a) \(5x+x=39-3^{11}:3^9\)    

\(\Leftrightarrow6x=39-3^2\)                    

\(\Leftrightarrow6x=30\)

\(\Leftrightarrow x=5\)

b) \(2^x:2^5=16\)

\(\Leftrightarrow2^x:2^5=2^4\)

\(\Leftrightarrow2^x=2^4.2^5\)

\(\Leftrightarrow2^x=2^9\)

\(\Leftrightarrow x=9\)

c) \(7x-x=5^{21}:5^{19}+3.2^2-7^0\)

\(\Leftrightarrow6x=5^2+3.4-1\)

\(\Leftrightarrow6x=36\)

\(\Leftrightarrow x=6\)

d) \(7x-2x=6^{17}:6^{15}+44:11\)

\(\Leftrightarrow5x=6^2+4\)

\(\Leftrightarrow5x=40\)

\(\Leftrightarrow x=8\)

a)⇔6x=39-3²                                             

   ⇔6x=30                                             

   ⇔ x=5                                                

b)2^x:2⁵=16

⇔2^x=2⁴.2⁵ 

⇔ 2^x=2⁹                                                   

⇔ x=9

c)⇔6x=5²+3.2²-1

  ⇔ 6x= 36

  ⇔ x=6

d)⇔5x=6²+4

   ⇔ 5x=40

   ⇔ x=8

a: \(\Leftrightarrow6x=30\)

hay x=5

b: \(\Leftrightarrow6x=25+12-1=36\)

hay x=6

A x=5

B x = 6 (36)

a: \(\Leftrightarrow8x=108+12=120\)

hay x=15

b: \(\Leftrightarrow6x=60\)

hay x=10

a) 5x + 3x = 120
    x . (5+3)= 120
    x . 8       = 120
    x            = 120 : 8
    x            = 15

chi chi mà cả nhả

b: Ta có: 7x-8=713

nên 7x=721

hay x=103

c: Ta có: x-36:18=12

nên x-2=12

hay x=14

d: Ta có: (x-36):18=12

nên x-36=216

hay x=252

\(4^3\cdot4^{x-1}=64\)

\(\Leftrightarrow4^{x-1}=1\)

\(\Leftrightarrow x-1=0\)

hay x=1

a: 3x=81

nên x=27

b: \(5\cdot4^x=80\)

\(\Leftrightarrow4^x=16\)

hay x=2

c: \(2^x=4^5:4^3\)

\(\Leftrightarrow2^x=2^4\)

hay x=4

\(=3^2-147:49=9-3=6\)

=3²-147:49

=9-3

=6

\(180:\left\{\left(-30\right)+\left[4^2.5-\left(14+3^{11}:3^9\right)\right]\right\}\)

\(=180:\left\{\left(-30\right)+\left[16.5-\left(14+3^{11}.\dfrac{1}{3^9}\right)\right]\right\}\)

\(=180:\left\{\left(-30\right)+\left[80-\left(14+3^2\right)\right]\right\}\)

\(=180:\left\{\left(-30\right)+\left[80-\left(14+9\right)\right]\right\}\)

\(=180:\left[\left(-30\right)+80-23\right]\)

\(=180:27\\ =\dfrac{20}{3}\)

Ta có: \(\widehat{xOy}+\widehat{yOx'}=180^0\)(kề bù)

\(\Rightarrow5x+4x=180^0\)

\(\Rightarrow9x=180^0\Rightarrow x=20^0\)

\(\Rightarrow\left\{{}\begin{matrix}\widehat{xOy}=5x=5.20^0=100^0\\\widehat{yOx'}=4x=4.20^0=80^0\end{matrix}\right.\)

Vì xOx'kề bù nên:xOy+yOx'=180 độ

=>5x+4x=180 độ

x.9=180 độ =>x=20 độ

=>xOy=20x5=100 độ

=>yOx'=20x4=80 độ

HT

câu C bn tự làm nha