Gọi 

- Hướng dẫn giải

$$\begin{gathered} Ta có \\ 4{\left(3x-5\right)}^2-9{\left(9x^2-25\right)}^2=0 \\ 4{\left(3x-5\right)}^2-9{\left[{\left(3x\right)}^2-5^2\right]}^2=0 \\ 4{\left(3x-5\right)}^2-9{\left[\left(3x-5\right)\left(3x + 5\right)\right]}^2=0 \\ 4{\left(3x-5\right)}^2-9{\left(3x-5\right)}^2{\left(3x+5\right)}^2=0 \\ {\left(3x-5\right)}^2\left[4-{\left(3\left(3x+5\right)\right)}^2\right]=0 \\ {\left(3x-5\right)}^2\left[4-{\left(3\left(3x+5\right)\right)}^2\right]=0 \\ {\left(3x-5\right)}^2\left(2^2-{\left(9x+15\right)}^2\right)=0 \\ {\left(3x-5\right)}^2\left(2 + 9x + 15\right)\left(2 – 9x – 15\right)=0 \\ {\left(3x-5\right)}^2\left(9x + 17\right)\left(-9x – 13\right)=0 \\ \Rightarrow \left[\begin{array}{l}3x-5=0\\ 9x+17=0\\ -9x-13=0\end{array}\right.\Rightarrow \left[\begin{array}{l}x=\frac{5}{3}\\ x=-\frac{17}{9}\\ x=-\frac{13}{9}\end{array}\right. \end{gathered}$$

$\Rightarrow x_1+x_2+x_3=\frac{5}{3}+\left(-\frac{17}{9}\right)+\left(-\frac{13}{9}\right)=-\frac{5}{3}$

Đáp án cần chọn là: C