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Bài 8: Tính bằng cách thuận tiện nhất:
a, 6 x 31 x 5 =(6x5)x31
=30x31
=930
b, 8 x 63 x 5 =(8x5)x63
=40x63
=2520
a.
6 x 31 x 5
= (6 x 5) x 31
= 30 x 31
= 930
b.
8 x 63 x 5
= (8 x 5) x 63
= 40 x 63
= 2520
a) \(6-\dfrac{2}{5}+\dfrac{3}{5}\) x \(3\)
\(=6-\dfrac{2}{5}+\dfrac{9}{5}\)
\(=\dfrac{30}{5}-\dfrac{2}{5}+\dfrac{9}{5}\)
\(=\dfrac{30-2+9}{5}=\dfrac{37}{5}\)
b) \(\left(\dfrac{31}{35}-\dfrac{4}{7}\right)\) x \(\dfrac{8}{7}:2\)
\(=\left(\dfrac{31}{35}-\dfrac{20}{35}\right)\) x \(\dfrac{8}{7}:2\)
\(=\dfrac{11}{35}\) x \(\dfrac{8}{7}:2\)
\(=\dfrac{11}{35}\) x \(\dfrac{8}{7}\) x \(\dfrac{1}{2}\)
\(=\dfrac{\text{11 x 8 x 1}}{\text{ 35 x 7 x 2}}=\dfrac{88}{490}=\dfrac{88}{245}\)
a) 6−25+356−25+35 x 33
=6−25+95=6−25+95
=305−25+95=305−25+95
=30−2+95=375=30−2+95=375
b) (3135−47)(3135−47) x 87:287:2
=(3135−2035)=(3135−2035) x 87:287:2
=1135=1135 x 8
Hướng dẫn giải:
a) 4 x 78 x 5 = (4 x 5) x 78 = 20 x 78 = 1560
b) 2 x 99 x 5 = (2 x 5) x 99 = 10 x 99 = 990
c) 6 x 31 x 5 = (6 x 5) x 31 = 30 x 31 = 930
d) 8 x 63 x 5 = (8 x 5) x 63 = 40 x 63 = 2520
\(\frac{6-2}{5}+\frac{3}{5.3}\)
\(=\frac{4}{5}+\frac{1}{5.1}\)
\(=\frac{4}{5}+\frac{1}{5}\)
\(=\frac{5}{5}=1\)
a. 6 x 3 - ( \(\frac{2}{5}+\frac{3}{5}\)) = 6 x 3 - \(\frac{5}{5}\)= 6 x 3 - 1 = 18 - 1 = 17
b.( \(\frac{31}{35}-\frac{4}{7}\)) x \(\frac{8}{7}\): 2 =\(\frac{11}{35}\)x \(\frac{8}{7}\): 2 =\(\frac{88}{245}\): 2 = \(\frac{88}{490}\)= \(\frac{44}{245}\)= ...........rút gọn đến khi hết
a) \(6-\frac{2}{5}+\frac{3}{5}\times3\)
\(=6-\frac{2}{5}+\frac{9}{5}\)
\(=\frac{30}{5}-\frac{2}{5}+\frac{9}{5}\)
\(=\frac{28}{5}+\frac{9}{5}\)
\(=\frac{37}{5}\)
b) \(\left(\frac{31}{35}-\frac{4}{7}\right)\times\frac{8}{7}:2\)
\(=\frac{11}{35}\times\frac{8}{7}:2\)
\(=\frac{88}{245}:2\)
\(=\frac{44}{245}\)
a) \(\dfrac{2}{5}+\dfrac{11}{15}=\dfrac{6}{15}+\dfrac{11}{15}=\dfrac{17}{15}\)
b) \(\dfrac{7}{8}-\dfrac{7}{9}=\dfrac{63}{72}-\dfrac{56}{72}=\dfrac{7}{72}\)
c) \(\dfrac{11}{13}\cdot\dfrac{26}{31}=\dfrac{26}{13}\cdot\dfrac{11}{31}=\dfrac{22}{31}\)
d) \(\dfrac{1}{2}:\dfrac{1}{3}\cdot\dfrac{2}{5}=\dfrac{1}{2}\cdot3\cdot\dfrac{2}{5}=\dfrac{3}{5}\)
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