a.
\(\Leftrightarrow8x+2x^2-12-3x=x^2-12\)
\(\Leftrightarrow x^2-5x=0\)
\(\Leftrightarrow x\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
b.
\(\Leftrightarrow x\left(3x^2+5x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3x^2+5x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=\dfrac{1}{3}\end{matrix}\right.\)
c.ĐK:\(x\ne-3;x\ne2\)
\(\Leftrightarrow\dfrac{2x+1}{x+3}-\dfrac{x-3}{2-x}=\dfrac{-3x+1}{\left(x-2\right)\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{\left(2x+1\right)\left(x-2\right)+\left(x-3\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\dfrac{-3x+1}{\left(x+3\right)\left(x-2\right)}\)
\(\Rightarrow2x^2-4x+x-2+x^2-9=-3x+1\)
\(\Leftrightarrow3x^2-12=0\)
\(\Leftrightarrow3\left(x^2-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(l\right)\\x=-2\left(n\right)\end{matrix}\right.\)
\(a,\\ \Leftrightarrow8x+2x^2-12-3x-x^2+12=0\\ \Leftrightarrow x^2+5x=0\\ \Rightarrow x\left(x+5\right)=0\\ \Rightarrow\left\{{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\\ b,\\ \Leftrightarrow x\left(3x^2+5x-2\right)=0\\ \Leftrightarrow x\left[\left(3x^2-x\right)+\left(6x-2\right)\right]=0\\ \Leftrightarrow x\left[x\left(3x-1\right)+2\left(3x-1\right)\right]=0\)
\(\Leftrightarrow x\left(3x-1\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-2\end{matrix}\right.\\ c,\\ Đk:x\ne2;x\ne-3\\ \Leftrightarrow\dfrac{\left(2x+1\right)\left(x-2\right)+\left(x-3\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\dfrac{-3x+1}{\left(x+3\right)\left(x-2\right)}\)
\(\Leftrightarrow2x^2-4x+x-2+x^2-9=-3x+1\\ \Leftrightarrow3x^2-3x-11+3x-1=0\\ \Leftrightarrow3x^2=12\\ \Leftrightarrow x^2=4\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(loại\right)\\x=-2\left(tm\right)\end{matrix}\right.\)
\(a,\left(2x-3\right)\left(4+x\right)=x^2-12\\ \Leftrightarrow8x-12+2x^2-3x-x^2+12=0\\ \Leftrightarrow x^2+5x=0\\ \Leftrightarrow x\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
\(b,3x^3+5x^2-2x=0\\ \Leftrightarrow x\left(3x^2+5x-2\right)=0\\ \Leftrightarrow x\left[\left(3x^2+6x\right)-\left(x+2\right)\right]=0\\ \Leftrightarrow x\left[3x\left(x+2\right)-\left(x+2\right)\right]=0\\ \Leftrightarrow x\left(3x-1\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-2\end{matrix}\right.\)
\(c,ĐKXĐ:\left\{{}\begin{matrix}x+3\ne0\\2-x\ne0\\x^2+x-6\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-3\\x\ne2\\\left(x^2+3x\right)-\left(2x+6\right)\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-3\\x\ne2\\\left(x-2\right)\left(x+3\right)\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne2\\x\ne-3\end{matrix}\right.\)
\(\dfrac{2x+1}{x+3}-\dfrac{x-3}{2-x}=\dfrac{-3x+1}{x^2+x-6}\\ \Leftrightarrow\dfrac{\left(2x+1\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}+\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\dfrac{-3x+1}{\left(x^2+3x\right)-\left(2x+6\right)}\)
\(\Leftrightarrow\dfrac{2x^2+x-4x-2}{\left(x+3\right)\left(x-2\right)}+\dfrac{x^2-9}{\left(x+3\right)\left(x-2\right)}=\dfrac{-3x+1}{x\left(x+3\right)-2\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{2x^2+x-4x-2+x^2-9}{\left(x+3\right)\left(x-2\right)}=\dfrac{-3x+1}{\left(x+3\right)\left(x-2\right)}\)
\(\Leftrightarrow3x^2-3x-11=-3x+1\)
\(\Leftrightarrow3x^2-12=0\\ \Leftrightarrow x^2-4=0\\ \Leftrightarrow x^2=4\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(ktm\right)\\x=-2\left(tm\right)\end{matrix}\right.\)
