Đáp án: $-\dfrac{31}{10}\pm\sqrt{\dfrac{2241}{100}}$
Giải thích các bước giải:
ĐKXĐ: $x\ne 2,-4$
Ta có:
$\dfrac{x-3}{x-2}+\dfrac{x-2}{x+4}=\dfrac{11}3$
$\to \dfrac{x-3}{x-2}\cdot \:3\left(x-2\right)\left(x+4\right)+\dfrac{x-2}{x+4}\cdot \:3\left(x-2\right)\left(x+4\right)=\dfrac{11}{3}\cdot \:3\left(x-2\right)\left(x+4\right)$
$\to 3\left(x-3\right)\left(x+4\right)+3\left(x-2\right)^2=11\left(x-2\right)\left(x+4\right)$
$\to 6x^2-9x-24=11x^2+22x-88$
$\to 5x^2+31x-64=0$
$\to 5\left(x+\dfrac{31}{10}\right)^2-\dfrac{2241}{20}=0$
$\to 5\left(x+\dfrac{31}{10}\right)^2=\dfrac{2241}{20}$
$\to \left(x+\dfrac{31}{10}\right)^2=\dfrac{2241}{100}$
$\to x+\dfrac{31}{10}=\pm\sqrt{\dfrac{2241}{100}}$
$\to x=-\dfrac{31}{10}\pm\sqrt{\dfrac{2241}{100}}$ thỏa mãn $ĐKXĐ$
