\(\frac{4x}{x^2+4x+3}-1=6\left(\frac{1}{x+3}-\frac{1}{2x+2}\right)\)đkxđ \(x\ne-1;-3\)
\(\Leftrightarrow\frac{4x}{x^2+4x+3}-\frac{6}{x+3}+\frac{3}{x+1}=1\)
\(\Leftrightarrow\frac{4x-6x-6+3x+9}{\left(x+1\right)\left(x+3\right)}=1\)
\(\Leftrightarrow\frac{x+3}{\left(x+3\right)\left(x+1\right)}=1\)
\(\Leftrightarrow\frac{1}{x+1}=1\)
\(\Leftrightarrow x+1=1\)
\(\Leftrightarrow x=0\left(tm\right)\)
\(\frac{4x}{x^2+4x+3}-1=6\left(\frac{1}{x+3}-\frac{1}{2x+2}\right)\)
\(\frac{4x}{\left(x+1\right)\left(x+3\right)}-1=6\left(\frac{1}{x+3}-\frac{1}{2x+2}\right)\)
\(\frac{4x}{\left(x+1\right)\left(x+3\right)}-1=6\left(\frac{1}{x+3}-\frac{1}{2x-2}\right)\)
\(4x-\left(x+1\right)\left(x+3\right)=6\left(\frac{1}{x+3}-\frac{1}{2\left(x+1\right)}\right)\left(x+1\right)\left(x+2\right)\)
\(-x^2-3=\frac{6x^2}{x+3}+\frac{24x}{x+3}+\frac{18}{x+3}-\frac{3x^2}{x+1}-\frac{12x}{x+1}-\frac{9}{x+1}\)
\(-x^4-4x^3-6x^2-12x=3x^3+9x^2-3x\)
\(-x^4-4x^3-6x^2-12x=3x^3+9x^2-3x\)
\(-x^4-4x^3-6x^2-12x-3x^3-9x^2+3x=0\)
\(x^4+7x^3+15x^2+9x=0\)
\(x\left(x^3+6x+9\right)\left(x+1\right)=0\)
\(x\left(x+3\right)^2\left(x+1\right)=0\)
\(x=0;-3;-1\)
