`a, 177 - 15x = 2^3 . 3^2`
`=> 177 - 15x = 8 . 9`
`=> 177 - 15x = 72`
`=> 15x = 177 - 72`
`=>15x=105`
`=>x=105:15`
`=>x=7`
Vậy: `x=7`
`b,` Ta có:
`x\vdots 6`
`-> x \in B(6)={0;6;12;18;24;30;...}`
Mà `x < 31` nên `x = {0;6;12;18;24;30}`
Vậy: `x = {0;6;12;18;24;30}`
`d, x(x-15)=0`
`=> [(x = 0),(x-15=0):}`
`=> [(x = 0),(x=0+15):}`
`=> [(x=0),(x=15):}`
Vậy: `x=0;x=15`
`a, 177 - 15x =2^3 * 3^2 `
`=> 177 - 15x = 8 * 9`
`=> 177 - 15x = 72`
`=> 15x = 177 - 72`
`=> 15x = 105`
`=> x = 105 : 15`
`=> x = 7`
Vậy `x = 7`
`b, x \vdots 6 `và `x < 31`
`x \vdots 6` thì `=> x \in B(6) = {0;6;12;18;24;30;36;...}`
Mà `x < 31`
`=> x \in {0;6;12;18;24;30}`
\(d,x\left(x-15\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-15=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=15\end{matrix}\right.\)
Vậy `x \in {0;15}`
`c,` Xem lại đề
