a) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=27\)

\(\Rightarrow x^3+3^3-x\left(x^2-4\right)=27\)

\(\Rightarrow x^3+27-x^3+4x=27\)

\(\Rightarrow27+4x=27\)

\(\Rightarrow4x=0\)

\(\Rightarrow x=0\)

b) \(2x^2+7x+3=0\)

\(\Rightarrow2x^2+x+6x+3=0\)

\(\Rightarrow x\left(2x+1\right)+3\left(2x+1\right)=0\)

\(\Rightarrow\left(2x+1\right)\left(x+3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x+1=0\\x+3=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=-1\\x=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-3\end{matrix}\right.\)

a: \(=\dfrac{x}{y\left(x-y\right)}+\dfrac{2x-y}{y\left(x-y\right)}=\dfrac{x+2x-y}{y\left(x-y\right)}=\dfrac{3x-y}{y\left(x-y\right)}\)

b: \(=\dfrac{x\left(x+3\right)}{\left(x+3\right)^2}+\dfrac{3}{x-3}-\dfrac{6x}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{x}{x+3}+\dfrac{3}{x-3}-\dfrac{6x}{\left(x-3\right)\left(x+3\right)}\)

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

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

c: \(=\dfrac{x+9}{\left(x-3\right)\left(x+3\right)}-\dfrac{3}{x\left(x+3\right)}\)

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

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

d: \(=\dfrac{x^2-1-x^2+4}{x+1}=\dfrac{3}{x+1}\)

\(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\left(x\ne1\right)\)

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

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

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

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

\(\Leftrightarrow-4x^2+3x+1=0\left(\frac{1}{\left(x-1\right)\left(x^2+x+1\right)}\ne0\right)\)

\(\Leftrightarrow-4x^2+4x-x+1=0\)

\(\Leftrightarrow-4x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(-4x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\-4x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\-4x=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\left(loại\right)\\x=\frac{-1}{4}\end{cases}}}\)

Vậy \(x=\frac{-1}{4}\)

3,

\(9\left(x-3\right)^2=4\left(x+2\right)^2\)

\(\Leftrightarrow3^2\left(x-3\right)^2=2^2\left(x+2\right)^2\)

\(\Leftrightarrow\left(3x-9\right)^2=\left(2x+4\right)^2\)

\(\Leftrightarrow\left(3x-9\right)^2-\left(2x+4\right)^2=0\)

\(\Leftrightarrow\left(3x-9-2x-4\right)\left(3x-9+2x+4\right)=0\)

\(\Leftrightarrow\left(x-13\right)\left(5x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-13=0\\5x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-13\\x=1\end{matrix}\right.\)

1, \(\left(x+1\right)^2=4\left(x^2-2x+1\right)^2\)

\(\Leftrightarrow\left(x+1\right)^2-4\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x+1\right)^2-\left(2x-2\right)^2=0\)

\(\Leftrightarrow\left(x+1-2x+2\right)\left(x+1+2x-2\right)=0\)

\(\Leftrightarrow\left(3-x\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3-x=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)

vậy tập nghiệm của phương trinh \(S=\left\{3;\dfrac{1}{3}\right\}\)

2, \(\left(x^2-9\right)^2-9\left(x-3\right)^2=0\)

\(\Leftrightarrow\left(x^2-9\right)^2-\left(3x-9\right)^2=0\)

\(\Leftrightarrow\left(x^2-9-3x+9\right)\left(x^2-9+3x-9\right)=0\)

\(\Leftrightarrow\left(x^2-3x\right)\left(x^2+3x-18\right)=0\)

\(\Leftrightarrow x\left(x-3\right)\left(x^2+6x-3x-18\right)=0\)

\(\Leftrightarrow x\left(x-3\right)\left(x+6\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-6\end{matrix}\right.\)

vậy tập nghiệm của phương trinh \(S=\left\{0;3;-6\right\}\)

3, \(9\left(x-3\right)^2=4\left(x+2\right)^2\)

\(\Leftrightarrow\left(3x-9\right)^2-\left(2x+2\right)^2=0\)

\(\Leftrightarrow\left(3x-9-2x-2\right)\left(3x-9+2x+2\right)=0\)

\(\Leftrightarrow\left(x-11\right)\left(5x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-11=0\\5x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=\dfrac{7}{5}\end{matrix}\right.\)

vậy tập nghiệm của phương trinh \(S=\left\{11;\dfrac{7}{5}\right\}\)

Ta thấy \(\left(x-3\right)\left(2x+3\right)=2x^2-3x-9.\)

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

ĐK: \(x\ne3\)và \(x\ne-\frac{3}{2}\)

\(\Rightarrow x\left(2x+3\right)-2x^2-9=x-3\)

\(\Leftrightarrow2x^2+3x-2x^2-9=x-3\Leftrightarrow2x=6\Leftrightarrow x=2\)

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