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H = 1/4 . 2/6 . 3/8 . 4/10. ... . 30/62 . 31/64
H= 1/2.2 . 2/2.3 . 3/2.4 . 4/2.5 . ... . 30/2.31 . 31/2.32
H = 1/2 . 1/2 . 1/2 . 1/2 . ... . 1/2 . 1/32 (31 số 1/2)
H = 1/2^31 . 1/2^5
H = 1/2^36
Vậy H = 1/2^36
\(H=\frac{1}{4}.\frac{2}{6}.\frac{3}{8}...\frac{30}{62}.\frac{31}{64}\)
\(\Rightarrow H=\frac{1.2.3...30.31}{2.2.2.3.2.4...2.31.2.32}=\frac{1}{2^{31}.2^5}=\frac{1}{2^{36}}\)
`#` `\text{dkhanhqlv}`
`4)`
`a)3.(-5/11)`
`=-15/11`
`b)3/5+4/7 . 14/6`
`=3/5 + 4/3`
`=9/15+20/15`
`=29/15`
`c) 10/21-3/8 . 4/15`
`=10/21-1/10`
`=100/210-21/210`
`=79/100`
`d)(2/3+3/4)(5/7+5/14)`
`=(8/12+9/12)(10/14+5/14)`
`=17/12 . 15/14`
`=85/56`
`5)`
`a)x-1/2=3/10 . 5/6`
`=>x-1/2=1/4`
`=>x=1/4+1/2`
`=>x=1/4+2/4`
`=>x=3/4`
`b)x/5 = -3/14`
`=>x : 5 = -3/14`
`=>x=-3/14 . 5`
`=>x=-15/14`
`c)x+2/3=9/15 . 5/27`
`=>x+2/3=1/9`
`=>x=1/9-2/3`
`=>x=1/9-6/9`
`=>x=-5/9`
Câu 1 xem kỉ đề
\(B,\frac{49^6.5-7^{11}}{\left(-7\right)^{10}.5-2.49^5}=\frac{7^{12}.5-7^{11}}{7^{10}.5-2.7^{10}}=\frac{7^{11}.\left(7.5-1\right)}{7^{10}.\left(5-2\right)}=\frac{7.34}{3}=\frac{238}{3}\)
a) A=212.35-\(\frac{2^{12}.3^6}{2^{12}}\)+93+84.35
=212.35-36+36+212.35
=213.35
b)B=496.5-5.\(\frac{7^{11}}{\left(-7\right)^{10}}-2.49^5\)
=496.5-7.5-2.495
=712.5-7.5-2.710
a, \(\dfrac{3^8}{3^4}+2^2.2^3\)
= 34 + 25
= 113
b, 3.42 - 2.32
= 3.(42 - 2.3)
= 3.(16 - 6)
= 3.10
= 30
c, \(\dfrac{4^6.3^4.9^5}{6^{12}}\)
= \(\dfrac{\left(2^2\right)^6.3^4.\left(3^2\right)^5}{\left(2.3\right)^{12}}\)
= \(\dfrac{2^{12}.3^4.3^{10}}{2^{12}.3^{12}}\)
= \(\dfrac{3^{14}}{3^{12}}\)
= 32
@Dương Tuyết Mai
1. 25 : 5,7 = 250/57
2. 30:2.8.4 = 480
3. 20:2^2.14= 70
4. 125:5^3.170= 170
5. 64:2^5.30.4=240
6. (25:5^2.30): 15.7=14
bạn à! Nhiều quá mình ko làm hết được. sorry nha.^-^
\(\dfrac{1}{2}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot...\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}=\dfrac{1}{2^x}\)
=>\(\dfrac{2}{2}\cdot\dfrac{3}{6}\cdot\dfrac{4}{8}\cdot...\cdot\dfrac{30}{60}\cdot\dfrac{31}{62}\cdot\dfrac{1}{64}=\dfrac{1}{2^x}\)
=>\(\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot...\cdot\dfrac{1}{2}\cdot\dfrac{1}{64}=\dfrac{1}{2^x}\)
=>\(\dfrac{1}{2^{29}}\cdot\dfrac{1}{2^6}=\dfrac{1}{2^x}\)
=>x=29+6=35
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