Question

B = {x ∈ ℤ| (x2 – 10x + 21)(x^3 – x) = 0}

Answer 1

(x^2-10x+21)(x^3-x)=0

=>(x-3)(x-7)*x*(x^2-1)=0

=>x thuộc {0;1;-1;3;7}

=>B={0;1;-1;3;7}

Answer 2

Ta có:

\(\left(x^2-10x+21\right)\left(x^2-x\right)=0\)

\(\Leftrightarrow\left(x^2-3x-7x+21\right)x\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-7\right)x\left(x-1\right)=0\) (ĐK: \(x\in Z\))

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=3\left(tm\right)\\x=7\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)

\(\Rightarrow B=\left\{1;3;7;0\right\}\)

Answer 3

(x² - 10x + 21)(x³ - x) = 0

⇔ x(x² - 3x - 7x + 21)(x² - 1) = 0

⇔ x[(x² - 3x) - (7x - 21)](x - 1)(x + 1) = 0

⇔ x[x(x - 3) - 7(x - 3)](x - 1)(x + 1) = 0

⇔ x(x - 3)(x - 7)(x - 1)(x + 1) = 0

⇔ x = 0; x = 3; x = 7; x = 1; x = -1

⇒ B = {-1; 0; 1; 3; 7}