3^-1 . 3^x + 5.3^{x - 1} = 162

=> 3^{x - 1} + 5.3^{x - 1} = 162

=> 3^{x - 1}. (1 + 5) = 162

=> 3^{x - 1} . 6 = 162

=> 3^{x - 1} = 162

=> 3^{x - 1} = 27

=> 3^{x - 1} = 3³

=> x - 1 = 3

=> x = 3 + 1

=> x =4

3^-1 . 3^x + 5.3^{x -1} = 162

=> 1/3 . 3^x + 5.3^x : 3 = 162

=> 1/3 . 3^x + 5/3 . 3^x = 162

=> 3^x . (1/3 + 5/3)     = 162

=> 3^x . 2                    = 162

=> 3^x                         = 162 : 2

=> 3^x                         = 81

=> 3^x                         = 3⁴

=>   x                        = 4

32^x . 16^x = 1024

=> (2⁵)^x . (2⁴)^x = 2¹⁰

=> 2^5x . 2^4x      = 2¹⁰

=> 2^{5x + 4x}        = 2¹⁰

=> 2^9x              = 2¹⁰

=>  9x              = 10

=>    x              = 10 : 9

=>    x              = 10/9

TL:

\(32^x.16^x=1024\) 

\(\left(2^5\right)^x.\left(2^4\right)^x=1024\) 

\(2^{9x}=2^{10}\) 

\(\Rightarrow9x=10\Leftrightarrow x=\frac{10}{9}\) 

Vậy............

3/1/2 - 1/2x = 2/3

=> 3/2 - 1/2x = 2/3

=>          1/2x = 3/2 - 2/3

=>          1/2x = 5/6

=>               x  = 5/6 : 1/2

=>               x  = 5/3

\(\frac{3}{2}-\frac{1}{2}x=\frac{2}{3}\)

\(\frac{1}{2}x=\frac{3}{2}-\frac{2}{3}\)

\(\frac{1}{2}x=\frac{5}{6}\)

\(\frac{5}{6}:\frac{1}{2}=x\)

\(\frac{5}{3}=x\)

Giải: Do ON là tia p/giác của \(\widehat{AOM}\) nên :

       \(\widehat{O_1}=\widehat{O_2}=\frac{\widehat{AOM}}{2}=25^0\)

=>  \(\widehat{AOM}=2.\widehat{O_1}=2.25^0=50^0\)

Do OM là tia p/giác của \(\widehat{AOB}\)nên

  \(\widehat{O_3}=\widehat{AOM}=\frac{\widehat{AOB}}{2}=50^0\)

=> \(\widehat{AOB}=2.\widehat{AOM}=2.50^0=100^0\)

Vậy ...

#)Giải :

Vì \(x^0=1\)

\(\Rightarrow x+3=0\Rightarrow x=-3\)

vì x⁰=1

=>x-3=0=>x=-3

giải em viết nhầm ah

\(280000-\left\{3^6.\left[370-\left(2^4:2^2.13\right)\right]\right\}\)

\(=280000-\left\{729.\left[370-\left(16:4.13\right)\right]\right\}\)

\(=280000-\left\{729.\left[370-52\right]\right\}\)

\(=280000-\left\{729.318\right\}\)

\(=280000-231822\)

\(=48178\)