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`.

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

<=> \(\frac{1.2.3....31}{4.6.8....64}=2^n\Rightarrow\frac{1.2.3....30.31}{2\left(2.3.4.5...31\right).32}=2^n\Leftrightarrow\frac{1}{2.32}=2^n\Leftrightarrow\frac{1}{2^6}=2^n\)

=> 2^6.2^n = 1 

=> 2^ (n + 6 ) = 2^0

=> n+ 6  = 0 

=> n = - 6 

\(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}....\frac{31}{64}=\frac{1.2.3....31}{4.6.8....64}=\frac{1.2.3....31}{2.3.2.4....2.32}=\frac{1.2.3....31}{2^{30}.\left(3.4....32\right)}=\frac{2}{2^{30}.32}=\frac{1}{2^{34}}=2^{-34}=2^n=>n=-34\)

2^n × 2³¹ = \(\dfrac{ }{ }\)2/4×4/6×6/8×...×62/64

2^n×2³¹=1/32=2^-5

2^n=2^-5 ÷ 2³¹=2^-36

=>n=-36

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 \(x=\dfrac{1}{128}\)

\(\frac{1.2.3.4....30.31}{2.2.2.3.2.3.....2.32}=\frac{2.3.4....30.31}{2^{31}\left(2.3...31\right).32}=\frac{1}{2^{31}.2^5}=\frac{1}{2^{36}}=2^{-36}\)

Vậy x=-36

ta có \(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}.....\frac{30}{62}\cdot\frac{31}{64}=2^x\)

=>\(\frac{1.2.3.4....31}{2\cdot2\cdot2\cdot3\cdot2\cdot3.....\cdot2\cdot3\cdot2}=\frac{2\cdot3\cdot4...30.31}{2^{31}\left(2\cdot3\cdot4...31\right)32}=\frac{1}{2^{31}\cdot2^5}=\frac{1}{2^{36}}=2^{-36}\)

\(=>x=-36\)

trong cái xã hội này có làm thì mới có ăn,ko lam mà ăn chỉ có ăn đầu b** ăn c**

\(\Rightarrow\frac{1}{2.2}.\frac{2}{2.3}.\frac{3}{2.4}.\frac{4}{2.5}.\frac{5}{2.6}...\frac{30}{2.31}.\frac{31}{2.32}=2^x\)

\(\Rightarrow\frac{1}{2^{31}}.\frac{1.2.3.4....31}{2.3.4...32}=2^x\)

\(\Rightarrow\frac{1}{2^{31}}.\frac{1}{32}=2^x\)

\(\Rightarrow\frac{1}{2^{31}.2^5}=2^x\)

\(\Rightarrow\frac{1}{2^{36}}=2^x\Rightarrow x=-36\)

\(A\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\\\Leftrightarrow \left(x-1\right)^{x+4}-\left(x-1\right)^{x+2}=0\\ \Leftrightarrow\left(x-1\right)^{x+2}\left(\left(x-1\right)^{x+2}+1\right)=0\\ \Leftrightarrow\left(x-1\right)^{x+2}=0hoac\left(x-1\right)^{x+2}+1=0\)

Giả tiếp đc x=1

\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}\cdot....\cdot\frac{30}{62}\cdot\frac{31}{64}=2^x\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot.....\cdot\frac{30}{31}\cdot\frac{31}{32}\right)=2^x\)

\(\Leftrightarrow\frac{1}{32}=2^{x+1}\)

Làm nốt.

ko làm được câu này hay câu b ib với tớ nha.khẳng định tối giải.