Tính F = (0.25)^-1 . (1/1/4)^2 + 25.[(4/3)^-2:(5/4)^3] :(-2/3)^-3
E = (2/3)^5.9^3 + (3/8)^8: (3/16)^4/ 3.2^6+2.3^4
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a\()\) 16/9 +3/5
=107/45
b\()\) 4/13--2/17
=51/221--26/221
=77/221
c\()\) -3/2+4/5
=-15/10+8/10
=-7/10
d\()\) 3/-4-1/4
=-1
e\()\) -1/5.5/7
=-1/7
f\()\) 7/8.64/49
=8/7
g\()\) 3/4.15/24
=15/32
2/3 + 7/1 = 2/3 + 7 = 2/3 + 21/3 = 23/3
25/48 + 11/24 = 25/48 + 22/48 = 47/48
5/7 + 3/8 = 40/56 + 21/56 = 61/56
15/24 + 12/6 = 5/8 + 2 = 5/8 + 16/8 = 21/8
5/6 + 4/3 = 5/6 + 8/6 = 13/6
3/8 + 7/12 = 9/24 + 14/24 = 23/24
a) 17.13+17.42-17.35
=17.(13+42-35)
=17.20=340
b) [25.(18-42)-10]:4+6
=(25.2-10):4+6
=40:4+6=16
c) 36:32+23.22-32.3
=34+25-33
=81+32-27=86
d) B=3.42-22.3
=3.(16-4)
=3.12=36
e)20220+3.[52.10-(23-13)2]
=1+3.(250-100)
=1+450=451
g) 27.77+24.27-27
=27.(77+24-1)
=27.100=2700
h) 5.23+79:77-12020
=40+72-1
=89-1=88
i) 120:{54[50:2+(32-2.4)]}
=120:[54(25+1)]
=120:1404=10/117
\(A=\frac{1\cdot2+2\cdot4+3\cdot6+4\cdot8+5\cdot10+6\cdot12}{3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20+18\cdot24}\)
\(A=\frac{2\cdot3\left[1\cdot2\right]+2\cdot3\left[2\cdot4\right]+2\cdot3\left[3\cdot6\right]+2\cdot3\left[4\cdot8\right]+2\cdot3\left[5\cdot10\right]}{3\cdot4\left[3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20\right]}\)
\(A=\frac{\left[3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20\right]}{2\cdot3\left[3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20\right]}=\frac{1}{2\cdot3}=\frac{1}{6}\)
a, 2/3+1/2+1/6
=4/6+3/6+1/6
=4/3
b, 5/12+5/6-3/4
=10/24+20/24-18/24
=1/2
c, 1/3*3/5*2/5
=(1*3*2)/(3*5*5)
=2/25
d, 15/16:3/8*3/4
= 15/16*8/3*3/4
= 15/8
a) \(\frac{2}{3}\)+\(\frac{1}{2}\)+\(\frac{1}{6}\) = \(\frac{4}{6}\)+\(\frac{3}{6}\)+\(\frac{1}{6}\) =\(\frac{8}{6}\) =\(\frac{4}{3}\)
b)\(\frac{5}{12}+\frac{5}{6}-\frac{3}{4}\)=\(\frac{5}{12}+\frac{10}{12}-\frac{9}{12}\)=\(\frac{6}{12}\)= \(\frac{1}{2}\)
c) \(\frac{1}{3}\cdot\frac{3}{5}\cdot\frac{2}{5}\) =\(\frac{6}{75}\)=\(\frac{2}{25}\)