tính biểu thức
\(\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}}{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}}:\frac{3+\frac{3}{2}+\frac{3}{3}+\frac{3}{4}}{2-\frac{2}{2}+\frac{2}{3}-\frac{2}{4}}\)
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a) Cách 1:
\(\begin{array}{l}(8 + 2\frac{1}{3} - \frac{3}{5}) - (5 + 0,4) - (3\frac{1}{3} - 2)\\ = (8 + \frac{7}{3} - \frac{3}{5}) - (5 + \frac{4}{{10}}) - (\frac{{10}}{3} - 2)\\ = 8 + \frac{7}{3} - \frac{3}{5} - 5 - \frac{2}{5} - \frac{{10}}{3} + 2\\ = (8 - 5 + 2) + (\frac{7}{3} - \frac{{10}}{3}) - (\frac{3}{5} + \frac{2}{5})\\ = 5 + \frac{{ - 3}}{3} - \frac{5}{5}\\ = 5 + ( - 1) - 1\\ = 3\end{array}\)
Cách 2:
\(\begin{array}{l}(8 + 2\frac{1}{3} - \frac{3}{5}) - (5 + 0,4) - (3\frac{1}{3} - 2)\\ = (8 + \frac{7}{3} - \frac{3}{5}) - (5 + \frac{4}{{10}}) - (\frac{{10}}{3} - 2)\\ = (\frac{{120}}{{15}} + \frac{{35}}{{15}} - \frac{9}{{15}}) - (\frac{{25}}{5} + \frac{2}{5}) - (\frac{{10}}{3} - \frac{6}{3})\\ = \frac{{146}}{{15}} - \frac{{27}}{5} - \frac{4}{3}\\ = \frac{{146}}{{15}} - \frac{{81}}{{15}} - \frac{{20}}{{15}}\\ = \frac{{45}}{{15}}\\ = 3\end{array}\)
b)
\(\begin{array}{l}(7 - \frac{1}{2} - \frac{3}{4}):(5 - \frac{1}{4} - \frac{5}{8})\\ = (\frac{{28}}{4} - \frac{2}{4} - \frac{3}{4}):(\frac{{40}}{8} - \frac{2}{8} - \frac{5}{8})\\ = \frac{{23}}{4}:\frac{{33}}{8}\\ = \frac{{23}}{4}.\frac{8}{{33}}\\ = \frac{{46}}{{33}}\end{array}\)
\(\begin{array}{l}a)A = (2 - \frac{1}{2} - \frac{1}{8}):(1 - \frac{3}{2} - \frac{3}{4})\\ = (\frac{{16}}{8} - \frac{4}{8} - \frac{1}{8}):(\frac{4}{4} - \frac{6}{4} - \frac{3}{4})\\ = \frac{{11}}{8}:\frac{{ - 5}}{4}\\ = \frac{{11}}{8}.\frac{4}{{ - 5}}\\ = \frac{{ - 11}}{{10}}\\b)B = 5 - \frac{{1 + \frac{1}{3}}}{{1 - \frac{1}{3}}}\\ = 5 - \frac{{\frac{3}{3} + \frac{1}{3}}}{{\frac{3}{3} - \frac{1}{3}}}\\ = 5 - \frac{{\frac{4}{3}}}{{\frac{2}{3}}}\\ = 5 - \frac{4}{3}:\frac{2}{3}\\ = 5 - \frac{4}{3}.\frac{3}{2}\\ = 5 - 2\\ = 3\end{array}\)
Chú ý:
Khi thực hiện phép cộng hai phân số, nếu phân số thu được chưa tối giản thì ta rút gọn thành phân số tối giản.
a) \(\frac{3}{4} + \left( {\frac{1}{2} - \frac{1}{3}} \right) = \frac{9}{{12}} + \left( {\frac{6}{{12}} - \frac{4}{{12}}} \right) = \frac{9}{{12}} + \frac{2}{{12}} = \frac{{11}}{{12}}\)
\(\frac{3}{4} + \frac{1}{2} - \frac{1}{3} = \frac{9}{{12}} + \frac{6}{{12}} - \frac{4}{{12}} = \frac{{15}}{{12}} - \frac{4}{{12}} = \frac{{11}}{{12}}\)
Vậy \(\frac{3}{4} + \left( {\frac{1}{2} - \frac{1}{3}} \right)\) = \(\frac{3}{4} + \frac{1}{2} - \frac{1}{3}\)
b)\(\frac{2}{3} - \left( {\frac{1}{2} + \frac{1}{3}} \right) = \frac{4}{6} - \left( {\frac{3}{6} + \frac{2}{6}} \right) = \frac{4}{6} - \frac{5}{6} = \frac{{ - 1}}{6}\)
\(\frac{2}{3} - \frac{1}{2} - \frac{1}{3} = \frac{4}{6} - \frac{3}{6} - \frac{2}{6} = \frac{1}{6} - \frac{2}{6} = \frac{{ - 1}}{6}\)
Vậy \(\frac{2}{3} - \left( {\frac{1}{2} + \frac{1}{3}} \right)\)=\(\frac{2}{3} - \frac{1}{2} - \frac{1}{3}\).
`#3107`
`a)`
`3/4 + (1/2 - 1/3)`
`= 3/4 + (3/6 - 2/6)`
`= 3/4 + 1/6`
`= 11/12`
`3/4 + 1/2 - 1/3`
`= 9/12 + 6/12 - 4/12`
`= (9 + 6 - 4)/12`
`= 11/12`
Vì `11/12 = 11/12`
`=> 3/4 + (1/2 - 1/3) = 3/4 + 1/2 - 1/3`
`b)`
`2/3 - (1/2 + 1/3)`
`= 2/3 - (3/6 + 2/6)`
`= 2/3 - 5/6`
`= -1/6`
`2/3 - 1/2 - 1/3`
`= 4/6 - 3/6 - 2/6`
`= (4 - 3 - 2)/6`
`= -1/6`
Vì `-1/6 = -1/6`
`=> 2/3 - (1/2 + 1/3) = 2/3 - 1/2 - 1/3`
giải:
ta có :
\(\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}}{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}}:\frac{3+\frac{3}{2}+\frac{3}{3}+\frac{3}{4}}{2-\frac{2}{2}+\frac{2}{3}-\frac{2}{4}}\)
\(\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}}{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}}.\frac{2\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}{3\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)}=\frac{2}{3}\)