1/4×2/6×3/8×4/10×5/12×.....×30/62×1/64=2^x
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NH
2
3 tháng 4 2021
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
TA
3 tháng 4 2021
`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`.
NT
1
2 tháng 4 2017
Có: \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}...\frac{30}{62}.\frac{31}{64}=\frac{1}{2.2}.\frac{2}{2.3}.\frac{3}{2.4}...\frac{30}{2.31}.\frac{31}{2.32}=\frac{1}{2}.\frac{1}{2}.\frac{1}{2}...\frac{1}{2}.\frac{1}{2}.\frac{1}{32}\)
\(=\frac{1}{2^{31}.2^5}=\frac{1}{2^{36}}=2^x\)\(\Rightarrow1=2^x.2^{36}=2^{36+x}\)\(\Rightarrow2^{36+x}=2^0\Rightarrow36+x=0\Rightarrow x=-36\)
13 tháng 7 2023
=>\(1\cdot\dfrac{2}{4}\cdot\dfrac{3}{6}\cdot...\cdot\dfrac{31}{62}\cdot\dfrac{1}{64}=2^x\)
=>\(2^x=\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot...\cdot\dfrac{1}{2}\cdot\dfrac{1}{64}=\left(\dfrac{1}{2}\right)^{30}\cdot\left(\dfrac{1}{2}\right)^6=\dfrac{1}{2^{36}}\)
=>x=-36
CN
3
6 tháng 8 2015
<=> \(\frac{1.2.3....31}{4.6.8....64}=2^n\Rightarrow\frac{1.2.3....30.31}{2\left(2.3.4.5...31\right).32}=2^n\Leftrightarrow\frac{1}{2.32}=2^n\Leftrightarrow\frac{1}{2^6}=2^n\)
=> 2^6.2^n = 1
=> 2^ (n + 6 ) = 2^0
=> n+ 6 = 0
=> n = - 6
DD
6 tháng 8 2015
\(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}....\frac{31}{64}=\frac{1.2.3....31}{4.6.8....64}=\frac{1.2.3....31}{2.3.2.4....2.32}=\frac{1.2.3....31}{2^{30}.\left(3.4....32\right)}=\frac{2}{2^{30}.32}=\frac{1}{2^{34}}=2^{-34}=2^n=>n=-34\)
NN
1
12 tháng 7 2021
a) Ta có: \(\left(x-1\right)^{x+2}-\left(x-1\right)^{x+4}=0\)
\(\Leftrightarrow\left(x-1\right)^x\cdot\left(x-1\right)^2-\left(x-1\right)^x\cdot\left(x-1\right)^4=0\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\cdot\left[1-\left(x-1\right)^2\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
b) Ta có: \(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot\dfrac{5}{15}\cdot...\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}=2x\)
\(\Leftrightarrow2x=\dfrac{1}{64}\)
hay \(x=\dfrac{1}{128}\)
4 tháng 7 2015
\(\frac{1.2.3.4....30.31}{2.2.2.3.2.3.....2.32}=\frac{2.3.4....30.31}{2^{31}\left(2.3...31\right).32}=\frac{1}{2^{31}.2^5}=\frac{1}{2^{36}}=2^{-36}\)
Vậy x=-36
TV
13 tháng 8 2017
ta có \(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}.....\frac{30}{62}\cdot\frac{31}{64}=2^x\)
=>\(\frac{1.2.3.4....31}{2\cdot2\cdot2\cdot3\cdot2\cdot3.....\cdot2\cdot3\cdot2}=\frac{2\cdot3\cdot4...30.31}{2^{31}\left(2\cdot3\cdot4...31\right)32}=\frac{1}{2^{31}\cdot2^5}=\frac{1}{2^{36}}=2^{-36}\)
\(=>x=-36\)
21 tháng 5 2017
\(\Rightarrow\frac{1}{2.2}.\frac{2}{2.3}.\frac{3}{2.4}.\frac{4}{2.5}.\frac{5}{2.6}...\frac{30}{2.31}.\frac{31}{2.32}=2^x\)
\(\Rightarrow\frac{1}{2^{31}}.\frac{1.2.3.4....31}{2.3.4...32}=2^x\)
\(\Rightarrow\frac{1}{2^{31}}.\frac{1}{32}=2^x\)
\(\Rightarrow\frac{1}{2^{31}.2^5}=2^x\)
\(\Rightarrow\frac{1}{2^{36}}=2^x\Rightarrow x=-36\)
LV
1
4 tháng 7 2018
\(A\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\\\Leftrightarrow \left(x-1\right)^{x+4}-\left(x-1\right)^{x+2}=0\\ \Leftrightarrow\left(x-1\right)^{x+2}\left(\left(x-1\right)^{x+2}+1\right)=0\\ \Leftrightarrow\left(x-1\right)^{x+2}=0hoac\left(x-1\right)^{x+2}+1=0\)
Giả tiếp đc x=1
21 tháng 1 2019
\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}\cdot....\cdot\frac{30}{62}\cdot\frac{31}{64}=2^x\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot.....\cdot\frac{30}{31}\cdot\frac{31}{32}\right)=2^x\)
\(\Leftrightarrow\frac{1}{32}=2^{x+1}\)
Làm nốt.
ko làm được câu này hay câu b ib với tớ nha.khẳng định tối giải.
chiều tau bày cho