Tính một cách hợp lí:
a) \(\left(-16\right).\left(-7\right).5;\) b) \(11.\left(-12\right)+11.\left(-18\right);\)
c) \(87.\left(-19\right)-37.\left(-19\right);\) d) \(41.81.\left(-451\right).0.\)
Tính một cách hợp lí:
a) \(\left( { - \frac{5}{6}} \right) - \left( { - 1,8} \right) + \left( { - \frac{1}{6}} \right) - 0,8\)
b) \(\left( { - \frac{9}{7}} \right) + \left( { - 1,23} \right) - \left( { - \frac{2}{7}} \right) - 0,77\)
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)
\(\begin{array}{l}\left( { - \frac{9}{7}} \right) + \left( { - 1,23} \right) - \left( { - \frac{2}{7}} \right) - 0,77\\ = \left[ {\left( { - \frac{9}{7}} \right) - \left( { - \frac{2}{7}} \right)} \right] + \left[ {\left( { - 1,23} \right) - 0,77} \right]\\ =\frac{-7}{7}+(-2)= - 1 + \left( { - 2} \right) = - 3\end{array}\)
Tính một cách hợp lí:
a) \(1,8 - \left( {\frac{3}{7} - 0,2} \right)\)
b) \(12,5 - \frac{{16}}{{13}} + \frac{3}{{13}}\)
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)
\(\begin{array}{l}12,5 - \frac{{16}}{{13}} + \frac{3}{{13}}\\ = 12,5 - \frac{{16}}{{13}} + \frac{3}{{13}}\\ = 12,5 + \left( { - \frac{{16}}{{13}} + \frac{3}{{13}}} \right)\\ = 12,5 + \left( { - 1} \right) = 11,5\end{array}\)
Tính một cách hợp lí:
a)\(\left( { - 0,4} \right) + \frac{3}{8} + \left( { - 0,6} \right)\);
b)\(\frac{4}{5} - 1,8 + 0,375 + \frac{5}{8}\).
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)
\(\frac{4}{5} - 1,8 + 0,375 + \frac{5}{8} = (0,8 - 1,8) + (0,375 + 0,625) = ( - 1) + 1 = 0\)
Tính một cách hợp lí:
a) \(\frac{4}{{15}} - \left( {2,9 - \frac{{11}}{{15}}} \right)\);
b) \(( - 36,75) + \left( {\frac{{37}}{{10}} - 63,25} \right) - ( - 6,3)\);
c) \(6,5 + \left( { - \frac{{10}}{{17}}} \right) - \left( { - \frac{7}{2}} \right) - \frac{7}{{17}}\);
d) \(( - 39,1) \cdot \frac{{13}}{{25}} - 60,9 \cdot \frac{{13}}{{25}}\).
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)
\(\begin{array}{l}( - 39,1) \cdot \frac{{13}}{{25}} - 60,9 \cdot \frac{{13}}{{25}}\\ = \frac{{13}}{{25}}.\left( { - 39,1 - 60,9} \right)\\ = \frac{{13}}{{25}}.\left( { - 100} \right)\\ = - 52\end{array}\).
Tính một cách hợp lí:
a)\(\frac{7}{3}.\left( { - 2,5} \right).\frac{6}{7};\)
b)\(0,8.\frac{{ - 2}}{9} - \frac{4}{5}.\frac{7}{9} - 0,2.\)
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)
\(\begin{array}{l}0,8.\frac{{ - 2}}{9} - \frac{4}{5}.\frac{7}{9} - 0,2\\ = \frac{4}{5}.\frac{{ - 2}}{9} - \frac{4}{5}.\frac{7}{9}-\frac{2}{10}\\ = \frac{4}{5}.\left( {\frac{{ - 2}}{9} - \frac{7}{9}} \right) -\frac{1}{5}\\ = \frac{4}{5}.\left( { - 1} \right)-\frac{1}{5} \\= \frac{{ - 4}}{5}-\frac{1}{5}\\=\frac{-5}{5}\\=-1.\end{array}\)
Tính một cách hợp lí:
a) \(\left(\dfrac{617}{191}+\dfrac{29}{33}-\dfrac{115}{117}\right).\left(\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{20}\right);\)
b) \(\dfrac{12}{5}\cdot\left(\dfrac{10}{3}-\dfrac{5}{12}\right);\)
c) 1,23 - 5,48 + 8,77 - 4,32;
d) 7.0,25 + 9.0,25.
a) Ta có: \(\left(\dfrac{617}{191}+\dfrac{29}{33}-\dfrac{115}{17}\right)\cdot\left(\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{20}\right)\)
\(=\left(\dfrac{617}{191}+\dfrac{29}{33}-\dfrac{115}{17}\right)\cdot\left(\dfrac{5}{20}-\dfrac{4}{20}-\dfrac{1}{20}\right)\)
\(=0\cdot\left(\dfrac{617}{191}+\dfrac{29}{33}-\dfrac{115}{17}\right)=0\)
b) Ta có: \(\dfrac{12}{5}\cdot\left(\dfrac{10}{3}-\dfrac{5}{12}\right)\)
\(=\dfrac{12}{5}\cdot\left(\dfrac{40}{12}-\dfrac{5}{12}\right)\)
\(=\dfrac{12}{5}\cdot\dfrac{35}{12}\)
=7
c) Ta có: \(1.23-5.48+8.77-4.32\)
\(=\left(1.23+8.77\right)-\left(5.48+4.32\right)\)
=10-10=0
Bài 1.( 2 điểm)Tính bằng cách hợp lí:
a) \(\dfrac{1}{2}+\dfrac{9}{10}+\dfrac{5}{6}-\dfrac{11}{14}-\dfrac{1}{3}+\dfrac{-4}{35}\)
b) \(\left(\dfrac{18}{23}+\dfrac{7}{12}\right)+\left(\dfrac{-13}{19}-\dfrac{3}{4}\right)+\left(\dfrac{-6}{19}+\dfrac{5}{23}\right)\)
c) \(\dfrac{4}{3}+\dfrac{-5}{6}+\dfrac{-1}{4}\)
d) \(\dfrac{5}{6}-\dfrac{7}{5}+\dfrac{17}{30}\)
1: \(\dfrac{1}{2}+\dfrac{9}{10}+\dfrac{5}{6}-\dfrac{11}{14}-\dfrac{1}{3}+\dfrac{-4}{35}\)
\(=\left(\dfrac{1}{2}+\dfrac{5}{6}-\dfrac{1}{3}\right)+\dfrac{9}{10}-\left(\dfrac{11}{14}+\dfrac{4}{35}\right)\)
\(=\dfrac{3+5-2}{6}+\dfrac{9}{10}-\dfrac{55+8}{70}\)
\(=1+\dfrac{9}{10}-\dfrac{9}{10}\)
=1
tính một cách hợp lí:
a) \(\dfrac{-5}{18}+\dfrac{32}{45}-\dfrac{9}{10}\)
b) \(\left(\dfrac{-1}{4}+\dfrac{51}{33}-\dfrac{5}{3}\right)-\left(\dfrac{-15}{12}+\dfrac{6}{11}-\dfrac{42}{29}\right)\)
c) \(1-\dfrac{1}{2}+2-\dfrac{2}{3}+3-\dfrac{3}{4}+4-\dfrac{1}{4}-3-\dfrac{1}{3}-2-\dfrac{1}{2}-1\)
giải chi tiết giúp mình nha
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: =−2590+6490−8190=−2590+6490−8190
(−14+5133−53)−(−1512+611−4229)(−14+5133−53)−(−1512+611−4229)
=−53+4229=−53+4229
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
Tính giá trị của biểu thức sau một cách hợp lí:
\(B = \left( {\frac{{ - 3}}{{13}}} \right) + \frac{{16}}{{23}} + \left( {\frac{{ - 10}}{{13}}} \right) + \frac{5}{{11}} + \frac{7}{{23}}\)
\(\begin{array}{l}B = \left( {\frac{{ - 3}}{{13}}} \right) + \frac{{16}}{{23}} + \left( {\frac{{ - 10}}{{13}}} \right) + \frac{5}{{11}} + \frac{7}{{23}}\\ = \left[ {\left( {\frac{{ - 3}}{{13}}} \right) + \left( {\frac{{ - 10}}{{13}}} \right)} \right] + \left[ {\frac{{16}}{{23}} + \frac{7}{{23}}} \right] + \frac{5}{{11}}\\ = - 1 + 1 + \frac{5}{{11}}\\ = \frac{5}{{11}}\end{array}\)
`B= ( (-3)/13 + (-10)/13) + (16/23 + 7/23 ) +5/11`
`B= -13/13 + 23/23 +5/11`
`B=-1+1+5/11`
`B=0+5/11`
`B=5/11`
\(B=\left(-\dfrac{3}{13}\right)+\dfrac{16}{23}+\left(-\dfrac{10}{13}\right)+\dfrac{5}{11}+\dfrac{7}{23}\)
\(B=\left[\left(-\dfrac{3}{13}\right)+\left(-\dfrac{10}{13}\right)\right]+\left(\dfrac{16}{23}+\dfrac{7}{23}\right)+\dfrac{5}{11}\)
\(B=\left(-1\right)+1+\dfrac{5}{11}\)
\(B=\dfrac{5}{11}\)