\(\Leftrightarrow x:3+367\cdot\left(-2\right)=-60\)

=>x:3=674

hay x=2022

x= 2022

  Bài 1:

2^\(x\) = 4

2\(^x\) = 2²

 \(x=2\)

Vậy \(x=2\)

Bài 2:

2\(^x\) = 8

2\(^x\) = 2³

\(x=3\)

Vậy \(x=3\)

Bài 3

2\(^x\) = 16

2\(^x\) = 2⁴

  \(x=4\)

Vậy \(x=4\)

   2^x + 2^x + 2 = 8² - 6² + 2² . 3

= 2^x + 2^x + 2 = 64 - 36 + 4 . 3

= 2^x + 2^x + 2 = 64 - 36 + 12

= 2^x + 2^x + 2 = 28 + 12

= 2^x + 2^x + 2 = 40

= 2^x + 2^x = 40 - 2

= 2^x + 2^x = 38

= 2^x . 2 = 38

= 2^x = 38 : 2

= 2^x = 19

=> x không có giá trị

\(a,\Rightarrow x=19-17=2\\ b,\Rightarrow x+8=28:2=14\\ \Rightarrow x=14-8=6\\ c,\Rightarrow42-x=5^2=25\\ \Rightarrow x=42-25=17\)

a)x=2

b)x+8=14

x=6

c)\(42-x=5^2\)

\(42-x=25\)

\(-x=-17\)

\(x=17\)

a)x+17=19

⇔x=2

b)2(x+8)=28

⇔x+8=14

⇔x=6

c)42-x=5⁸:5⁶

⇔42-x=5²

⇔42-x=25

⇔x=17

a) x = 4 

b) x = 0 vì số nào nhân với 0 cũng bằng 0

OwO

a 

\(x^3=8^2\)   

\(x^3=64\)   

\(x^3=4^3\)   

\(\Rightarrow x=4\)   

b 

\(x^{20}=x\)   

\(x^{20}-x=0\)   

\(x\left(x^{19}-1\right)=0\)   

\(\orbr{\begin{cases}x=0\\x^{19}=1\end{cases}}\)   

\(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

\(a,A=2024=2^3\times11\times23\\B=8^5\times 125^6=\left(2^3\right)^5\times\left(5^3\right)^6=2^{15}\times5^{18}\\ b,Ư\left(84\right)=\left\{1;2;3;4;6;7;12;14;21;28;42;84\right\}\\\Rightarrow x\in\left\{1;2;3;4;6;7;12;14;21;28;42;84\right\}\\ x\in B\left(21\right)=\left\{0;21;42;63;84;105;126;147;168;189;210;....\right\}\)

\(a,\Rightarrow x=30-18=12\\ b,\Rightarrow x+6=45:5=9\\ \Rightarrow x=9-6=3\\ c,\Rightarrow38-3x=4^2=16\\ \Rightarrow3x=38-16=22\\ \Rightarrow x=\dfrac{22}{3}\)