Tính hợp lý giá trị của biểu thức sau:
\(A=49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)
\(B=71\frac{38}{45}-\left(43\frac{8}{45}-1\frac{17}{57}\right)\)
\(C=\frac{-3}{7}.\frac{5}{9}+\frac{4}{9}.\frac{-3}{7}+2\frac{3}{7}\)
\(D=\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right).\frac{4}{5}\)
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e) \(E=0,7.2\frac{2}{3}.20.0,375.\frac{5}{28}\)
\(=\left(0,7.20\right)\left(2\frac{2}{3}.0,375\right)\frac{5}{28}\)
\(=14.1.\frac{5}{28}\)
\(=14.\frac{5}{28}\)
\(=\frac{5}{2}\)
f) \(F=\left(9,75.21\frac{3}{7}+\frac{39}{4}.18\frac{4}{7}\right)\frac{15}{78}\)
\(=\left(\frac{39}{4}.21\frac{3}{7}+\frac{39}{4}.18\frac{4}{7}\right)\frac{15}{78}\)
\(=\frac{39}{4}\left(21\frac{3}{7}+18\frac{4}{7}\right)\frac{15}{78}\)
\(=\frac{39}{4}.40.\frac{15}{78}\)
\(=390.\frac{15}{78}\)
\(=78\)
Chúc bạn học tốt môn Toán!!!
Mình ko ghi đề đâu
\(A=49\frac{8}{23}-5\frac{7}{32}-14\frac{8}{23}=\left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}=35-5\frac{7}{32}=29\frac{25}{32}\)
\(B=71\frac{38}{45}-43\frac{8}{45}+1\frac{17}{57}=\left(71\frac{38}{45}-43\frac{8}{45}\right)+1\frac{17}{57}=28\frac{2}{3}+1\frac{17}{57}=29\frac{55}{57}\)
\(C=-\frac{3}{7}.\left(\frac{5}{9}+\frac{4}{9}\right)+2\frac{3}{7}=-\frac{3}{7}.1+2\frac{3}{7}=-\frac{3}{7}+2\frac{3}{7}=2\)
\(A=49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)
\(A=49\frac{8}{23}-5\frac{7}{32}+14\frac{8}{23}\)
\(A= \left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}\)
\(A=\left[\left(49-14\right)-\left(\frac{8}{23}-\frac{8}{23}\right)\right]-5\frac{7}{32}\)
\(A=\left[35-0\right]-5\frac{7}{32}\)
\(A=35-5\frac{7}{32}\)
\(A=\frac{953}{32}\)
\(B=71\frac{38}{45}-\left(43\frac{38}{45}-1\frac{17}{57}\right)\)
\(B=71\frac{38}{45}-\frac{36377}{855}\)
\(B=\frac{1670}{57}\)
\(C=\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right):\frac{4}{5}\)
\(C=\left[\left(19\frac{5}{8}-13\frac{1}{4}\right):\frac{7}{12}\right]:\frac{4}{5}\)
\(C=\left[\frac{51}{8}:\frac{7}{12}\right]:\frac{4}{5}\)
\(C=\frac{153}{14}:\frac{4}{5}\)
\(C=\frac{765}{56}\)
\(D=\left[\left(\frac{10}{15}-\frac{2}{3}\right):\frac{1}{7}\right]\cdot0,15-\frac{1}{4}\)
\(D=\left[0:\frac{1}{7}\right]\cdot\frac{3}{20}-\frac{1}{4}\)
\(D=0\cdot\frac{3}{20}-\frac{1}{4}\)
\(D=0-\frac{1}{4}\)
\(D=-\frac{1}{4}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot2\frac{1}{2}-\left[\left(\frac{1}{2}+\frac{1}{3}\right):\frac{53}{90}\right]:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\left[\frac{5}{6}:\frac{53}{90}\right]:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\frac{75}{53}:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{14}{9}-\frac{3}{2}\)
\(\)\(E=\frac{22}{45}\)
CHUC BAN HOC TOT >.<
1135/23-167/32+330/23
=(1135/23-330/23)+167/32
=805/23+167/32
=35+167/32
=1120/32+167/32
=1287/32
a: \(A=49+\dfrac{8}{23}-14-\dfrac{8}{23}-5-\dfrac{7}{32}=30-\dfrac{7}{32}=\dfrac{953}{32}\)
b:
Sửa đề: \(B=71\dfrac{38}{45}-\left(43\dfrac{8}{45}-1\dfrac{17}{51}\right)\)
\(B=71+\dfrac{38}{45}-43-\dfrac{8}{45}+1+\dfrac{17}{51}\)
\(=71-43+1+1\)
=28+2=30
\(=49\frac{8}{23}-5\frac{7}{32}-14\frac{8}{23}\)
\(=\left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}\)
\(=35\frac{0}{23}-5\frac{7}{32}\)
\(=\frac{805}{23}-\frac{167}{32}\)
\(\frac{25760}{736}-\frac{3841}{736}\)
\(=\frac{21919}{736}=\frac{953}{32}\)
A= \(49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)
=\(\frac{1135}{32}-\left(\frac{167}{32}+\frac{330}{23}\right)\)
=\(\frac{1135}{23}-\frac{167}{32}-\frac{330}{23}\)
=\(\left(\frac{1135}{23}-\frac{330}{23}\right)-\frac{167}{32}\)
=\(\frac{805}{23}-\frac{167}{32}\)
=\(\frac{953}{32}\)