Question

[Christy Devil]

Giúp mình nhé !

Tìm x thuộc Z

a) x^2 + 11x = 0

b) ( x^2 - 1 ) . ( x^2 - 9 ) = 0

c) ( | x + 1 | - 5 ) . ( x^2 - 9 ) = 0

d) 3x - 16 chia hết cho x + 2

( Lưu ý : x^2 là x mũ 2.
11x là 11 nhân x
Dấu ". " là dấu nhân )

Giúp mình nhé!

Answer 1

[bảo nam trần]

a, x² + 11x = 0

x(x + 11) = 0

\(\Rightarrow\left[\begin{matrix}x=0\\x+11=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=0\\x=-11\end{matrix}\right.\)

b, (x² - 1)(x² - 9) = 0

\(\Rightarrow\left[\begin{matrix}x^2-1=0\\x^2-9=0\end{matrix}\right.\)\(\Rightarrow\left[\begin{matrix}x^2=1\\x^2=9\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=1\\x=-1\\x=3\\x=-3\end{matrix}\right.\)

Vậy \(x\in\left\{1;-1;3;-3\right\}\)

c, ( |x + 1| - 5)(x² - 9) = 0

\(\Rightarrow\left[\begin{matrix}\left|x+1\right|-5=0\\x^2-9=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}\left|x+1\right|=5\\x^2=9\end{matrix}\right.\)

\(\Rightarrow\left[\begin{matrix}x+1=5\\x+1=-5\\x=3\\x=-3\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=4\\x=-6\\x=3\\x=-3\end{matrix}\right.\)

Vậy \(x\in\left\{4;-6;3;-3\right\}\)

d, \(3x-16⋮x+2\)

\(\Rightarrow3x+6-22⋮x+2\)

\(\Rightarrow3\left(x+2\right)-22⋮x+2\)

Vì \(3\left(x+2\right)⋮x+2\) nên để \(3\left(x+2\right)-22⋮x+2\) thì \(22⋮x+2\)

\(\Rightarrow x+2\inƯ\left(22\right)=\left\{\pm1;\pm2;\pm11;\pm22\right\}\)

x + 2	1	-1	2	-2	11	-11	22	-22
x	-1	-3	0	-4	9	-13	20	-24

Vậy x = {-1;-3;0;-4;9;-13;20;-24}