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`.

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 \(x=\dfrac{1}{128}\)

`1/4. 2/6. 3/8. 4/10........ 30/62. 31/64=2^x`

`=>\underbrace{1/2. 1/2. 1/2. 1/2..............1/2}_{\text{30 số 2}}. 1/64=2^x`

`=>(1/2)^{30}.(1/2)^{6}=2^x`

`=>(1/2)^{36}=2^x`

`=>2^{-36}=2^x`

`=>x=-36`

Vậy `x=-36`

các bạn thêm bằng 2^x hộ mk nha , mk quên

\(E=\frac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{4\cdot6\cdot8\cdot10\cdot...\cdot62\cdot64}=\frac{1\cdot1\cdot1\cdot1\cdot.....\cdot1\cdot1}{2\cdot2\cdot2\cdot....\cdot2\cdot64}=\frac{1}{2\cdot30\cdot64}=\frac{1}{3840}\)

Có: \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}...\frac{30}{62}.\frac{31}{64}=\frac{1}{2.2}.\frac{2}{2.3}.\frac{3}{2.4}...\frac{30}{2.31}.\frac{31}{2.32}=\frac{1}{2}.\frac{1}{2}.\frac{1}{2}...\frac{1}{2}.\frac{1}{2}.\frac{1}{32}\)

\(=\frac{1}{2^{31}.2^5}=\frac{1}{2^{36}}=2^x\)\(\Rightarrow1=2^x.2^{36}=2^{36+x}\)\(\Rightarrow2^{36+x}=2^0\Rightarrow36+x=0\Rightarrow x=-36\)

=>\(1\cdot\dfrac{2}{4}\cdot\dfrac{3}{6}\cdot...\cdot\dfrac{31}{62}\cdot\dfrac{1}{64}=2^x\)

=>\(2^x=\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot...\cdot\dfrac{1}{2}\cdot\dfrac{1}{64}=\left(\dfrac{1}{2}\right)^{30}\cdot\left(\dfrac{1}{2}\right)^6=\dfrac{1}{2^{36}}\)

=>x=-36

\(\dfrac{1}{2.2}.\dfrac{2}{2.3}.....\dfrac{31}{64}=2^x\\ =>\dfrac{1}{2.2.2.....2.64}=2^x\\ \dfrac{1}{2^{30}.26}=2^x\\ =>\dfrac{1}{2^{36}}=2^x\\ =>2^{-36}=2^x\\ =>x=-36\)

\(n=-36\)

Ta có:  \(2^n=\dfrac{1}{4}.\dfrac{2}{6}.\dfrac{3}{8}.\dfrac{4}{10}.\dfrac{5}{12}....\dfrac{30}{62}.\dfrac{31}{64}\)

⇔ \(2^n=\dfrac{1.2.3.4....31}{2.\left(2.3.4.....1\right).64}=\)

⇔ \(2^n=\dfrac{1}{2}.\dfrac{1}{64}=\dfrac{1}{128}\)  \(\Leftrightarrow\)   \(2^n=\dfrac{1}{2^6}\)

⇔ \(2^{x+6}=1\)

⇔ \(x+6=0\)

⇒  \(\left\{{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)