\(\Leftrightarrow\frac{x^2\left(x+3\right)^2+9x^2}{\left(x+3\right)^2}=40\)

\(\Leftrightarrow x^2\left(x^2+6x+9\right)+9x^2=40\left(x^2+6x+9\right)\)

\(\Leftrightarrow x^4+6x^3-22x^2-240x-360=0\)

\(\Leftrightarrow x^4-6x^3+12x^3-72x^2+50x^2-300x+60x-360=0\)

\(\Leftrightarrow x^3\left(x-6\right)+12x^2\left(x-6\right)+50x\left(x-6\right)+60\left(x-6\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x^3+12x^2+50x+60\right)=0\Rightarrow x=6\)

Làm

\(x^2+\frac{9x^2}{\left(x+3\right)^2}=40\)\(\left(x\ne-3,x\ne0\right)\)

(=) \(\left(x-\frac{3x}{x+3}\right)^2+\frac{6x^2}{x+3}-40=0\)

(=)\(\left(\frac{x^2}{x+3}\right)^2+6.\frac{x^2}{x+3}-40=0\)

(=) a²+6a-40=0 (với a = \(\frac{x^2}{x+3}\))

(=) (a-4)(a+10)=0

(=)\(\left[{}\begin{matrix}a-4=0\\a+10=0\end{matrix}\right.\)

(=)\(\left[{}\begin{matrix}x^2-4x-12=0\left(1\right)\\x^2+10x+30=0\left(2\right)\end{matrix}\right.\)

(1) (=) \(\left[{}\begin{matrix}x=-2\left(chọn\right)\\x=6\left(chọn\right)\end{matrix}\right.\)

(2)(=) x²+10x+30 =0

Mà x²+10x+30 >0 => Pt vô ng

Kl: Vậy...........................

\(x^2+\frac{9x^2}{\left(x+3\right)^2}=40\) (ĐK x ≠ -3)

\(\Leftrightarrow\left[x^2+\frac{6x^2}{x+3}+\frac{9x^2}{\left(x+3\right)^2}\right]-\frac{6x^2}{x+3}-40=0\)

\(\Leftrightarrow\left(x-\frac{3x}{x+3}\right)-\frac{6x^2}{x+3}-40=0\)

\(\Leftrightarrow\left(\frac{x^2}{x+3}\right)^2+6.\frac{x^2}{x+3}-40=0\left(1\right)\)

Đặt \(\frac{x^2}{x+3}=a\)

\(\left(1\right)\Leftrightarrow a^2+6a-40=0\Leftrightarrow\left[{}\begin{matrix}a=4\\a=-10\end{matrix}\right.\)

\(\left[{}\begin{matrix}\frac{x^2}{x+3}=4\\\frac{x^2}{x+3}=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2-4x-12=0\\x^2+10x+30=0\left(vn\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)

Vậy x = 6, x= -2

= \(\left[\frac{x.\left(x+3\right)}{\left(x+3\right).\left(x^2+9\right)}+\frac{3}{x+9}\right]:\left[\frac{1}{x-3}-\frac{6x}{\left(x-3\right)\left(x^2+9\right)}\right]\) ]

 \(=\frac{x+3}{x^2-9}.\frac{\left(x-3\right).\left(x^2+9\right)}{x^2+9-6x}\)

= \(\frac{\left(x-3\right).\left(x+3\right)}{\left(x-3\right)^2}\)

= \(\frac{x+3}{x-3}\)

k mik nhé. Plssss~

  \(\left(\frac{x^2+3x}{x^3+3x^2+9x+27}\right)\): \(\left(\frac{1}{x-3}-\frac{6x}{x^3-3x^2+9x-27}\right)\)

=\(\left[\frac{x\left(x+3\right)}{x^2\left(x+3\right)+9\left(x+3\right)}\right]\):\(\left[\frac{1}{x-3}-\frac{6x}{x^2\left(x-3\right)+9\left(x-3\right)}\right]\)

=\(\left[\frac{x\left(x-3\right)}{\left(x^2+9\right)\left(x-3\right)}\right]\):\(\left[\frac{1}{x-3}-\frac{6x}{\left(x^2+9\right)\left(x-3\right)}\right]\)

=\(\frac{x}{x^2+9}\):\(\left[\frac{x^2+9}{\left(x-3\right)\left(x^2+9\right)}-\frac{6x}{\left(x-3\right)\left(x^2+9\right)}\right]\)

=\(\frac{x}{x^2+9}\):\(\frac{\left(x-3\right)^2}{\left(x-3\right)\left(x^2+9\right)}\)

=\(\frac{x}{x^2+9}\):\(\frac{x-3}{x^2+9}\)

=\(\frac{x}{x^2+9}\).\(\frac{x^2+9}{x-3}\)

=\(\frac{x}{x-3}\)

\(ĐKXĐ:x\ne\pm3\)

\(P=\left(\frac{x^2-3x}{x^3+3x^2+9x+27}+\frac{3}{x^2+9}\right):\left(\frac{1}{x-3}-\frac{6x}{x^3-3x^2+9x-27}\right)\)

\(\Leftrightarrow P=\left(\frac{x^2-3x}{\left(x+3\right)\left(x^2+9\right)}+\frac{3}{x^2+9}\right):\left(\frac{1}{x-3}-\frac{6x}{\left(x-3\right)\left(x^2+9\right)}\right)\)

\(\Leftrightarrow P=\frac{\left(x^2-3x\right)+3\left(x+3\right)}{\left(x+3\right)\left(x^2+9\right)}:\frac{x^2+9-6x}{\left(x-3\right)\left(x^2+9\right)}\)

\(\Leftrightarrow P=\frac{x^2+9}{\left(x+3\right)\left(x^2+9\right)}:\frac{\left(x-3\right)^2}{\left(x-3\right)\left(x^2+9\right)}\)

\(\Leftrightarrow P=\frac{1}{x+3}:\frac{x-3}{x^2+9}\)

\(\Leftrightarrow P=\frac{x^2+9}{\left(x+3\right)\left(x-3\right)}\)

\(x^2+\frac{9x^3}{\left(x+3\right)^2}=40\left(x\ne-3\right)\)

\(\Leftrightarrow x^2+\left(x+3\right)^2+9x^2=40\left(x+3\right)^2\)

\(\Leftrightarrow x^4+6x^3+18x^2=40x^2+240x+360\)

\(\Leftrightarrow x^4+6x^3-22x^2-240x-360=0\)

\(\Leftrightarrow\left(x^3+10x+30\right)\left(x-6\right)\left(x+2\right)=0\)

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............................