Question

[Lê Ngọc Hân]

a) ( x + 1 ) (x^2 - 4 ) = 0 

b)(x-2) ( x^2+1 ) = 0

c) 13. ( x - 5 ) = -169

d) x. ( x - 2 ) = 0

Answer 1

[Nguyễn Thị Thương Hoài]

a; (\(x+1\))(\(x^2\) - 4) = 0

      \(\left[{}\begin{matrix}x+1=0\\x^2-4=0\end{matrix}\right.\)

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

       Vậy \(x\in\) {-1; -2; 2}

b; (\(x\) - 2).(\(x^2\) + 1) = 0

    Vì \(x^2\) ≥ 0 ∀ \(x\); \(x\)² + 1 ≥ 1 > 0 ∀ \(x\)

 ⇒ \(x-2\) = 0 ⇒ \(x\) = 2

Vậy \(x=2\)

c; 13.(\(x-5\)) = - 169

          \(x-5\) = 169 : 13

           \(x-5\) = -13

           \(x=-13+5\)

Vậy \(x=-8\); 

d; \(x.\left(x-2\right)\) = 0

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

         \(\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

     Vậy \(x\in\) {0; 2}

Answer 2

[NGUYỄN HOÀNG KHÁNH LINH]

khó quá