\(ĐKXĐ:x\ne2\)
\(\frac{9x^2}{x^3-8}+\frac{6}{x^2+2x+4}=\frac{3}{x-2}\)
\(\Leftrightarrow\frac{9x^2}{x^3-8}+\frac{6\left(x-2\right)}{x^3-8}=\frac{3\left(x^2+2x+4\right)}{x^3-8}\)
\(\Rightarrow9x^2+6x-12=3x^2+6x+12\)
\(\Leftrightarrow9x^2-3x^2+6x-6x-12-12=0\)
\(\Leftrightarrow6x^2-24=0\)
\(\Leftrightarrow6\left(x^2-4\right)=0\)
\(\Leftrightarrow x^2-4=0\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow x=\pm2\) (loại x = 2)
vậy x = -2
\(\frac{9x^2}{x^3-8}+\frac{6}{x^2+2x+4}=\frac{3}{x-2}\)
=>\(\frac{9x^2}{x^3-8}+\frac{6\left(x-2\right)}{x^3-8}-\frac{3\left(x^2+2x+4\right)}{x^3+8}=0\)
=>\(9x^2+6x-12-3x^2-6x-24=0\)
=>\(6x^2-36\)\(6x^2-6\)
=>\(\left(6x-6\right)\left(6x+6\right)\)
=> \(6\left(x-1\right)6\left(x+1\right)\)
=>\(\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
#kenz
\(ĐKXĐ:x\ne2\)
\(pt\Leftrightarrow\frac{9x^2+6x-12}{x^3-8}=\frac{3\left(x^2+2x+4\right)}{x^3-8}\)
\(\Leftrightarrow9x^2+6x-12=3x^2+6x+12\)
\(\Leftrightarrow6x^2=24\Leftrightarrow x=\pm2\)
Vậy x = -2
cảm ơn bạn số1 vs số2
