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=>\(1\cdot\dfrac{2}{4}\cdot\dfrac{3}{6}\cdot...\cdot\dfrac{31}{62}\cdot\dfrac{1}{64}=2^x\)
=>\(2^x=\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot...\cdot\dfrac{1}{2}\cdot\dfrac{1}{64}=\left(\dfrac{1}{2}\right)^{30}\cdot\left(\dfrac{1}{2}\right)^6=\dfrac{1}{2^{36}}\)
=>x=-36
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`.
<=> \(\frac{1.2.3....31}{4.6.8....64}=2^n\Rightarrow\frac{1.2.3....30.31}{2\left(2.3.4.5...31\right).32}=2^n\Leftrightarrow\frac{1}{2.32}=2^n\Leftrightarrow\frac{1}{2^6}=2^n\)
=> 2^6.2^n = 1
=> 2^ (n + 6 ) = 2^0
=> n+ 6 = 0
=> n = - 6
\(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}....\frac{31}{64}=\frac{1.2.3....31}{4.6.8....64}=\frac{1.2.3....31}{2.3.2.4....2.32}=\frac{1.2.3....31}{2^{30}.\left(3.4....32\right)}=\frac{2}{2^{30}.32}=\frac{1}{2^{34}}=2^{-34}=2^n=>n=-34\)
\(\Rightarrow\frac{1}{2.2}.\frac{2}{2.3}.\frac{3}{2.4}.\frac{4}{2.5}.\frac{5}{2.6}...\frac{30}{2.31}.\frac{31}{2.32}=2^x\)
\(\Rightarrow\frac{1}{2^{31}}.\frac{1.2.3.4....31}{2.3.4...32}=2^x\)
\(\Rightarrow\frac{1}{2^{31}}.\frac{1}{32}=2^x\)
\(\Rightarrow\frac{1}{2^{31}.2^5}=2^x\)
\(\Rightarrow\frac{1}{2^{36}}=2^x\Rightarrow x=-36\)
\(\dfrac{1}{2.2}.\dfrac{2}{2.3}.....\dfrac{31}{64}=2^x\\ =>\dfrac{1}{2.2.2.....2.64}=2^x\\ \dfrac{1}{2^{30}.26}=2^x\\ =>\dfrac{1}{2^{36}}=2^x\\ =>2^{-36}=2^x\\ =>x=-36\)
Ta có: \(\frac{1}{2.2}.\frac{2}{2.3}.\frac{3}{2.4}.\frac{4}{2.5}.\frac{5}{2.6}.......\frac{30}{2.31}.\frac{31}{64}=4^x\)
\(\frac{1}{2^{30}.64}=4^x\Leftrightarrow4^x.2^{30}.2^6=1\)
\(\Leftrightarrow2^{2x+36}2^0\)
\(\Leftrightarrow2x+36=0\)
\(\Leftrightarrow2x=-36\)
\(\Leftrightarrow x=-18\)
Vậy ........
$4^x.64=1$\(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}.\frac{5}{12}.....\frac{30}{62}.\frac{31}{64}=4^x\)
\(\Leftrightarrow\frac{1}{2.2}.\frac{2}{2.3}.\frac{3}{2.4}.\frac{4}{2.5}.\frac{5}{2.6}.....\frac{30}{2.31}.\frac{31}{2.32}=4^x\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.....\frac{30}{31}.\frac{31}{32}\right)=4^x\)
\(\Leftrightarrow\frac{1}{2}.\frac{1.2.3.4.5.....30.31}{2.3.4.5.6.....31.32}=4^x\)
\(\Leftrightarrow\frac{1}{2}.\frac{1}{32}=4^x\)
\(\Leftrightarrow4^x=\frac{1}{64}\)
\(\Leftrightarrow4^x.64=1\)
\(\Leftrightarrow4^x.4^3=1\Leftrightarrow4^{x+3}=4^0\Leftrightarrow x+3=0\Leftrightarrow x=-3\)
Vậy x = -3
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 \(x=\dfrac{1}{128}\)
b)
\(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}....\frac{30}{62}.\frac{31}{64}=2^x\)
\(\frac{1.2.3.4.....30.31}{4.6.8.10....62.64}=2^x\)
\(\frac{1.2.3.4.5....30.31}{2.2.2.3.2.4.2.5.....2.31.64}=2^x\)
\(\frac{1.2.3.4.5.....30.31}{\left(2.2.2....2.2\right).\left(2.3.4.5....30.31\right).64}=2^x\)
\(2.2.2.2.2.....2.64=2^x\)
\(2^{31}.2^6=2^x\)
\(2^{37}=2^x\)
=> \(x=37\)
Có: \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}...\frac{30}{62}.\frac{31}{64}=\frac{1}{2.2}.\frac{2}{2.3}.\frac{3}{2.4}...\frac{30}{2.31}.\frac{31}{2.32}=\frac{1}{2}.\frac{1}{2}.\frac{1}{2}...\frac{1}{2}.\frac{1}{2}.\frac{1}{32}\)
\(=\frac{1}{2^{31}.2^5}=\frac{1}{2^{36}}=2^x\)\(\Rightarrow1=2^x.2^{36}=2^{36+x}\)\(\Rightarrow2^{36+x}=2^0\Rightarrow36+x=0\Rightarrow x=-36\)