\(\dfrac{24}{x^2-2x+4}=\dfrac{3x}{x+2}+\dfrac{72}{x^2+8}\) (1)
đkxđ: \(x\ne-2\)
\(\Leftrightarrow\dfrac{24}{x^2-2x+4}-\dfrac{3}{x+2}-\dfrac{72}{\left(x+2\right)\left(x^2-2x+4\right)}=0\)\(\Leftrightarrow\dfrac{24\left(x+2\right)}{x^3+8}-\dfrac{3\left(x^2-2x+4\right)}{x^3+8}-\dfrac{72}{x^3+8}=0\)\(\Rightarrow24x+48-3x^2+6x-12-72=0\)
\(\Leftrightarrow-3x^2+30x-36=0\)
\(\Leftrightarrow-3\left(x^2-10x+12\right)=0\)
\(\Leftrightarrow x^2-10x+12=0\)
\(\Leftrightarrow\left(x^2-10x+25\right)=13\)
\(\Leftrightarrow\left(x-5\right)^2=13\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=\sqrt{13}\\x-5=-\sqrt{13}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{13}+5\\x=-\sqrt{13}+5\end{matrix}\right.\)
\(\dfrac{24}{x^2-2x+4}=\dfrac{3x}{x+2}+\dfrac{72}{x^3+8}\) ĐK: x>0
\(\Leftrightarrow\dfrac{24\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=\dfrac{3x\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}+\dfrac{72}{\left(x+2\right)\left(x^2-2x+4\right)}\)
\(\Rightarrow24\left(x+2\right)=3x\left(x^2-2x+4\right)+72\\ \Leftrightarrow24x+48-3x^3+6x^2-12x-216x-72=0\)
\(\Leftrightarrow-3x^3+6x^2-204x-24=0\)
Tới đây là cách giải phương trình bậc 3 thì mk xin chịu :3
Cách giải Phương trình bậc 3 bạn xem tham khải tại đây: https://vi.wikipedia.org/wiki/Ph%C6%B0%C6%A1ng_tr%C3%ACnh_b%E1%BA%ADc_ba