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a)
\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right).\\A = \left( {\frac{{30}}{{15}} + \frac{5}{{15}} - \frac{6}{{15}}} \right) - \left( {\frac{{105}}{{15}} - \frac{9}{{15}} - \frac{{20}}{{15}}} \right) - \left( {\frac{3}{{15}} + \frac{{25}}{{15}} - \frac{{60}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} - \left( {\frac{{ - 32}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} + \frac{{32}}{{15}}\\A = \frac{{ - 15}}{{15}}\\A = - 1\end{array}\)
b)
\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right)\\A = 2 + \frac{1}{3} - \frac{2}{5} - 7 + \frac{3}{5} + \frac{4}{3} - \frac{1}{5} - \frac{5}{3} + 4\\A = \left( {2 - 7 + 4} \right) + \left( {\frac{1}{3} + \frac{4}{3} - \frac{5}{3}} \right) + \left( { - \frac{2}{5} + \frac{3}{5} - \frac{1}{5}} \right)\\A = - 1 + 0 + 0 = - 1\end{array}\)
C1: dễ nên tự làm nhé
C2: \(A=\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{2}+\frac{5}{2}\right)\)
\(=6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)
\(=6-5-3+\left(\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\right)-\left(\frac{2}{3}+\frac{5}{3}-\frac{7}{3}\right)\)
\(=-2-\frac{1}{2}=\frac{-4}{2}-\frac{1}{2}=\frac{-5}{2}\)
Cách 1:
A = \(\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)
A = \(\frac{35}{6}-\frac{31}{6}-\frac{19}{6}\)
A = \(\frac{-15}{6}=\frac{-5}{2}\)
Cách 2:
A = \(\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)
A = \(6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)
A = \(6-5-3-\frac{2}{3}-\frac{5}{3}+\frac{7}{3}+\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\)
A = \(\left(6-5-3\right)-\left(\frac{2}{3}+\frac{5}{3}-\frac{7}{3}\right)+\left(\frac{1}{2}-\frac{3}{2}-\frac{5}{2}\right)\)
A = \(-2-0+\left(2-\frac{5}{2}\right)\)
A = \(-2+\left(2-\frac{5}{2}\right)\)
A = \(-2+2-\frac{5}{2}\)
A = \(0-\frac{5}{2}\)
A = \(\frac{-5}{2}\)
\(\begin{array}{l}a)\left( {\frac{2}{3} + \frac{1}{6}} \right):\frac{5}{4} + \left( {\frac{1}{4} + \frac{3}{8}} \right):\frac{5}{2}\\ = \left( {\frac{4}{6} + \frac{1}{6}} \right).\frac{4}{5} + \left( {\frac{2}{8} + \frac{3}{8}} \right).\frac{2}{5}\\ = \frac{5}{6}.\frac{4}{5} + \frac{5}{8}.\frac{2}{5}\\ = \frac{2}{3} + \frac{1}{4}\\ = \frac{8}{{12}} + \frac{3}{{12}}\\ = \frac{{11}}{{12}}\\b)\frac{5}{9}:\left( {\frac{1}{{11}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{2}{7}} \right)\\ = \frac{5}{9}:\left( {\frac{2}{{22}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{4}{{14}}} \right)\\ = \frac{5}{9}:\frac{{ - 3}}{{22}} + \frac{7}{4}.\frac{{ - 3}}{{14}}\\ = \frac{5}{9}.\frac{{ - 22}}{3} + \frac{{ - 3}}{8}\\ = \frac{{ - 110}}{{27}} + \frac{{ - 3}}{8}\\ = \frac{{ - 880}}{{216}} + \frac{{ - 81}}{{216}}\\ = \frac{{ - 961}}{{216}}\end{array}\)
Cách 1: A= \(\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)\)\(-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)
= \(\frac{35}{6}-\frac{31}{6}-\frac{19}{6}=\frac{-15}{6}\)=\(\frac{-5}{2}\)
Cách 2: A= \(\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)\)\(-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)
= \(6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)
= \(\left(6-5-3\right)\)\(+\left(-\frac{2}{3}-\frac{5}{3}+\frac{7}{3}\right)\)\(+\left(\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\right)\)
= \(-2+0+\left(\frac{-1}{2}\right)\)=\(\frac{-5}{2}\)
Cách 1:A=(6−2/3+1/2)−(5+5/3−3/2)−(3−7/3+5/2)
= ( 36/6 - 4/6 + 3/6)-(30/6 + 10/6 - 9/6) -(18/6 -14/6 +15/6)
= 35/6 - 31/6 - 19/6
= -15/2
Cách 2: A=(6−2/3+1/2)−(5+5/3−3/2)−(3−7/3+5/2)
= 6- 2/3 +1/2 - 5 - 5/3 + 3/2 - 3 + 7/3 -5/2
= (6-5-3) + ( -2/3-5/3+7/3) + (1/2+3/2-5/2)
= -2 + 0 - 1/2
= -2 - 1/2 = -4/2 - 1/2 = -5/2
Cách 1:
\(A=\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)
\(=\left(\frac{36}{6}-\frac{4}{6}+\frac{3}{6}\right)-\left(\frac{30}{6}+\frac{10}{6}-\frac{9}{6}\right)-\left(\frac{18}{6}-\frac{14}{6}+\frac{15}{6}\right)\)
\(=\frac{35}{6}-\frac{31}{6}-\frac{19}{6}\)
\(=-\frac{15}{6}\)
\(=-\frac{5}{2}\)
Cách 2:
\(A=\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)
\(=6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)
\(=\left(6-5-3\right)+\left(-\frac{2}{3}-\frac{5}{3}+\frac{7}{3}\right)+\left(\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\right)\)
\(=-2+0-\frac{1}{2}\)
\(=-\frac{4}{2}-\frac{1}{2}\)
\(=-\frac{5}{2}\)
C1: \(A=\frac{35}{6}-\frac{31}{6}-\frac{19}{6}\)
\(A=\frac{35-31-19}{6}=\frac{-15}{6}=\frac{-5}{2}\)
C2: \(A=6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)
\(A=\left(6-5-3\right)+\left(-\frac{2}{3}-\frac{5}{3}+\frac{7}{3}\right)+\left(\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\right)\)
\(A=-2-\frac{1}{2}=\frac{-5}{2}\)
C1:
\(A=\left(\frac{36-4+3}{6}\right)-\left(\frac{30+10-9}{6}\right)-\left(\frac{18-14+15}{6}\right)\)
\(A=\frac{35}{6}-\frac{31}{6}-\frac{19}{6}=\frac{-15}{6}=\frac{-5}{2}\)
C2: \(A=6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)
\(A=\left(6-5-3\right)-\left(\frac{2}{3}+\frac{5}{3}-\frac{7}{3}\right)+\left(\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\right)\)
\(A=-2-0+\frac{-1}{2}=\frac{-5}{2}\)
Trên đây là bài giải của mình, bạn hãy tính lại bằng máy tính xem Kq có đúng ko nhé!
\(\begin{array}{l}A = \left( {7 - \frac{2}{5} + \frac{1}{3}} \right) - \left( {6 - \frac{4}{3} + \frac{6}{5}} \right) - \left( {2 - \frac{8}{5} + \frac{5}{3}} \right)\\A = 7 - \frac{2}{5} + \frac{1}{3} - 6 + \frac{4}{3} - \frac{6}{5} - 2 + \frac{8}{5} - \frac{5}{3}\\A = \left( {7 - 6 - 2} \right) + \left( { - \frac{2}{5} - \frac{6}{5} + \frac{8}{5}} \right) + \left( {\frac{1}{3} + \frac{4}{3} - \frac{5}{3}} \right)\\A = - 1 + 0 + 0 = - 1\end{array}\)
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