\(x^2+6x+9=0\\ \Leftrightarrow x^2+2.3.x+3^2=0\\ \Leftrightarrow\left(x+3\right)^2=0\\ \Leftrightarrow x+3=0\\ \Leftrightarrow x=-3\)

Vậy \(x=-3\)

\(\Leftrightarrow12x^3-108x^2-7x^2+63x-10x+90=0\\ \Leftrightarrow\left(x-9\right)\left(12x^2-7x-10\right)=0\\ \Leftrightarrow\left(x-9\right)\left(12x^2-15x+8x-10\right)=0\\ \Leftrightarrow\left(x-9\right)\left(4x-5\right)\left(3x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9\\x=\dfrac{5}{4}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left(x-9\right)\left(4x-5\right)\left(3x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=\dfrac{5}{4}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

5^{3x - 3}=5\(^0\)

⇒3x-3=0

3x=0+3

3x=3

x=3:3

x=1

Đáp án A

B P T ⇔ 3 x − 2 x − 1 3 x − 9 ≥ 0 ⇔ 3 x ≥ 2 x + 1 3 x ≥ 9 3 x ≤ 2 x + 1 3 x ≤ 9 ⇔ 3 x ≥ 2 x + 1 x ≥ 2 3 x ≤ 2 x + 1 x ≤ 2 1 .  

P T    3 x = 2 x + 1  có hai nghiệm  x = 0 , x = 1.

Suy ra 1 ⇔ x ≥ 1 x ≤ 0 0 ≤ x ≤ 1 x ≤ 2 ⇔ x ≥ 2 0 ≤ x ≤ 1 ⇒ S = 0 ; 1 ∪ 2 ; + ∞ .

S = 0