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Tính nhanh:
912 + [88 + (-453) + (-547)]
= 912 + 88 - 453 - 547
= (912 + 88) - (453 + 547)
= 1000 - 1000
= 0
Thực hiện các phép tính sau:
6-8 = 6 + (-8) = -2
3-(-9) = 3 + 9 = 12
(-5)-10 = (-5) + (-10) = -15
0-7 = -7
4-0 = 4
(-2)-(-10) = (-2) + 10 = 8
Tính:
-2+(-6)+7- 3 = -4
-5-7-3+7 = -8
(-6)-4+5-7+12 = 0
2-(-8)+10-30 = -10
Bỏ dấu ngoặc rồi tính:
a) (4+32+6)+(10-36-6)
= 4 + 32 + 6 + 10 - 36 - 6
= (4 + 32 - 36) + (6 - 6) + 10
= 0 + 0 + 10
= 10
b) (77+22+-65)-(67+12-75)
= 77 + 22 - 65 - 67 - 12 + 75
= (77 - 67) + (22 - 12) - (65 - 75)
= 10 + 10 - (-10)
= 30
c)-(-21+43+7)-(11-53-17).
= 21 - 43 - 7 - 11 + 53 + 17
= (21 - 11) - (43 - 53) - (7 - 17)
= 10 - (-10) - (-10)
= 30
Tham khảo nha!!!
a, \(\frac{2}{5}.\frac{1}{3}-\frac{2}{15}:\frac{1}{5}+\frac{3}{5}.\frac{1}{3}\)
\(=\frac{1}{3}.\left(\frac{2}{5}+\frac{3}{5}\right)-\frac{2}{15}.5\)
\(=\frac{1}{3}.1-\frac{2}{3}\)
\(=\frac{1}{3}-\frac{2}{3}\)
\(=\frac{-1}{3}\)
b, \(\left(6-2\frac{4}{5}\right).3\frac{1}{8}+1\frac{3}{8}:\frac{1}{4}\)
\(=\left(6-\frac{14}{5}\right).\frac{25}{8}+\frac{11}{8}.4\)
\(=\frac{16}{5}.\frac{25}{8}+\frac{11}{2}\)
\(=10+\frac{11}{2}\)
\(=\frac{31}{2}\)
1/3×(3/5+2/5)-2/15×1/5
1/3×1-2/15×1/5
1/3-2/15×1/5
1/3-2/75
25/75-2/75
23/75
(6-14/5)×25/8-11/8:4/1
16/5×25/8-11/8:4/1
10/1-11/8:4/1
10/1-11/8×1/4
10/1-11/32
320/32-11/32
309/32
M=3/1.2-5/2.3+7/3.4-9/4.5+11/5.6-13/6.7+15/7.8+17/8.9
=(1/1.1+2/1.2)-(2/2.3+3/2.3)+(3/3.4+4/3.4)-(4/4.5+5/4.5)+...+(8/8.9+9/8.9)(phần ... là làm tương tự nhé)
=1/2+1-(1/3+1/2)+(1/4+1/3)-(1/5+1/4)+...+(1/9+1/8)(phần ... là làm tương tự nhé)
=1+(1/2-1/2)+(1/3-1/3)+(1/4-1/4)+...+(1/8-1/8)-1/9
=1+0+0+0+...+0-1/9
=1-1/9
=8/9
a)
i.Ta có: BCNN(12, 30) = 60
60 : 12 = 5; 60 : 30 = 2. Do đó:
\(\frac{5}{{12}} = \frac{{5.5}}{{12.5}} = \frac{{25}}{{60}}\) và \(\frac{7}{{30}} = \frac{{7.2}}{{30.2}} = \frac{{14}}{{60}}.\)
ii.Ta có: BCNN(2, 5, 8) = 40
40 : 2 = 20; 40 : 5 = 8; 40 : 8 = 5. Do đó:
\(\frac{1}{2} = \frac{{1.20}}{{2.20}} = \frac{{20}}{{40}}\)
\(\frac{3}{5} = \frac{{3.8}}{{5.8}} = \frac{{24}}{{40}}\)
\(\frac{5}{8} = \frac{{5.5}}{{8.5}} = \frac{{25}}{{40}}\).
b)
i.Ta có: BCNN(6, 8) = 24
24 : 6 = 4; 24: 8 = 3. Do đó
\(\begin{array}{l}\frac{1}{6} + \frac{5}{8} = \frac{{1.4}}{{6.4}} + \frac{{5.3}}{{8.3}}\\ = \frac{4}{{24}} + \frac{{15}}{{24}} = \frac{{19}}{{24}}.\end{array}\)
ii. Ta có: BCNN(24, 30) = 120
120: 24 = 5; 120: 30 = 4. Do đó:
\(\begin{array}{l}\frac{{11}}{{24}} - \frac{7}{{30}} = \frac{{11.5}}{{24.5}} - \frac{{7.4}}{{30.4}}\\ = \frac{{55}}{{120}} - \frac{{28}}{{120}} = \frac{{27}}{{120}} = \frac{9}{{40}}\end{array}\)
1) 5 + (-4) = 1
2) (-8) + 2 = -6
3) 8 + (-2) = 6
4) 11 + (-3) = 8
5) (-11) + 2 = -9
6) (-7) + 3 = -4
7) (-5) + 5 = 0
8) 11 + (-12) = -1
9) (-18) + 20 = 2
10) (15) + (-12) = 3
11) (-17) + 17 = 0
12) 16 + (-2) = 14
13) (30) + (-14) = 16
14) (-19) + 20 = 1
15) (-18) + 15 = -3
16) (10) + (-6) = 4
17) (-28) + 14 = -14
18) 15 + (-30) = -15
19) (15) + (-4) = 11
20) (-21) + 11 = -10
21) 8 + (-22) = -14
22) (-15) + 4 = -11
23) (-3) + 2 = -1
24) 17 + (-14) = 3
25) 17 + (-14) = 3
a. \(\frac{2}{3}+\frac{1}{3}.\left(\frac{-4}{9}+\frac{5}{6}\right):\frac{7}{12}\)
\(=1.\frac{7}{12}:\frac{7}{12}\)
\(=1\)
b.
\(\frac{5}{9}.\frac{8}{11}+\frac{5}{9}.\frac{9}{11}-\frac{5}{9}.\frac{6}{11}\)
\(=\frac{5}{9}.\left(\frac{8}{11}+\frac{9}{11}-\frac{6}{11}\right)\)
\(=\frac{5}{9}.1\)
\(=\frac{5}{9}\)
Tk mk nha!
b) \(=\frac{5}{9}.\left(\frac{8}{11}+\frac{9}{11}-\frac{6}{11}\right)\)
\(=\frac{5}{9}.1\)
\(=\frac{5}{9}\)
a: \(=\dfrac{216+3\cdot36+27}{13}=\dfrac{351}{13}=27\)
b: \(=\dfrac{1140-200-40}{9}+3^4=900+81=981\)
\(\frac{2^{30}.9^8}{6^2.8^6}\) =\(\frac{2^8.9^8.2^{32}}{\left(2.3\right)^2.\left(2^3\right)^6}\) =\(\frac{2^{10}.\left(3^2\right)^8}{3^2}\) =\(2^{10}.9^7\)