Bỏ dấu ngoặc rồi tính
a) (5,1-3,5) - (3,5+5,1)
b) -(271,5+381)+5,261-(1-381)
c) (3/4+3/4)-(-3/4+2/5)
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Giải:
a) \(\dfrac{56.2,7+56.3,5}{0,62}\)
\(=\dfrac{56.6,2}{0,62}\)
\(=56.10=560\)
b) \(371.65+371.14+629.79\)
\(=371\left(65+14\right)+629.79\)
\(=371.79+629.79\)
\(=79\left(371+629\right)\)
\(=79.1000=79000\)
c) \(0,48.700-2,4.45.2\)
\(=4,8.70-4,8.45\)
\(=4,8\left(70-45\right)\)
\(=4,8.25=120\)
e) \(3,2.950+8.175.4+320.73\)
\(=32.95+32.175+32.730\)
\(=32\left(95+175+730\right)\)
\(=32.1000=32000\)
Vậy ...
Giải:
a) 56.2,7+56.3,50,6256.2,7+56.3,50,62
=56.6,20,62=56.6,20,62
=56.10=560=56.10=560
b) 371.65+371.14+629.79371.65+371.14+629.79
=371(65+14)+629.79=371(65+14)+629.79
=371.79+629.79=371.79+629.79
=79(371+629)=79(371+629)
=79.1000=79000=79.1000=79000
c) 0,48.700−2,4.45.20,48.700−2,4.45.2
=4,8.70−4,8.45=4,8.70−4,8.45
=4,8(70−45)=4,8(70−45)
=4,8.25=120=4,8.25=120
e) 3,2.950+8.175.4+320.733,2.950+8.175.4+320.73
=32.95+32.175+32.730=32.95+32.175+32.730
=32(95+175+730)=32(95+175+730)
=32.1000=32000=32.1000
__________
9,6 x 0,8 - 3,5 x 4/5 - 2/5 x 2 x 5,1 - 0,8
= 9,6 x 0,8 - 3,5 x 0,8 - 0,4 x 2 x 5,1 - 0,8 x 1
= 9,6 x 0,8 - 3,5 x 0,8 - 0,8 x 5,1 - 0,8 x 1
= (9,6 - 3,5 - 5,1 - 1) x 0,8
= 0 x 0,8
= 0
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: \(\left(3965-2378\right)-\left(437-1378\right)-528\)
\(=3965-2378-437+1378-528\)
\(=\left(3965-437-528\right)-\left(2378-1378\right)\)
\(=3000-1000=2000\)
b: \(\left(-457\right)+\left(-123\right)+23+237\)
\(=\left(-457+237\right)+\left(-123+23\right)\)
\(=-220-100=-320\)
c: \(\left(-329\right)-\left(15-101\right)-\left(25-440\right)\)
\(=-329-15+101-25+440\)
\(=\left(440-15-25\right)+\left(-329+101\right)\)
\(=400-228=172\)
Cách 1 : Trước hết tính giá trị từng biểu thức trong ngoặc
\(-\left(4-\dfrac{1}{3}+\dfrac{3}{5}\right)-\left(3+\dfrac{2}{3}-\dfrac{4}{5}\right)+\left(5+\dfrac{4}{3}-\dfrac{4}{5}\right)\)
\(=-\left(\dfrac{60}{15}-\dfrac{5}{15}+\dfrac{9}{15}\right)-\left(\dfrac{45}{15}+\dfrac{10}{15}-\dfrac{20}{15}\right)+\left(\dfrac{75}{15}+\dfrac{20}{15}-\dfrac{12}{15}\right)\)
\(=-\left(\dfrac{64}{15}\right)-\left(\dfrac{35}{15}\right)+\left(\dfrac{83}{15}\right)\)
\(=-\dfrac{64}{15}-\dfrac{35}{15}+\dfrac{83}{15}\)
\(=-\dfrac{16}{15}\)
\(=-\dfrac{8}{5}\)
Cách 2 : Bỏ dấu ngoặc rồi nhóm các số hạng thích hợp
\(-\left(4-\dfrac{1}{3}+\dfrac{3}{5}\right)-\left(3+\dfrac{2}{3}-\dfrac{4}{5}\right)+\left(5+\dfrac{4}{3}-\dfrac{4}{5}\right)\)
\(=-4+\dfrac{1}{3}-\dfrac{3}{5}-3-\dfrac{2}{3}+\dfrac{4}{5}+5+\dfrac{4}{3}-\dfrac{4}{5}\)
\(=\left(-4-3+5\right)+\left(\dfrac{1}{3}-\dfrac{2}{3}+\dfrac{4}{3}\right)+\left(-\dfrac{3}{5}+\dfrac{4}{5}-\dfrac{4}{5}\right)\)
\(=-2+1-\dfrac{3}{5}\)
\(=-\dfrac{8}{5}\)
(135,4 - 84,3) :2,5 - 10,46
= 51,1 :2,5 - 10,46
= 20,44 - 10,46 = 9,98
Tương tự nha
a,\(\left(5,1-3,5\right)-\left(3,5+5,1\right)\)
\(=5,1-3,5-3,5-5,1\)
\(=\left[5,1-5,1\right]\)\(-\)\(\left[3,5-3,5\right]\)
\(=0-0\)
\(=0\)
b,c bn tự làm nha