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

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

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

=> 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

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

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

=>x=-36

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 \(x=\dfrac{1}{128}\)

a)   x² + x = 0

=>   x( x+ 1 ) = 0

=>  x  = 0 

hoặc x = -1 

b)  b, (x-1)^x+2 = (x-1)^x+4

=>  x + 2    =   x  + 4 

=> 0x = 2 ( ktm)

Vậy ko có giá trị x nào thoả mãn đk 

d) Ta có: x-1/x+5 = 6/7

=>(x-1).7 = (x+5).6

=>7x-7 = 6x+ 30

=> 7x-6x = 7+30

=> x = 37

Vậy x = 37

e, x²/ 6= 24/25

=>  x²  . 25 = 6 . 24

 ⇒${}^{}.25=144$

$\Rightarrow {}^{}=144÷25$

$\Rightarrow {}^{}=5,76=2,{}^{}=\left(-2,{}^{}\right)$

$\Rightarrow x\in \left\{2,4;-2,4\right\}$

Vậy $x\in \left\{2,4;-2,4\right\}$

g, x-2/ x-1= x+4/ x+7

$\iff \left(x-2\right)\times \left(x+7\right)=\left(x+4\right)\times \left(x-1\right)$

$\iff {}^{}+7x-2x-14={}^{}-x+4x-4$

$\iff {}^{}-{}^{}+7x-2x+x-4x-14+4=0$

$\iff$$2x-10=0$

$\Rightarrow x=5$

Câu b thôi các bạn nhé, câu a mình ko cần nx với cả mình ghi sai dữ liệu câu a r

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

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

\(\Leftrightarrow\frac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{2\cdot2\cdot3\cdot2\cdot4\cdot2\cdot5\cdot2\cdot....\cdot31\cdot2\cdot32\cdot2}=2x\)

\(\Leftrightarrow\frac{1}{2\cdot2\cdot2\cdot2\cdot....\cdot2\cdot2\cdot32}=2x\)

Có  : (31 - 1) : 1 + 1 = 31 (thừa số 2) 

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

\(\Rightarrow x=\frac{1}{2^{31}.32}\div2\)

b, \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)

\(\Leftrightarrow x+1=x+4\)

\(\Leftrightarrow0=3\text{ (vô lý) }\)

Tìm x biết :             \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}.\frac{5}{12}...\frac{30}{62}.\frac{31}{64}=2x\) 

 Biến đổi mẫu thức của vế trái :     \(\frac{1}{2.2}.\frac{2}{2.3}.\frac{3}{2.4}.\frac{4}{2.5}.\frac{5}{2.6}...\frac{30}{2.31}.\frac{31}{2.32}=2x\)   Giản ước vế trái đực :

                                               \(\frac{1}{2}.\frac{1}{2}.\frac{1}{2}.\frac{1}{2}.\frac{1}{2}...\frac{1}{2}.\frac{1}{2.32}=2x\)        (*)   

                                                  \(\Leftrightarrow\frac{1}{2^{36}}=2x\Leftrightarrow x=\frac{1}{2^{37}}\)  

a)\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\Leftrightarrow x\left(x-1\right)^{x+2}\left(x-2\right)=0\)

Do đó \(x\in\left\{0;1;2\right\}\)

b)

\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot...\cdot\frac{31}{64}=2^x\Leftrightarrow\frac{1\cdot2\cdot3\cdot...\cdot31}{4\cdot6\cdot8\cdot...\cdot64}=2^x\Leftrightarrow\frac{31!}{\left(2\cdot2\right)\cdot\left(2\cdot3\right)\cdot\left(2\cdot4\right)\cdot...\cdot\left(2\cdot31\right)\cdot64}=2^x\)

\(\frac{31!}{2^{30}\cdot31!\cdot2^6}=2^x\Leftrightarrow\frac{1}{2^{36}}=2^x\Leftrightarrow2^{-36}=2^x\Rightarrow x=-36\)

a)5^x+2=625

<=>5^x+2=5⁴

<=>x+2=4

<=>x=2

b)

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

  \(\frac{1.2.3.4.....30.31}{4.6.8.10....62.64}=2^x\)

 \(\frac{1.2.3.4.5....30.31}{2.2.2.3.2.4.2.5.....2.31.64}=2^x\)

 \(\frac{1.2.3.4.5.....30.31}{\left(2.2.2....2.2\right).\left(2.3.4.5....30.31\right).64}=2^x\)

 \(2.2.2.2.2.....2.64=2^x\)

 \(2^{31}.2^6=2^x\)

\(2^{37}=2^x\)

=> \(x=37\)

a)5^x+2=625

<=>5^x+2=5⁴

<=>x+2=4

<=>x=2