2 x 17 x 12 + 4 x 2016 + 8 x 62 x 3
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2x17x12+4x6x21+8x3x62
=2x12x17+4x6x21+8x3x62
=24x17+24x21+24x62
=24x(17+21+62)
=24x100
=2400
2 x 17 x 12 + 4 x 6 x 21 + 8 x 3 x 62
= 24 x 17 + 24 x 21 + 24 x 62
= 24 x (17 + 21 + 62)
= 24 x 100
= 2400
2 x 17 x 12 + 4 x 6 x 21 + 8 x 3 x 62
= 2 x 12 x 17 + 24 x 21 + 24 x 62
= 24 x 17 + 24 x 21 + 24x 62
= 24 x ( 17 + 21 + 62)
= 24 x 100
= 2400
ban vt sai dau bai roi phai la
1/2+1/6+1/12+....+1/(x+1).x=2016/2017
1/1.2+1/2.3+1/3.4+.....+1/x.(x+1)=2016/2017
1-1/2+1/2-1/3+......+1/x-1/x+1=2016/2017
1-1/x+1=2016/2017
1/x+1=1-2016/2017
1/x+1=1/2017
=>x+1=2017
=>x=2016
b)
85/8:x+(-4)/17:x+15:51=4/11
85/8:x+(-4)/17:x=13/187
1413/136:x=13/187
x=15543/104
`#3107.101107`
A,
\(2\times32\times12+4\times6\times41+8\times27\times3\\ =24\times32+24\times41+24\times27\\ =24\times\left(32+41+27\right)\\ =24\times100\\ =2400\)
B,
\(\left(2006\times2005^{2016}-2005^{2016}\right)\div2005^{2017}\\ =\left[2005^{2016}\times\left(2006-1\right)\right]\div2005^{2017}\\ =\left(2005^{2016}\times2005\right)\div2005^{2017}\\ =2005^{2017}\div2005^{2017}\\ =1\)
Ta có: \(2^x=\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot\dfrac{5}{12}\cdot...\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}\)
\(\Leftrightarrow2^x=\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot31}{2\cdot\left(2\cdot3\cdot4\cdot...\cdot31\right)\cdot64}\)
\(\Leftrightarrow2^x=\dfrac{1}{2}\cdot\dfrac{1}{64}=\dfrac{1}{128}\)
\(\Leftrightarrow2^x=\dfrac{1}{2^6}\)
\(\Leftrightarrow2^{x+6}=1\)
\(\Leftrightarrow x+6=0\)
hay x=-6
Vậy: x=-6
`1/4 . 2/6 . 3/8 ... . 30/62 .31/64 =2^x`
`-> (1.2.3....30.31)/(4.6.8....62.64)=2^x`
`-> (1.(2.3...31))/(2.(2.3.4...31).32)=2^x`
`-> 1/(2.32)=2^x`
`-> 1/64=2^x`
`-> 1/(2^6)=2^x`
`-> x=-6`.
=(2x12)x17+(4x2016)+(8x3)+16
=24x17+8064+24x16
=408+8064+384
=8856
2 x 17 x 12 + 4 x 2016 + 8 x 62 x 3
= 24 x 17 + 336 x 24 + 24 x 62
= 24 x ( 17 + 336 + 62 )
= 24 x 415
= 9960