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a)\(\left( { - 0,4} \right) + \frac{3}{8} + \left( { - 0,6} \right) = \left[ {\left( { - 0,4} \right) + \left( { - 0,6} \right)} \right] + \frac{3}{8} = - 1 + \frac{3}{8} = \frac{{ - 5}}{8}\).
b)
\(\frac{4}{5} - 1,8 + 0,375 + \frac{5}{8} = (0,8 - 1,8) + (0,375 + 0,625) = ( - 1) + 1 = 0\)
a)\(\frac{7}{3}.\left( { - 2,5} \right).\frac{6}{7} = \frac{7}{3}.\frac{6}{7}.\left( { - 2,5} \right) = 2.\left( { - 2,5} \right) = - 5\)
b)
\(\begin{array}{l}0,8.\frac{{ - 2}}{9} - \frac{4}{5}.\frac{7}{9} - 0,2\\ = \frac{4}{5}.\frac{{ - 2}}{9} - \frac{4}{5}.\frac{7}{9}-\frac{2}{10}\\ = \frac{4}{5}.\left( {\frac{{ - 2}}{9} - \frac{7}{9}} \right) -\frac{1}{5}\\ = \frac{4}{5}.\left( { - 1} \right)-\frac{1}{5} \\= \frac{{ - 4}}{5}-\frac{1}{5}\\=\frac{-5}{5}\\=-1.\end{array}\)
a)
\(\begin{array}{l}\left( { - \frac{5}{6}} \right) - \left( { - 1,8} \right) + \left( { - \frac{1}{6}} \right) - 0,8\\ = \left( { - \frac{5}{6}} \right) + 1,8 + \left( { - \frac{1}{6}} \right) - 0,8\\ = \left[ {\left( { - \frac{5}{6}} \right) + \left( { - \frac{1}{6}} \right)} \right] + \left[ {1,8 - 0,8} \right]\\ =\frac{-6}{6}+1= - 1 + 1 = 0\end{array}\)
b)
\(\begin{array}{l}\left( { - \frac{9}{7}} \right) + \left( { - 1,23} \right) - \left( { - \frac{2}{7}} \right) - 0,77\\ = \left[ {\left( { - \frac{9}{7}} \right) - \left( { - \frac{2}{7}} \right)} \right] + \left[ {\left( { - 1,23} \right) - 0,77} \right]\\ =\frac{-7}{7}+(-2)= - 1 + \left( { - 2} \right) = - 3\end{array}\)
a)
\(\begin{array}{l}1,8 - \left( {\frac{3}{7} - 0,2} \right)\\ = 1,8 - \frac{3}{7} + 0,2\\ = \left( {1,8 + 0,2} \right) - \frac{3}{7}\\ = 2 - \frac{3}{7} =\frac{{14}}{7}-\frac{{3}}{7}= \frac{{11}}{7}\end{array}\)
b)
\(\begin{array}{l}12,5 - \frac{{16}}{{13}} + \frac{3}{{13}}\\ = 12,5 - \frac{{16}}{{13}} + \frac{3}{{13}}\\ = 12,5 + \left( { - \frac{{16}}{{13}} + \frac{3}{{13}}} \right)\\ = 12,5 + \left( { - 1} \right) = 11,5\end{array}\)
a) Ta có: \(\dfrac{-5}{18}+\dfrac{32}{45}-\dfrac{9}{10}\)
\(=\dfrac{-25}{90}+\dfrac{64}{90}-\dfrac{81}{90}\)
\(=\dfrac{-42}{90}=-\dfrac{7}{15}\)
b) Ta có: \(\left(-\dfrac{1}{4}+\dfrac{51}{33}-\dfrac{5}{3}\right)-\left(-\dfrac{15}{12}+\dfrac{6}{11}-\dfrac{42}{29}\right)\)
\(=\dfrac{-1}{4}+\dfrac{17}{11}-\dfrac{5}{3}+\dfrac{5}{4}-\dfrac{6}{11}+\dfrac{42}{29}\)
\(=\dfrac{-5}{3}+\dfrac{42}{29}\)
\(=\dfrac{-145}{87}+\dfrac{126}{87}=\dfrac{-19}{87}\)
c) Ta có: \(1-\dfrac{1}{2}+2-\dfrac{2}{3}+3-\dfrac{3}{4}+4-\dfrac{1}{4}-3-\dfrac{1}{3}-2-\dfrac{1}{2}-1\)
\(=\left(1-1\right)-\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(2-2\right)-\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\left(3-3\right)-\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+4\)
\(=-1-1-1+4\)
=1
a) Ta có: −518+3245−910−518+3245−910
=−2590+6490−8190=−2590+6490−8190
=−4290=−715=−4290=−715
b) Ta có: (−14+5133−53)−(−1512+611−4229)(−14+5133−53)−(−1512+611−4229)
=−14+1711−53+54−611+4229=−14+1711−53+54−611+4229
=−53+4229=−53+4229
=−14587+12687=−1987=−14587+12687=−1987
c) Ta có: 1−12+2−23+3−34+4−14−3−13−2−12−11−12+2−23+3−34+4−14−3−13−2−12−1
=(1−1)−(12+12)+(2−2)−(23+13)+(3−3)−(34+14)+4=(1−1)−(12+12)+(2−2)−(23+13)+(3−3)−(34+14)+4
=−1−1−1+4=−1−1−1+4
=1
a)4,25+3,15+3,75+5,85=(4,25+3,75)+(3,15+5,85)=8+9=17.
b)(3/5+4/7)+2/5=(3/5+2/5)+4/7=1+4/7=11/7.
c)6,24+5,36+3,76=(6,24+3,76)+5,36=10+5,36=15,36.
d)(19/15+3/5)+11/15=(19/15+11/15)+3/5=2+3/5=13/5.
ủng hộ nha có j kb vs mk
Tính bằng cách thuận tiện nhất:
a ) 4,25 + 3,15 + 3,75 + 5,85
= ( 4,25 + 3,75 ) + ( 3,15 + 5, 85 )
= 8 + 9
= 17
tks giúp mk vs ạ!
a) Cách 1:
\(\begin{array}{l}\left( {\frac{{ - 2}}{{ - 5}} + \frac{{ - 5}}{{ - 6}}} \right) + \frac{4}{5} = \frac{2}{5} + \frac{5}{6} + \frac{4}{5}\\ = \frac{{12}}{{30}} + \frac{{25}}{{30}} + \frac{{24}}{{30}} = \frac{{61}}{{30}}\end{array}\)
Cách 2:
\(\begin{array}{l}\left( {\frac{{ - 2}}{{ - 5}} + \frac{{ - 5}}{{ - 6}}} \right) + \frac{4}{5} = \left( {\frac{2}{5} + \frac{4}{5}} \right) + \frac{5}{6}\\ = \frac{6}{5} + \frac{5}{6} = \frac{{36}}{{30}} + \frac{{25}}{{30}} = \frac{{61}}{{30}}\end{array}\)
b) Cách 1:
\(\begin{array}{l}\frac{{ - 3}}{{ - 4}} + \left( {\frac{{11}}{{ - 15}} + \frac{{ - 1}}{2}} \right) = \frac{3}{4} + \frac{{ - 11}}{{15}} + \frac{{ - 1}}{2}\\ = \frac{{45}}{{60}} + \frac{{ - 44}}{{60}} + \frac{{ - 30}}{{60}}\\ = \frac{{ - 29}}{{60}}\end{array}\).
Cách 2:
\(\begin{array}{l}\frac{{ - 3}}{{ - 4}} + \left( {\frac{{11}}{{ - 15}} + \frac{{ - 1}}{2}} \right) = \frac{3}{4} + \frac{{ - 11}}{{15}} + \frac{{ - 1}}{2}\\ = \left( {\frac{3}{4} + \frac{{ - 1}}{2}} \right) + \frac{{ - 11}}{{15}}\\ = \left( {\frac{3}{4} + \frac{{ - 2}}{4}} \right) + \frac{{ - 11}}{{15}}\\ = \frac{1}{4} + \frac{{ - 11}}{{15}}\\ = \frac{{15}}{{60}} + \frac{{ - 44}}{{60}}\\ = \frac{{ - 29}}{{60}}\end{array}\)
a)
\(\begin{array}{l}\frac{4}{{15}} - \left( {2,9 - \frac{{11}}{{15}}} \right)\\ = \frac{4}{{15}} - 2,9 + \frac{{11}}{{15}}\\ = \left( {\frac{4}{{15}} + \frac{{11}}{{15}}} \right) - 2,9\\=\frac{15}{15}-2,9 \\= 1 - 2,9 = - 1,9\end{array}\)
b)
\(\begin{array}{l}( - 36,75) + \left( {\frac{{37}}{{10}} - 63,25} \right) - ( - 6,3)\\ = ( - 36,75) + 3,7 - 63,25 + 6,3\\ = \left( { - 36,75 - 63,25} \right) + \left( {3,7 + 6,3} \right)\\ = - 100 + 10 = - 90\end{array}\)
c)
\(\begin{array}{l}6,5 + \left( { - \frac{{10}}{{17}}} \right) - \left( { - \frac{7}{2}} \right) - \frac{7}{{17}}\\ = \frac{{65}}{{10}} - \frac{{10}}{{17}} + \frac{7}{2} - \frac{7}{{17}}\\ = \left( {\frac{{65}}{{10}} + \frac{7}{2}} \right) - \left( {\frac{{10}}{{17}} + \frac{7}{{17}}} \right)\\ = \left( {\frac{{65}}{{10}} + \frac{{35}}{{10}}} \right) - \frac{17}{17}\\ = \frac{100}{10}-1\\=10 - 1 = 9\end{array}\)
d)
\(\begin{array}{l}( - 39,1) \cdot \frac{{13}}{{25}} - 60,9 \cdot \frac{{13}}{{25}}\\ = \frac{{13}}{{25}}.\left( { - 39,1 - 60,9} \right)\\ = \frac{{13}}{{25}}.\left( { - 100} \right)\\ = - 52\end{array}\).