[tex]\Rightarrow f'(x)=9(2x-3)-\frac{9(3x+1)^2}{\sqrt{(3x+1)^3}}[/tex]
Đề bài thì: [tex]9(2x-3)-\frac{9(3x+1)^2}{\sqrt{(3x+1)^3}}=\frac{9}{x}\\\Leftrightarrow 2x-3-\sqrt{3x+1}=\frac{1}{x}\\\Leftrightarrow (2x-3)x-x\sqrt{3x+1}-1=0\\\Leftrightarrow 2x^2-x\sqrt{3x+1}-(3x+1)=0\\\Leftrightarrow 2x^2-xt-t^2=0 (t=\sqrt{3x+1}>0)\\\Leftrightarrow (2x+t)(x-t)=0\\TH1:x=t\\\Leftrightarrow x=\sqrt{3x+1}\Leftrightarrow x=\frac{3+\sqrt{13}}{2}(Do:x>0)\\TH2:2x=-t(Loai(Do:x>0\Rightarrow VT>0,VP<0))[/tex]
Vậy $x=\frac{3+\sqrt{13}}{2}$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
