Tính:
a) \(\left( { - 12} \right):\left( { - 6} \right)\);
b) \(\left( { - 64} \right):\left( { - 8} \right)\).
K
Khách
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HQ
Hà Quang Minh
Giáo viên
10 tháng 10 2023
a) \(\left( { - 3} \right).7 = - \left( {3.7} \right) = - 21\)
b) \(\left( { - 8} \right).\left( { - 6} \right) = 8.6 = 48\)
c) \(\left( { + 12} \right).\left( { - 20} \right) = - \left( {12.20} \right) = - 240\)
d) \(24.\left( { + 50} \right) = 24.50 = 1200\)
HQ
Hà Quang Minh
Giáo viên
9 tháng 10 2023
a)
Cách 1.
\(\begin{array}{l}\left( {4 + 32 + 6} \right) + \left( {10 - 36 - 6} \right)\\ = 4 + 32 + 6 + 10 - 36 - 6\\ = 52 + \left( { - 36} \right) + \left( { - 6} \right)\\ = 52 + \left( { - 42} \right) = 52 - 42 = 10\end{array}\)
Cách 2.
\(\begin{array}{l}\left( {4 + 32 + 6} \right) + \left( {10 - 36 - 6} \right)\\ = 4 + 32 + 6 + 10 - 36 - 6\\ = 36 + 6 + 10 + \left( { - 36} \right) + \left( { - 6} \right)\\ = 36 + \left( { - 36} \right) + 6 + \left( { - 6} \right) + 10\\ = 0 + 0 + 10 = 10\end{array}\)
b) \(\left( {77 + 22 - 65} \right) - \left( {67 + 12 - 75} \right)\)
\(\begin{array}{l} = 77 + 22 - 65 - 67 - 12 + 75\\ = 77 - 67 + 22 - 12 + 75 - 65\\ = 10 + 10 + 10 = 30\end{array}\)
c) \( - \left( { - 21 + 43 + 7} \right) - \left( {11 - 53 - 17} \right)\)
\(\begin{array}{l} = 21 - 43 - 7 - 11 + 53 + 17\\ = 21 - 11 + 53 - 43 + 17 - 7\\ = 10 + 10 + 10 = 30\end{array}\)
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
a)
\(\begin{array}{l}\frac{3}{7}.\left( { - \frac{1}{9}} \right) + \frac{3}{7}.\left( { - \frac{2}{3}} \right)\\ = \frac{3}{7}.\left( { - \frac{1}{9} + \frac{-2}{3}} \right)\\ = \frac{3}{7}.\left( { - \frac{1}{9} - \frac{6}{9}} \right)\\ = \frac{3}{7}.\frac{{ - 7}}{9} = \frac{{ - 1}}{3}\end{array}\)
b)
\(\begin{array}{l}\left( {\frac{{ - 7}}{{13}}} \right).\frac{5}{{12}} + \left( {\frac{{ - 7}}{{13}}} \right).\frac{7}{{12}} + \left( {\frac{{ - 6}}{{13}}} \right)\\ = \frac{{ - 7}}{{13}}.\left( {\frac{5}{{12}} + \frac{7}{{12}}} \right) + \left( {\frac{{ - 6}}{{13}}} \right)\\ = \frac{{ - 7}}{{13}}.1 + \left( {\frac{{ - 6}}{{13}}} \right)\\ = \frac{{ - 7}}{{13}} + \left( {\frac{{ - 6}}{{13}}} \right)\\ = \frac{{ - 13}}{{13}}\\ = -1\end{array}\)
c)
\(\begin{array}{l}\left[ {\left( {\frac{{ - 2}}{3} + \frac{3}{7}} \right)} \right]:\frac{5}{9} + \left( {\frac{4}{7} - \frac{1}{3}} \right):\frac{5}{9}\\ = \left[ {\left( {\frac{{ - 2}}{3} + \frac{3}{7}} \right)} \right].\frac{9}{5} + \left( {\frac{4}{7} - \frac{1}{3}} \right).\frac{9}{5}\\ = \left( {\frac{{ - 2}}{3} + \frac{3}{7} + \frac{4}{7} - \frac{1}{3}} \right).\frac{9}{5}\\ = \left[ {\left( {\frac{{ - 2}}{3} - \frac{1}{3}} \right) + \left( {\frac{3}{7} + \frac{4}{7}} \right)} \right].\frac{9}{5}\\ = \left( { - 1 + 1} \right).\frac{9}{5}\\ = 0.\frac{9}{5} = 0\end{array}\)
d)
\(\begin{array}{l}\frac{5}{9}:\left( {\frac{1}{{11}} - \frac{5}{{22}}} \right) + \frac{5}{9}:\left( {\frac{1}{{15}} - \frac{2}{3}} \right)\\ = \frac{5}{9}:\left( {\frac{2}{{22}} - \frac{5}{{22}}} \right) + \frac{5}{9}:\left( {\frac{1}{{15}} - \frac{{10}}{{15}}} \right)\\ = \frac{5}{9}:\frac{{ - 3}}{{22}} + \frac{5}{9}:\frac{{ - 9}}{15}\\= \frac{5}{9}:\frac{{ - 3}}{{22}} + \frac{5}{9}:\frac{{ - 3}}{5}\\ = \frac{5}{9}.\frac{{ - 22}}{3} + \frac{5}{9}.\frac{{ - 5}}{3}\\ = \frac{5}{9}.\left( {\frac{{ - 22}}{3} - \frac{5}{3}} \right)\\ = \frac{5}{9}.\frac{-27}{3}= \frac{5}{9}.\left( { - 9} \right) = - 5\end{array}\)
e)
\(\begin{array}{l}\frac{3}{5} + \frac{3}{{11}} - \left( {\frac{{ - 3}}{7}} \right) + \left( {\frac{{ - 2}}{{97}}} \right) - \frac{1}{{35}} - \frac{3}{4} + \left( {\frac{{ - 23}}{{44}}} \right)\\ = \frac{3}{5} + \frac{3}{{11}} + \frac{3}{7} - \frac{2}{{97}} - \frac{1}{{35}} - \frac{3}{4} - \frac{{23}}{{44}}\\ = \left( {\frac{3}{5} + \frac{3}{7} - \frac{1}{{35}}} \right) + \left( {\frac{3}{{11}} - \frac{3}{4} - \frac{{23}}{{44}}} \right) - \frac{2}{{97}}\\ = \left( {\frac{{21}}{{35}} + \frac{{15}}{{35}} - \frac{1}{{35}}} \right) + \left( {\frac{{12}}{{44}} - \frac{{33}}{{44}} - \frac{{23}}{{44}}} \right) - \frac{2}{{97}}\\ = \frac{35}{{35}}+ \frac{-44}{{44}}- \frac{2}{{97}}\\= 1 + \left( { - 1} \right) - \frac{2}{{97}}\\ = - \frac{2}{{97}}\end{array}\)
26 tháng 8 2021
a: Ta có: \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}-11\right)\)
\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}-11\right)\)
\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}-11\right)\)
\(=127-22\sqrt{6}\)
b: Ta có: \(\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
\(=\left(1-\sqrt{5}\right)\left(-1-\sqrt{5}\right)\)
=-1+5
=4
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
\(a)\frac{2}{{15}} + \left( {\frac{{ - 5}}{{24}}} \right) = \frac{{16}}{{120}} + \left( {\frac{{ - 25}}{{120}}} \right) = \frac{{ - 9}}{{120}} = \frac{{ - 3}}{{40}}\)
b) \(\left( {\frac{{ - 5}}{9}} \right) - \left( { - \frac{7}{{27}}} \right) = \left( {\frac{{ - 15}}{{27}}} \right) + \frac{7}{{27}} = \frac{{ - 8}}{{27}}\)
c)\(\left( { - \frac{7}{{12}}} \right) + 0,75 = \left( { - \frac{7}{{12}}} \right) + \frac{75}{100} \\= \left( { - \frac{7}{{12}}} \right) + \frac{3}{4} \\= \left( { - \frac{7}{{12}}} \right) + \frac{9}{{12}} = \frac{2}{{12}} = \frac{1}{6}\)
d)\(\left( {\frac{{ - 5}}{9}} \right) - 1,25 =\left( {\frac{{ - 5}}{9}} \right) - \frac{125}{100} = \left( {\frac{{ - 5}}{9}} \right) - \frac{5}{4}\\ = \left( {\frac{{ - 20}}{{36}}} \right) - \frac{{45}}{{36}} = \frac{{ - 65}}{{36}}\)
e)\(0,34.\frac{{ - 5}}{{17}} =\frac{{34}}{{100}}.\frac{{ - 5}}{{17}} = \frac{{17}}{{50}}.\frac{{ - 5}}{{17}} = \frac{{ - 1}}{{10}}\)
g) \(\frac{4}{9}:\left( { - \frac{8}{{15}}} \right) = \frac{4}{9}.\left( { - \frac{{15}}{8}} \right) = \frac{{ - 5}}{6}\)
h)\(\left( {1\frac{2}{3}} \right):\left( {2\frac{1}{2}} \right) = \frac{5}{3}:\frac{5}{2} = \frac{5}{3}.\frac{2}{5} = \frac{2}{3}\)
i) \(\frac{2}{5}.\left( { - 1,25} \right) = \frac{2}{5}.\frac{{ - 125}}{100} = \frac{2}{5}.\frac{{ - 5}}{4} = \frac{{ - 1}}{2}\)
k) \(\left( {\frac{{ - 3}}{5}} \right).\left( {\frac{{15}}{{ - 7}}} \right).3\frac{1}{9} = \left( {\frac{{ - 3}}{5}} \right).\left( {\frac{{15}}{{ - 7}}} \right).\frac{{28}}{9}\\ = \frac{{ - 3.3.5.7.4}}{{5.\left( { - 7} \right).3.3}} = 4\)
OY
25 tháng 7 2021
a) \(\dfrac{-5}{9}+\dfrac{8}{15}+\dfrac{-2}{11}+\dfrac{4}{-9}+\dfrac{7}{15}\)
=\(\left(\dfrac{-5}{9}+\dfrac{-4}{9}\right)+\left(\dfrac{8}{15}+\dfrac{7}{15}\right)+\dfrac{-2}{11}\)
=\(\left(-1\right)+1+\dfrac{-2}{11}\)
=\(\dfrac{-2}{11}\)
b) \(\left(\dfrac{-5}{12}+\dfrac{6}{11}\right)+\left(\dfrac{7}{17}+\dfrac{5}{11}+\dfrac{5}{12}\right)\)
=\(\dfrac{-5}{12}+\dfrac{6}{11}+\dfrac{7}{17}+\dfrac{5}{11}+\dfrac{5}{12}\)
=\(\left(\dfrac{-5}{12}+\dfrac{5}{12}\right)+\left(\dfrac{6}{11}+\dfrac{5}{11}\right)+\dfrac{7}{17}\)
=\(0+0+\dfrac{7}{17}\)
=\(\dfrac{7}{17}\)
c) A= \(49\dfrac{8}{23}-\left(5\dfrac{7}{32}+14\dfrac{8}{23}\right)\)
A=\(49\dfrac{8}{23}-5\dfrac{7}{32}-14\dfrac{8}{23}\)
A=\(\left(49\dfrac{8}{23}-14\dfrac{8}{23}\right)-5\dfrac{7}{32}\)
A=\(35-5\dfrac{7}{32}\)
A=\(35-\dfrac{167}{32}=\dfrac{953}{32}\)
d) C=\(\dfrac{-3}{7}.\dfrac{5}{9}+\dfrac{4}{9}.\dfrac{-3}{7}+2\dfrac{3}{7}\)
C=\(\dfrac{-3}{7}.\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{17}{7}\)
C=\(\dfrac{-3}{7}.1+\dfrac{17}{7}\)
C=\(\dfrac{-3}{7}+\dfrac{17}{7}=2\)
HD
25 tháng 7 2021
a, `(-5)/9+8/15+(-2)/11+4/(-9)+7/15`
`=-5/9+8/15-2/11-4/9+7/15`
`=(-5/9-4/9)+(8/15+7/15)-2/11`
`=-9/9+15/15-2/11`
`=-1+1-2/11`
`=-2/11`
b, `((-5)/12+6/11)+(7/17+5/11+5/12)`
`=-5/12+6/11+7/17+5/11+5/12`
`=(-5/12+5/12)+(6/11+5/11)+7/17`
`=0+11/11+7/17`
`=1+7/17`
`=17/17+7/17`
`=24/17`
c, `A=49 8/23 - (5 7/32 + 14 8/23)`
`A=49 8/23 - 5 7/32 - 14 8/23`
`A=(49 8/23 - 14 8/23)-5 7/32`
`A=35 - 167/32`
`A=953/32`
d, `C=(-3)/7.5/9+4/9.(-3)/7+2 3/7`
`C=-3/7 . 5/9-4/9 . 3/7+17/7`
`C=-3/7.(5/9+4/9)+17/7`
`C=-3/7 . 1+17/7`
`C=2`
HQ
Hà Quang Minh
Giáo viên
16 tháng 9 2023
a)\(5,75.\frac{{ - 8}}{9} =\frac{{575}}{100}.\frac{{ - 8}}{9}= \frac{{23}}{4}.\frac{{ - 8}}{9} = \frac{{ - 46}}{9}\)
b)\(2\frac{3}{8}.\left( { - 0,4} \right) = \frac{{19}}{8}.\frac{{ - 4}}{10} =\frac{{19}}{8}.\frac{{ - 2}}{5} = \frac{{ - 19}}{{20}}\);
c)\(\frac{{ - 12}}{5}:\left( { - 6,5} \right) = \frac{{ - 12}}{5}:\frac{{ - 65}}{10} =\frac{{ - 12}}{5}:\frac{{ - 13}}{2} = \frac{{ - 12}}{5}.\frac{{ - 2}}{{13}} = \frac{{24}}{{65}}\).
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 5 2023
\(5\sqrt{\left(-2\right)^4}=5\sqrt{4^2}=5.4=20\)
\(-4\sqrt{\left(-3\right)^6}=-4\sqrt{27^2}=-4.27=-108\)
\(\sqrt{\sqrt{\left(-5\right)^8}}=\sqrt{\sqrt{\left(5^4\right)^2}}=\sqrt{5^4}=\sqrt{25^2}=25\)
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
a)
\(\begin{array}{l}1\frac{1}{2} + \frac{1}{5}.\left[ {\left( { - 2\frac{5}{6} + \frac{1}{3}} \right)} \right]\\ = \frac{3}{2} + \frac{1}{5}.\left[ {\left( { - \frac{{17}}{6} + \frac{2}{6}} \right)} \right]\\ = \frac{3}{2} + \frac{1}{5}.\frac{{ - 15}}{6}\\ = \frac{3}{2} + \frac{{ - 1}}{2}\\ = \frac{2}{2}\\=1\end{array}\)
b)
\(\begin{array}{l}\frac{1}{3}.\left( {\frac{2}{5} - \frac{1}{2}} \right):{\left( {\frac{1}{6} - \frac{1}{5}} \right)^2}\\ = \frac{1}{3}.\left( {\frac{4}{{10}} - \frac{5}{{10}}} \right):{\left( {\frac{5}{{30}} - \frac{6}{{30}}} \right)^2}\\ = \frac{1}{3}.\frac{{ - 1}}{{10}}:{\left( {\frac{{ - 1}}{{30}}} \right)^2}\\ = \frac{{ - 1}}{{30}}:\frac{1}{{{{30}^2}}}\\ = \frac{{ - 1}}{{30}}{.30^2}\\ = - 30\end{array}\)
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
a)
\(\begin{array}{l}\left( {\frac{{ - 3}}{7}} \right) + \left( {\frac{5}{6} - \frac{4}{7}} \right)\\ = \left( {\frac{{ - 3}}{7}} \right) + \frac{5}{6} - \frac{4}{7}\\ = \left[ {\left( {\frac{{ - 3}}{7}} \right) - \frac{4}{7}} \right] + \frac{5}{6}\\ =\frac{-7}{7}+\frac{5}{6}\\= - 1 + \frac{5}{6}\\ = \frac{{ - 1}}{6}\end{array}\)
b)
\(\begin{array}{l}\frac{3}{5} - \left( {\frac{2}{3} + \frac{1}{5}} \right)\\ = \frac{3}{5} - \frac{2}{3} - \frac{1}{5}\\ = (\frac{3}{5} - \frac{1}{5}) - \frac{2}{3}\\ = \frac{2}{5} - \frac{2}{3}\\ = \frac{6}{{15}} - \frac{{10}}{{15}}\\ = \frac{{ - 4}}{{15}}\end{array}\)
c)
\(\begin{array}{l}\left[ {\left( {\frac{{ - 1}}{3}} \right) + 1} \right] - \left( {\frac{2}{3} - \frac{1}{5}} \right)\\ = \left( {\frac{{ - 1}}{3}} \right) + 1 - \frac{2}{3} + \frac{1}{5}\\ = \left( {\frac{{ - 1}}{3} - \frac{2}{3}} \right) + 1 + \frac{1}{5}\\ = \frac{-3}{3}+1+\frac{1}{5}\\= - 1 + 1 + \frac{1}{5}\\ = \frac{1}{5}\end{array}\)
d)
\(\begin{array}{l}1\frac{1}{3} + \left( {\frac{2}{3} - \frac{3}{4}} \right) - \left( {0,8 + 1\frac{1}{5}} \right)\\ = 1 + \frac{1}{3} + \frac{2}{3} - \frac{3}{4} - \left( {\frac{4}{5} + 1 + \frac{1}{5}} \right)\\=1+\frac{3}{3}-\frac{3}{4}-(\frac{5}{5}+1)\\ = 1 + 1 - \frac{3}{4} - (1+1)\\ = - \frac{3}{4}\end{array}\).
a) (- 12) : (- 6) = 12 : 6 = 2.
b) (- 64) : (- 8) = 64 : 8 = 8