1+2+3+4+5+6=
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1: =1/8*9/4=9/32
2: =8/27*243/32=9/4
3: =(5/4*4/5)^5*(4/5)^2=16/25
4: \(=\left(-\dfrac{5}{6}\cdot\dfrac{6}{5}\right)^2\cdot\left(\dfrac{6}{5}\right)^2=\dfrac{36}{25}\)
5: \(=\left(-\dfrac{4}{3}\right)^3\cdot\left(\dfrac{3}{4}\right)^{10}=\left(-1\right)\left(\dfrac{3}{4}\right)^7=-\left(\dfrac{3}{4}\right)^7\)
6: \(=\left(\dfrac{1}{3}\cdot\dfrac{-9}{2}\right)^4\left(-\dfrac{9}{2}\right)^2=\left(-\dfrac{3}{2}\right)^4\cdot\dfrac{81}{4}=\dfrac{9}{4}\cdot\dfrac{81}{4}=\dfrac{729}{16}\)
8: =(0,2*5)^4*5^2=25
10: =-0,5^5*2^10
=-0,5^5*2^5*2^5
=-32
13: =(0,5*2)^2*2^2=4
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\(\dfrac{2}{5}+\dfrac{2}{3}+\dfrac{2}{4}\)
= \(\dfrac{24}{60}\) + \(\dfrac{40}{60}\) + \(\dfrac{30}{60}\)
= \(\dfrac{64}{60}\) + \(\dfrac{30}{60}\)
= \(\dfrac{47}{30}\)
\(\dfrac{2}{6}+\dfrac{3}{12}\)
= \(\dfrac{4}{12}\) + \(\dfrac{3}{12}\)
= \(\dfrac{7}{12}\)
\(\dfrac{5}{6}\) + \(\dfrac{1}{3}\)
= \(\dfrac{5}{6}\) + \(\dfrac{2}{6}\)
= \(\dfrac{7}{6}\)
\(\dfrac{1}{3}\) + \(\dfrac{5}{12}\) + \(\dfrac{5}{6}\)
= \(\dfrac{4}{12}\) + \(\dfrac{5}{12}\) + \(\dfrac{10}{12}\)
= \(\dfrac{9}{12}\) + \(\dfrac{10}{12}\)
= \(\dfrac{19}{12}\)
\(\dfrac{5}{8}\) + \(\dfrac{4}{7}\)
= \(\dfrac{35}{56}\) + \(\dfrac{32}{56}\)
= \(\dfrac{67}{56}\)
\(\dfrac{7}{3}\) + \(\dfrac{8}{7}\)
= \(\dfrac{49}{21}\) + \(\dfrac{24}{21}\)
= \(\dfrac{73}{21}\)
\(\dfrac{1}{5}+\dfrac{5}{35}\)
= \(\dfrac{7}{35}\) + \(\dfrac{5}{35}\)
= \(\dfrac{12}{35}\)
a) \(12\dfrac{1}{3}-\left(3\dfrac{3}{4}+4\dfrac{3}{4}\right)=\dfrac{37}{3}-\left(\dfrac{15}{4}+\dfrac{19}{4}\right)\)
\(=\dfrac{37}{3}-\dfrac{34}{4}=\dfrac{37}{3}-\dfrac{17}{2}=\dfrac{74}{6}-\dfrac{51}{6}=\dfrac{23}{6}\)
b) \(3\dfrac{5}{6}+2\dfrac{1}{6}.6=\dfrac{23}{6}+\dfrac{13}{6}.6=\dfrac{23}{6}+\dfrac{78}{6}=\dfrac{101}{6}\)
c) \(3\dfrac{1}{2}+4\dfrac{5}{7}-5\dfrac{5}{14}=\dfrac{7}{2}+\dfrac{33}{7}-\dfrac{75}{14}=\dfrac{49}{14}+\dfrac{66}{14}-\dfrac{75}{14}=-\dfrac{92}{14}=-\dfrac{46}{7}\)
d) \(4\dfrac{1}{2}+\dfrac{1}{2}:5\dfrac{1}{2}=\dfrac{9}{2}+\dfrac{1}{2}:\dfrac{11}{2}=\dfrac{9}{2}+\dfrac{1}{2}.\dfrac{2}{11}=\dfrac{9}{2}+\dfrac{1}{11}=\dfrac{99}{22}+\dfrac{2}{22}=\dfrac{101}{22}\)
a. \(12\dfrac{1}{3}-\left(3\dfrac{3}{4}+4\dfrac{3}{4}\right)=\dfrac{37}{3}-\left(\dfrac{15}{4}+\dfrac{19}{4}\right)\)
\(=\dfrac{37}{3}-\dfrac{34}{4}=\dfrac{23}{6}\)
\(b.3\dfrac{5}{6}+2\dfrac{1}{6}.6=\dfrac{23}{6}+13=\dfrac{101}{6}\)
\(c.3\dfrac{1}{2}+4\dfrac{5}{7}-5\dfrac{5}{14}=\dfrac{7}{2}+\dfrac{33}{7}-\dfrac{75}{14}=\dfrac{20}{7}\)
d \(4\dfrac{1}{2}+\dfrac{1}{2}:5\dfrac{1}{2}\)
\(=\dfrac{9}{2}+\dfrac{1}{2}:\dfrac{11}{2}\)
\(=\dfrac{9}{2}+\dfrac{1}{11}\)
\(=\dfrac{101}{22}\)
5: \(=3-\dfrac{1}{4}+\dfrac{2}{3}-5+\dfrac{1}{3}+\dfrac{6}{5}-6+\dfrac{7}{4}-\dfrac{3}{2}\)
\(=3-5-6+\dfrac{-1}{4}+\dfrac{7}{4}+\dfrac{2}{3}+\dfrac{1}{3}+\dfrac{6}{5}-\dfrac{3}{2}\)
\(=-8+\dfrac{3}{2}+1+\dfrac{-3}{10}\)
\(=-7+\dfrac{15-3}{10}=-7+\dfrac{6}{5}=-\dfrac{29}{5}\)
6: \(=6-\dfrac{2}{3}+\dfrac{1}{2}-5-\dfrac{5}{3}+\dfrac{3}{2}-3+\dfrac{7}{3}-\dfrac{5}{2}\)
\(=6-5-3-\dfrac{2}{3}-\dfrac{5}{3}+\dfrac{7}{3}+\dfrac{1}{2}+\dfrac{3}{2}-\dfrac{5}{2}\)
\(=-2-\dfrac{1}{2}=-\dfrac{5}{2}\)
7: \(=\dfrac{5}{3}-\dfrac{3}{7}+9-2-\dfrac{5}{7}+\dfrac{2}{3}+\dfrac{8}{7}-\dfrac{4}{3}-10\)
\(=9-2-10+\dfrac{5}{3}+\dfrac{2}{3}-\dfrac{4}{3}+\dfrac{-3}{7}-\dfrac{5}{7}+\dfrac{8}{7}\)
=-3+1
=-2
8: \(=8-\dfrac{9}{4}+\dfrac{2}{7}+6+\dfrac{3}{7}-\dfrac{5}{4}-3-\dfrac{2}{4}+\dfrac{9}{7}\)
\(=8+6-3+\dfrac{2}{7}+\dfrac{3}{7}+\dfrac{9}{7}-1-\dfrac{2}{4}\)
\(=11+2-1-\dfrac{1}{2}\)
=11+1/2
=11,5
Lời giải chi tiết:
3 + 4 > 6 | 6 + 1 > 6 | 5 + 1 = 6 |
3 + 4 = 7 | 5 + 2 > 6 | 7 – 5 > 3 |
7 – 4 < 4 | 7 – 2 = 5 | 7 – 6 = 1 |
\(1+2+3+4+5+6=21\)
1+2+3+4+5+6=21