\(x^2+\frac{9x^3}{\left(x+3\right)^2}=40\left(x\ne-3\right)\)
\(\Leftrightarrow x^2+\left(x+3\right)^2+9x^2=40\left(x+3\right)^2\)
\(\Leftrightarrow x^4+6x^3+18x^2=40x^2+240x+360\)
\(\Leftrightarrow x^4+6x^3-22x^2-240x-360=0\)
\(\Leftrightarrow\left(x^3+10x+30\right)\left(x-6\right)\left(x+2\right)=0\)
Khi x-6=0 hoặc x+2=0 <=> x=6 hoặc x=-2
Khi \(x^3+10x+30=0\)
\(x=\frac{-10+2\sqrt{5}}{2};x=\frac{-10-2\sqrt{5}}{2}\)
Hơi khó hiểu 1 chút, bạn cố gắng nhé
\(x^2+\frac{9x^2}{\left(x+3\right)^2}=40^{\left(1\right)}\)
\(ĐKXĐ:x\ne-3\)
\(\left(1\right)\Leftrightarrow x^2-2.x.\frac{3x}{x+3}+\frac{\left(3x\right)^2}{\left(x+3\right)^2}+\frac{6x^2}{x+3}=40\)
\(\Leftrightarrow\left(x-\frac{3x}{x+3}\right)^2+\frac{6x^2}{x+3}=40\)
\(\Leftrightarrow\left(\frac{x^2}{x+3}\right)^2+6.\frac{x^2}{x+3}=40\)
Đặt \(t=\frac{x^2}{x+3}\)ta có
\(t^2+6t=40\)
\(\Leftrightarrow\left(t-4\right)\left(t+10\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}t-4=0\\t+10=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}t=4\\t=-10\end{cases}}\)
+) Với t =4 ta có
\(\frac{x^2}{x+3}=4\)
\(\Rightarrow4\left(x+3\right)=x^2\)
\(\Leftrightarrow x^2-4x-12=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x+2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=6\left(tm\right)\\x=-2\left(tm\right)\end{cases}}\)
+) với x=-10 ta có
\(\frac{x^2}{x+3}=-10\)
\(\Rightarrow-10\left(x+3\right)=x^2\)
\(\Leftrightarrow x^2+10x+30=0\)
\(\Leftrightarrow\left(x+5\right)^2=-5\)
Phương trình vô nghiệm
Vậy............................
