8}{23}-\frac{8}{23}\right)\right]-5\frac{7}{32}\)

\(A=\left[35-0\right]-5\frac{7}{32}\)

\(A=35-5\frac{7}{32}\)

\(A=\frac{953}{32}\)

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

\(B=71\frac{38}{45}-\frac{36377}{855}\)

\(B=\frac{1670}{57}\)

\(C=\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right):\frac{4}{5}\)

\(C=\left[\left(19\frac{5}{8}-13\frac{1}{4}\right):\frac{7}{12}\right]:\frac{4}{5}\)

\(C=\left[\frac{51}{8}:\frac{7}{12}\right]:\frac{4}{5}\)

\(C=\frac{153}{14}:\frac{4}{5}\)

\(C=\frac{765}{56}\)

\(D=\left[\left(\frac{10}{15}-\frac{2}{3}\right):\frac{1}{7}\right]\cdot0,15-\frac{1}{4}\)

\(D=\left[0:\frac{1}{7}\right]\cdot\frac{3}{20}-\frac{1}{4}\)

\(D=0\cdot\frac{3}{20}-\frac{1}{4}\)

\(D=0-\frac{1}{4}\)

\(D=-\frac{1}{4}\)

\(E=\frac{13}{30}+\frac{28}{45}\cdot2\frac{1}{2}-\left[\left(\frac{1}{2}+\frac{1}{3}\right):\frac{53}{90}\right]:\frac{50}{53}\)

\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\left[\frac{5}{6}:\frac{53}{90}\right]:\frac{50}{53}\)

\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\frac{75}{53}:\frac{50}{53}\)

\(E=\frac{13}{30}+\frac{14}{9}-\frac{3}{2}\)

\(\)\(E=\frac{22}{45}\)

CHUC BAN HOC TOT >.<

bai de the ma cung hoi

a)

\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right).\\A = \left( {\frac{{30}}{{15}} + \frac{5}{{15}} - \frac{6}{{15}}} \right) - \left( {\frac{{105}}{{15}} - \frac{9}{{15}} - \frac{{20}}{{15}}} \right) - \left( {\frac{3}{{15}} + \frac{{25}}{{15}} - \frac{{60}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} - \left( {\frac{{ - 32}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} + \frac{{32}}{{15}}\\A = \frac{{ - 15}}{{15}}\\A =  - 1\end{array}\)

b)

\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right)\\A = 2 + \frac{1}{3} - \frac{2}{5} - 7 + \frac{3}{5} + \frac{4}{3} - \frac{1}{5} - \frac{5}{3} + 4\\A = \left( {2 - 7 + 4} \right) + \left( {\frac{1}{3} + \frac{4}{3} - \frac{5}{3}} \right) + \left( { - \frac{2}{5} + \frac{3}{5} - \frac{1}{5}} \right)\\A =  - 1 + 0 + 0 =  - 1\end{array}\)

(-2).-38/21.-7/4.-3/8

=19/8

bài nãy dễ mà bạn

\(\left(-2\right).\frac{-38}{21}.\frac{-7}{4}.\frac{\left(-3\right)}{8}=\frac{-2.\left(-38\right).\left(-7\right).\left(-3\right)}{21.4.8}=\frac{4.19.21}{21.4.8}=\frac{19}{8}\)