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`.

\(\frac{1}{4}.\frac{2}{6}............\frac{31}{64}=2^x\)

\(\Rightarrow\frac{1.2........31}{2.2.2.3...........2.31.64}=2^x\)

\(\Rightarrow\frac{1}{2^{30}.2^4}=2^x\)

\(\Rightarrow\frac{1}{2^{34}}=2^x\)

\(\Rightarrow x=-34\)

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$

$\Rightarrow x=5$

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

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

câu 1 : x=4

câu 2 : x= 57

1: =>\(5^{x-2}-9=2^4-\left(6^2-6^2\right)\)

=>\(5^{x-2}=16+9=25\)

=>x-2=2

=>x=4

2: \(\Leftrightarrow3^x+16=19^6:19^5-3=19-3=16\)

=>3^x=0

=>x=0

3: \(\Leftrightarrow2^x+2^x\cdot16=272\)

=>2^x*17=272

=>2^x=16

=>x=4

4: \(\Leftrightarrow2^{x-1}+3=24-\left(4^2-2^2+1\right)=24-\left(16-4+1\right)\)

=>\(2^{x-1}+3=24-16+4-1=8+4-1=12-1=11\)

=>2^x-1=8

=>x-1=3

=>x=4

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

=> 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