\(x\inℤ\) hả 

x thuộc Z

(6x - 11)³ = (-3)² . 15 + 208

(6x - 11)³ =  9 . 15 + 208

(6x - 11)³ = 135 + 208

(6x - 11)³ = 343

(6x - 11)³ = 7³ (cùng số mũ)

⇒6x - 11 = 7

6x = 7 + 11

6x = 18

x  =3

\(\left(7x-11\right)^3=2^5.5^2+200\)

\(\Leftrightarrow\left(7x-11\right)^3=1000\)

\(\Leftrightarrow7x-11=10\)

\(\Leftrightarrow7x=21\)

\(\Leftrightarrow x=3\)

\(\left(7x-11\right)^3=2^5.5^2+200\)

\(\Rightarrow\left(7x-11\right)^3=32.25+200\)

\(\Rightarrow\left(7x-11\right)^3=800+200\)

\(\Rightarrow\left(7x-11\right)^3=1000=10^3\)

\(\Rightarrow7x-11=10\)

\(\Rightarrow7x=21\)

\(\Rightarrow x=3\)

(7x-11)³ = 2⁵.5²+ 2³.5²

(7x-11)³ = 2³.5²( 2²+1)

(7x-11)³ = 2³.5³=(2.5)³ 

7x-11=10

x=3

ban anhduc1501 hinh nhu lam chua dung to nghi vay

\(164:\left\{192-\left[2^3\cdot15-\left(40-37\right)^2\right]+2021^0\right\}\)

\(=164:\left\{192-120+9+1\right\}\)

\(=164:82=2\)

1. \(x⋮12,x⋮10\Rightarrow x\in BC(12,10)\)và -200 < x < 200

Theo đề bài , ta có :

\(12=2^2\cdot3\)

\(10=2\cdot5\)

\(\Rightarrow BCNN(10,12)=2^2\cdot3\cdot5=60\)

\(\Rightarrow BC(10,12)=B(60)=\left\{0;60;-60;120;-120;180;-180;240;...\right\}\)

Mà \(x\in BC(10,12)\)và -200 < x < 200 => \(x\in\left\{0;60;-60;120;-120;180;-180\right\}\)

Học tốt

\(\frac{47}{12}\left|x\right|=1\frac{11}{31}\cdot4\frac{3}{7}-\left(1,5-3\frac{1}{3}\right)\)

=> \(\frac{47}{12}\left|x\right|=\frac{42}{31}\cdot\frac{31}{7}-\left(\frac{3}{2}-\frac{10}{3}\right)\)

=> \(\frac{47}{12}\left|x\right|=6-\left(-\frac{11}{6}\right)\)

=> \(\frac{47}{12}\left|x\right|=\frac{47}{6}\)

=> \(\left|x\right|=\frac{47}{6}:\frac{47}{12}\)

=> \(\left|x\right|=\frac{47}{6}\cdot\frac{12}{47}\)

=> \(\left|x\right|=2\)

=> \(x\in\left\{2;-2\right\}\)