Nhớ Vote nha mọi người !
Đk: $x\ge \frac{1}{3}$Ta có: $pt\iff \frac{9x^2-9x-10}{x^2-4x+7}=x\sqrt{3x-1}-2x$
$\iff \frac{(3x-5)(3x+2)}{x^2-4x+7}=\frac{x(3x-5)}{\sqrt{3x-1}+2}$
$\iff x=\frac{5}{3}.v. \frac{3x+2}{x^2-4x+7}=\frac{x}{\sqrt{3x-1}+2}(1)$
$pt(1)\iff (3x+2)\sqrt{3x-1}+2(3x+2)=x(x^2-4x+7)$
$\iff (3x+2)\sqrt{3x-1}=x^3-4x^2+x-4$
$\iff x^3-4x^2+x-4-(3x+2)\sqrt{3x-1}=0$
$\iff x^3-4x^2+x-4-(3x+2)(x-2)+(3x+2)(x-2-\sqrt{3x-1})=0$
$\iff x^3-7x^2+5x+(3x+2)(x-2-\sqrt{3x-1})=0$
$\iff x(x-2-\sqrt{3x-1})(x-2+\sqrt{3x-1})+(3x+2)(x-2-\sqrt{3x-1})=0$
$\iff (x-2-\sqrt{3x-1})[x^2-2x+x\sqrt{3x-1}+3x+2]=0$
$\iff (x-2-\sqrt{3x-1})[x^2+x+2+x\sqrt{3x-1}]=0$
Do $x\ge \frac{1}{3}$ nên $[...]>0$
Suy ra $x-2-\sqrt{3x-1}=0\iff (x-2)^2=3x-1\iff x^2-7x+5=0$(x>=2)
$\iff x=\frac{7+\sqrt{29}}{2}.v.x=\frac{7-\sqrt{29}}{2}(l)$
Bạn tự KL nhé. Nhớ Vote! và chấm đúng nha!