GIẢI PHƯƠNG TRÌNH SAU 12= (2X-20)(6/(X-8)+0.5) 05/11/2021 Bởi Bella GIẢI PHƯƠNG TRÌNH SAU 12= (2X-20)(6/(X-8)+0.5)
Cách giải: $ĐKXĐ:x \neq 8$ $12=(2x-20).(\dfrac{6}{x-8}+\dfrac{1}{2})$ $\to 6=(x-10).\dfrac{12+x-8}{2(x-8)}$ $\to 6=(x-10).\dfrac{x+4}{2(x-8)}$ $\to 12(x-8)=(x-10)(x+4)$ $\to 12x-96=x^2-6x-40$ $\to x^2-18x+56=0$ $\to x^2-2.x.9+81-27=0$ $\to (x-9)^2=27$ $\to \left[ \begin{array}{l}x-9=3\sqrt{3}\\x-9=-3\sqrt{3}\end{array} \right.$ $\to \left[ \begin{array}{l}x=3\sqrt{3}+9\\x=-3\sqrt{3}+9\end{array} \right.$ Vậy $\left[ \begin{array}{l}x=3\sqrt{3}+9\\x=-3\sqrt{3}+9\end{array} \right.$ Bình luận
Cách giải:
$ĐKXĐ:x \neq 8$
$12=(2x-20).(\dfrac{6}{x-8}+\dfrac{1}{2})$
$\to 6=(x-10).\dfrac{12+x-8}{2(x-8)}$
$\to 6=(x-10).\dfrac{x+4}{2(x-8)}$
$\to 12(x-8)=(x-10)(x+4)$
$\to 12x-96=x^2-6x-40$
$\to x^2-18x+56=0$
$\to x^2-2.x.9+81-27=0$
$\to (x-9)^2=27$
$\to \left[ \begin{array}{l}x-9=3\sqrt{3}\\x-9=-3\sqrt{3}\end{array} \right.$
$\to \left[ \begin{array}{l}x=3\sqrt{3}+9\\x=-3\sqrt{3}+9\end{array} \right.$
Vậy $\left[ \begin{array}{l}x=3\sqrt{3}+9\\x=-3\sqrt{3}+9\end{array} \right.$