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

=> 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

22222222222222222

  Ta có :  

\(\frac{x^3}{8}\)= \(\frac{y^3}{64}\)= \(\frac{z^3}{216}\) \(\Rightarrow\)\(\frac{x^3}{2^3}\)= \(\frac{y^3}{4^3}\)= \(\frac{z^3}{6^3}\)\(\Rightarrow\)\(\frac{x^2}{2^2}\)=\(\frac{y^2}{4^2}\)=\(\frac{z^2}{6^2}\)

và có : \(^{x^2+y^2+z^2=224}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{x^2}{4}=\frac{y^2}{16}=\frac{z^2}{36}=\frac{x^2+y^2+z^2}{4+16+36}=\frac{224}{56}=4\)

=>    \(\frac{x^2}{4}=4\Rightarrow x^2=16\Rightarrow x\in4;-4\)​

\(\frac{y^2}{16}=4\Rightarrow y^2=64\Rightarrow y\in8:-8\)

\(\frac{z^2}{36}=4\Rightarrow z^2=144\Rightarrow z\in12:-12\)

Vì \(\frac{x^3}{8}=\frac{y^3}{64}=\frac{z^3}{216}\)nên x,y,z cùng dấu 

Vậy \(x,y,z\in\left(4;8;12\right);\left(-4;-8;-12\right)\)

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`.

(x - 5)² = 16

=> (x - 5)² = 4²

=> \(\orbr{\begin{cases}x-5=4\\x-5=-4\end{cases}}\)

=> \(\orbr{\begin{cases}x=9\\x=1\end{cases}}\)

(2x - 1)³ = -64

=> (2x - 1)³ = -4³

=> 2x - 1 = -4

=> 2x = -4 + 1

=> 2x = -3

=> x = -3/2

( x - 5)² = 16

=> (x - 5)² = 4²

=> \(\orbr{\begin{cases}x-5=4\\x-5=-4\end{cases}}\)

=> \(\orbr{\begin{cases}x=9\\x=1\end{cases}}\)

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

5 + ( x + 27 ) = 64

a) \(\Rightarrow x+27=59\Rightarrow x=32\)

b) \(\Rightarrow x-2=39\Rightarrow x=41\)

c) \(\Rightarrow x+5=-322\Rightarrow x=-327\)

d) \(\Rightarrow5x=35\Rightarrow x=7\)

e) \(\Rightarrow4\left(x-5\right)=56\Rightarrow x-5=14\Rightarrow x=19\)

f) \(\Rightarrow15+x=37\Rightarrow x=22\)

g) \(\Rightarrow7\left(13-x\right)=35\Rightarrow13-x=5\Rightarrow x=8\)

h) \(\Rightarrow10\left(x+1\right)=100\Rightarrow x+1=10\Rightarrow x=9\)