Question

Answer 1

3b;

-2x - 2 = (x + 1)² ⇔ -2.(x+1) = (x+1)²

⇔ 0=2(x+1) + (x+1)²⇔ (x+1).(x+1+2) =0

⇔ (x+1) = 0 hoặc (x+3) = 0 ⇔ x = -1 hoặc x = -3

3c;

4x³ + 16 = 0

⇔ x³ + 4 = 0

⇔ x³ = -4 

 

Answer 2

Bài 3: 

b: Ta có: \(-2x-2=\left(x+1\right)^2\)

\(\Leftrightarrow\left(x+1\right)^2+2\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+3\right)=0\)

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

c: Ta có: \(4x^3+16=0\)

\(\Leftrightarrow x^3+4=0\)

\(\Leftrightarrow x^3=-4\)

hay \(x=\sqrt[3]{-4}\)