a, \(x^2+\frac{9x^2}{\left(x-3\right)^2}=16\) (đk: \(x\ne3\))
<=> \(x^2+\frac{6x^2}{x-3}+\frac{9x^2}{\left(x-3\right)^2}-\frac{6x^2}{x-3}-16=0\)
<=>\(\left(x+\frac{3x}{x-3}\right)^2-\frac{6x^2}{x-3}-16=0\)
<=>\(\left(\frac{x^2}{x-3}\right)^2-\frac{6x^2}{x-3}-16=0\)
Đặt a=\(\frac{x^2}{x-3}\)
Có: \(a^2-6a-16=0\)
<=> (a+2)(a-8)=0
<=> \(\left[{}\begin{matrix}a=-2\\a=8\end{matrix}\right.\)
=> \(\frac{x^2}{x-3}=-2\) hoặc \(\frac{x^2}{x-3}=8\)
Tại \(\frac{x^2}{x-3}=-2\) <=> \(x^2+2x-6=0\)
\(\Delta=2^2-4\left(-6\right)=28>0\)
=> \(\sqrt{\Delta}=\sqrt{28}\)
=> \(\left[{}\begin{matrix}x_1=\frac{-2+2\sqrt{7}}{2}=-1+\sqrt{7}\\x_2=\frac{-2-2\sqrt{7}}{2}=-1-\sqrt{7}\end{matrix}\right.\)(tm)
Tại \(\frac{x^2}{x-3}=8\) <=> \(x^2-8x+24=0\) <=> (x-4)2+8=0( vô nghiệm)
Vậy....
