\(ĐKXĐ:x\ne\frac{5-\sqrt{13}}{2};x\ne\frac{5+\sqrt{13}}{2}\)
\(\frac{4x}{x^2+x+3}+\frac{5x}{x^2-5x+3}=-\frac{3}{2}\)
*) Xét x = 0 thì \(\frac{4x}{x^2+x+3}+\frac{5x}{x^2-5x+3}=0\)(Loại)
*) Xét \(x\ne0\)thì phương trình tương đương \(\frac{4}{x+\frac{3}{x}+1}+\frac{5}{x+\frac{3}{x}-5}=-\frac{3}{2}\)
Đặt \(x+\frac{3}{x}=t\)thì phương trình trở thành \(\frac{4}{t+1}+\frac{5}{t-5}=-\frac{3}{2}\)
\(\Leftrightarrow\frac{4t-20+5t+5}{\left(t+1\right)\left(t-5\right)}=-\frac{3}{2}\Leftrightarrow\frac{9t-15}{t^2-4t-5}=-\frac{3}{2}\)
\(\Leftrightarrow18t-30=-3t^2+12t+15\Leftrightarrow3t^2+6t-45=0\)
\(\Leftrightarrow3\left(t-3\right)\left(t+5\right)=0\Leftrightarrow\orbr{\begin{cases}t=3\\t=-5\end{cases}}\)
+) t = 3 thì \(x+\frac{3}{x}=3\Leftrightarrow\frac{x^2+3}{x}=3\Leftrightarrow x^2-3x+3=0\)
Mà \(x^2-3x+3=\left(x-\frac{3}{2}\right)^2+\frac{3}{4}>0\forall x\)nên loại trường hợp t = 3
+) t = -5 thì \(x+\frac{3}{x}=-5\Leftrightarrow\frac{x^2+3}{x}=-5\Leftrightarrow x^2+5x+3=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-5+\sqrt{13}}{2}\\x=\frac{-5-\sqrt{13}}{2}\end{cases}}\)
Vậy phương trình có 2 nghiệm \(\left\{\frac{-5+\sqrt{13}}{2};\frac{-5-\sqrt{13}}{2}\right\}\)
Bài làm:
đkxđ: \(x\ne\left\{\frac{5+\sqrt{13}}{2};\frac{5-\sqrt{13}}{2}\right\}\)
+ Nếu x = 0:
\(Pt\Leftrightarrow0=-\frac{3}{2}\)(vô nghiệm)
+ Nếu x khác 0:
\(Pt\Leftrightarrow\frac{4x}{x\left(x+\frac{3}{x}+1\right)}+\frac{5x}{x\left(x+\frac{3}{x}-5\right)}=-\frac{3}{2}\)
\(\Leftrightarrow\frac{4}{x+\frac{3}{x}+1}+\frac{5}{x+\frac{3}{x}-5}=-\frac{3}{2}\)
Đặt \(x+\frac{3}{x}=y\)
\(Pt\Leftrightarrow\frac{4}{y+1}+\frac{5}{y-5}=-\frac{3}{2}\)
\(\Leftrightarrow\frac{8\left(y-5\right)+10\left(y+1\right)}{2\left(y+1\right)\left(y-5\right)}=-\frac{3\left(y-5\right)\left(y+1\right)}{2\left(y+1\right)\left(y-5\right)}\)
\(\Rightarrow8y-40+10y+10=-3\left(y^2-4y-5\right)\)
\(\Leftrightarrow18y-30=-3y^2+12y+15\)
\(\Leftrightarrow3y^2+6y-45=0\)
\(\Leftrightarrow y^2+2y-15=0\)
\(\Leftrightarrow\left(y-3\right)\left(y+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}y-3=0\\y+5=0\end{cases}}\Leftrightarrow\Leftrightarrow\orbr{\begin{cases}y=3\\y=-5\end{cases}}\)
Nếu: \(y=3\Leftrightarrow x+\frac{3}{x}=3\Leftrightarrow\frac{x^2+3}{x}=3\Leftrightarrow x^2+3=3x\)
\(\Leftrightarrow x^2-3x+3=0\)
\(\Leftrightarrow\left(x^2-3x+\frac{9}{4}\right)+\frac{3}{4}=0\)
\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2=-\frac{3}{4}\)(vô lý)
=> không tồn tại x thỏa mãn
Nếu: \(y=-5\Leftrightarrow x+\frac{3}{x}=-5\Leftrightarrow\frac{x^2+3}{x}=-5\Leftrightarrow x^2+3=-5x\)
\(\Leftrightarrow x^2+5x+3=0\)
\(\Leftrightarrow\left(x^2+5x+\frac{25}{4}\right)-\frac{13}{4}=0\)
\(\Leftrightarrow\left(x+\frac{5}{2}\right)^2-\left(\frac{\sqrt{13}}{2}\right)^2=0\)
\(\Leftrightarrow\left(x+\frac{5}{2}-\frac{\sqrt{13}}{2}\right)\left(x+\frac{5}{2}+\frac{\sqrt{13}}{2}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{5-\sqrt{13}}{2}=0\\x+\frac{5+\sqrt{13}}{2}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{13}-5}{2}\\x=\frac{-5-\sqrt{13}}{2}\end{cases}}\)(thỏa mãn)
Vậy tập nghiệm của PT \(S=\left\{\frac{-5-\sqrt{13}}{2};\frac{\sqrt{13}-5}{2}\right\}\)
