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b)(1/3+12/67+13/41)-(79/67-28/41)
=1/3+12/67+13/41-79/67+28/41
=1/3+(12/67-79/67)+(13/41+28/41)
=1/3+(-67/67)+41/41
=1/3+(-1)+1
=1/3+0
=1/3.
\(\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{30}{31}\right)\times\left(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}\right)\)
\(=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{30}{31}\right)\times\left(\frac{-5}{12}+\frac{3}{12}+\frac{2}{12}\right)\)
\(=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{30}{31}\right)\times\left(\frac{-5+3+2}{12}\right)\)
\(=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{30}{31}\right)\times0\)
\(=0\)
_Chúc bạn học tốt_
\(\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{30}{31}\right)\times\left(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}\right)\)
\(=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{30}{31}\right)\times\left(\frac{-5}{12}+\frac{3}{12}+\frac{2}{12}\right)\)
\(=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{30}{31}\right)\times\left(\frac{-5+3+2}{12}\right)\)
\(=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{30}{31}\right)\times0\)
\(=0\)
_Chúc bạn học tốt_
a,\(\frac{31}{1000}\)
b,\(\frac{8}{108}\)
c,0
a,\(\frac{49}{97}\)
b,\(\frac{-1}{4751}\)
Câu 1 có vế sau = 0
Câu 2 có vế trước = 0
Một biểu thức nhân với 0 thì = 0, nên:
=> kết quả hai bài đều = 0
\(C=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right)\cdot\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(C=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right)\cdot0\)
\(C=0\)
\(C=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right)x\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
Ta thấy 1/2-1/3-1/6=1/6-1/6=0
\(\Rightarrow C=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right)x0\)
\(\Rightarrow C=0\)
Vậy...............
\(17\frac{2}{31}-\left(\frac{15}{17}+6\frac{2}{31}\right)\)
\(17\frac{2}{31}=\frac{529}{31};6\frac{2}{31}=\frac{188}{31}\)
\(17\frac{2}{31}-\left(\frac{15}{17}+6\frac{2}{31}\right)\)
\(=\frac{529}{31}-\left(\frac{15}{17}+\frac{188}{31}\right)\)
\(=\frac{529}{31}-\frac{3661}{527}\)
\(=\frac{172}{17}\)
\(\left(\frac{29}{31}-\frac{7}{8}\right)-\left(\frac{28}{31}-4\right)\)
\(=\frac{15}{248}-\left(-\frac{96}{31}\right)\)
\(=\frac{783}{248}\)
\(A=49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)
\(A=49\frac{8}{23}-5\frac{7}{32}+14\frac{8}{23}\)
\(A= \left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}\)
\(A=\left[\left(49-14\right)-\left(\frac{8}{23}-\frac{8}{23}\right)\right]-5\frac{7}{32}\)
\(A=\left[35-0\right]-5\frac{7}{32}\)
\(A=35-5\frac{7}{32}\)
\(A=\frac{953}{32}\)
\(B=71\frac{38}{45}-\left(43\frac{38}{45}-1\frac{17}{57}\right)\)
\(B=71\frac{38}{45}-\frac{36377}{855}\)
\(B=\frac{1670}{57}\)
\(C=\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right):\frac{4}{5}\)
\(C=\left[\left(19\frac{5}{8}-13\frac{1}{4}\right):\frac{7}{12}\right]:\frac{4}{5}\)
\(C=\left[\frac{51}{8}:\frac{7}{12}\right]:\frac{4}{5}\)
\(C=\frac{153}{14}:\frac{4}{5}\)
\(C=\frac{765}{56}\)
\(D=\left[\left(\frac{10}{15}-\frac{2}{3}\right):\frac{1}{7}\right]\cdot0,15-\frac{1}{4}\)
\(D=\left[0:\frac{1}{7}\right]\cdot\frac{3}{20}-\frac{1}{4}\)
\(D=0\cdot\frac{3}{20}-\frac{1}{4}\)
\(D=0-\frac{1}{4}\)
\(D=-\frac{1}{4}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot2\frac{1}{2}-\left[\left(\frac{1}{2}+\frac{1}{3}\right):\frac{53}{90}\right]:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\left[\frac{5}{6}:\frac{53}{90}\right]:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\frac{75}{53}:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{14}{9}-\frac{3}{2}\)
\(\)\(E=\frac{22}{45}\)
CHUC BAN HOC TOT >.<
a) \(\frac{31}{23}-\left(\frac{7}{32}+\frac{8}{23}\right)=\frac{31}{23}-\frac{7}{32}-\frac{8}{23}=1-\frac{7}{32}=\frac{25}{32}\)
b) \(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
\(=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)
\(=\frac{1}{3}-\left(\frac{79}{67}-\frac{12}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\)
\(=\frac{1}{3}-1+1=\frac{1}{3}\)
d) \(\frac{1}{7}.\frac{1}{3}+\frac{1}{7}.\frac{-1}{3}+\frac{17}{19}=\frac{1}{7}.\left(\frac{1}{3}-\frac{1}{3}\right)+\frac{17}{19}=\frac{17}{19}\)
e) \(\frac{3}{5}.\frac{7}{9}+\frac{7}{5}.\frac{2}{9}=\frac{7}{5}.\left(\frac{3}{9}+\frac{2}{9}\right)=\frac{7}{5}.\frac{5}{9}=\frac{7}{9}\)
a,b you cứ tính bt nhé
c)\(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\)
\(=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(=\frac{1}{4}-\frac{1}{11}\)
\(=\frac{7}{44}\)
d) \(\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}+\frac{5^2}{26.31}\)
\(=5\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}+\frac{5}{26.31}\right)\)
\(=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}+\frac{1}{26}-\frac{1}{31}\right)\)
\(=5\left(1-\frac{1}{31}\right)\)
\(=5.\frac{30}{31}\)
\(=\frac{150}{31}\)
\(\left( \frac{17}{28} + \frac{18}{29} - \frac{19}{30} - \frac{20}{31} \right) \left( \frac{-5}{12} + \frac{1}{4} + \frac{1}{6} \right)\)
ta có \(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}=\frac{-5}{12}+\frac{3}{12}+\frac{2}{12}\)
\(=\frac{-5 + 3 + 2}{12}=\frac{0}{12}=0\)
\(\Rightarrow\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right)\cdot0=0\)
bạn tính cái ngoặc thứ hai á,xong một hồi nó ra 0 thì mình có cái quy luật là cái lồ lố lô j nhân với 0 cũng=0-->tích trên =0
mình có đáp án từ vừa nãy r nhưng vẫn c ơn bn nhé!