Question

tìm x thuộc Z, biết:

a, (x+2).(3-x)=0

b, (2.x-5)^2=9

c,(1-3.x)^3=8

d,(x^2+1).(49.x^2)=0

Answer 1

a. (x + 2)(3 - x) = 0

\(\Rightarrow\left\{\begin{matrix}x+2=0\\3-x=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=-2\\x=3\end{matrix}\right.\)

b. (2x - 5)² = 9

(2x - 5)² = 3²

=> 2x - 5 = 3

2x = 3 + 5

2x = 8

x = 8 : 2

x = 4

c. (1 - 3x)³ = 8

(1 - 3x)³ = 2³

=> 1 - 3x = 2

3x = 1 - 2

3x = -1

x = \(-\frac{1}{3}\)

Mà x \(\in\) Z

=> x \(\in\phi\)

d. (x² + 1)(49.x²) = 0

\(\Rightarrow\left\{\begin{matrix}x^2+1=0\\49.x^2=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x^2=1\Rightarrow x\in\phi\\x^2=0\Rightarrow x=0\end{matrix}\right.\)

Vậy x = 0