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Sure=)
Simple task:)
Tính bằng cách thuận tiện nhất :
3/7 + 1/8 + 4/7 + 7/8
a. 2
b. 1 + 1
c. 2 + 1
d. ( 3/7 + 1/8 ) + ( 4/7 + 7/8 )
e. 3
g. 1 + 2
h. ( 3/7 + 4/7 ) + ( 1/8 + 7/8 )
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1+2+3+4+5+6+7+8+9+1+2+3+4+5+6+7+8+9+1+2+3+4+5+6+7+8+9=135
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a) \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}\)
\(=1-\frac{1}{4}\)
\(=\frac{3}{4}\)
b) \(\frac{4}{7}\times\frac{8}{9}+\frac{4}{7}\times\frac{1}{9}\)
\(=\frac{4}{7}\times\left(\frac{8}{9}+\frac{1}{9}\right)\)
\(=\frac{4}{7}\times1\)
\(=\frac{4}{7}\)
c) \(\frac{13}{6}\times\frac{3}{8}-\frac{3}{8}\times\frac{7}{6}\)
\(=\frac{3}{8}\times\left(\frac{13}{6}-\frac{7}{6}\right)\)
\(=\frac{3}{8}\times1\)
\(=\frac{3}{8}\)
d) \(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{9x10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
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\(a)9+\frac{24}{25}=\frac{9\cdot25+24}{25}=\frac{249}{25}\)
\(b)\frac{1}{20}+\frac{1}{4}+\frac{2}{5}=\frac{1}{20}+\frac{5}{20}+\frac{8}{20}=\frac{1+5+8}{20}=\frac{14}{20}=\frac{7}{10}\)
\(c)\frac{7}{12}+4=\frac{7}{12}+\frac{4}{1}=\frac{7+48}{12}=\frac{55}{12}\)
\(d)10+\frac{9}{8}=\frac{10}{1}+\frac{9}{8}=\frac{80+9}{8}=\frac{89}{8}\)
\(e)\frac{1}{4}+\frac{2}{25}+\frac{9}{100}=\frac{25}{100}+\frac{8}{100}+\frac{9}{100}=\frac{25+8+9}{100}=\frac{42}{100}=\frac{21}{50}\)
Còn 3 câu làm tương tự
Ba câu còn lại mình làm luôn :D
\(f)\frac{28}{90}+6=\frac{14}{45}+6=\frac{14}{45}+\frac{6}{1}=\frac{14+270}{45}=\frac{284}{45}\)
\(g)\frac{3}{7}+\frac{4}{9}+\frac{4}{7}+\frac{7}{11}=(\frac{3}{7}+\frac{4}{7})+\frac{4}{9}+\frac{4}{11}=1+\frac{4}{9}+\frac{4}{11}\)
\(=\frac{7}{7}+\frac{4}{9}+\frac{4}{11}=\frac{693+308+252}{693}=\frac{1253}{693}=\frac{179}{99}\)
\(h)\frac{1}{8}+\frac{1}{12}+\frac{3}{8}+\frac{5}{12}=(\frac{1}{8}+\frac{3}{8})+(\frac{1}{12}+\frac{5}{12})=\frac{4}{8}+\frac{6}{12}=\frac{1}{2}+\frac{1}{2}=1\)
`7/8-1/2-1/4`
`=7/8-4/8-2/8`
`=(7-4-2)/8`
`=1/8`