• # question_answer Solve for $x:\frac{1}{x+1}+\frac{2}{x+2}=\frac{4}{x+4},x\ne -1,-2,-4$

 We have, $\frac{1}{x+1}+\frac{2}{x+2}=\frac{4}{x+4},x\ne -1,-2,-4$ $(x+2)(x+4)+2(x+1)(x+4)=4(x+1)(x+2)$ ${{x}^{2}}+2x+4x+8+2({{x}^{2}}+x+4x+4)=4({{x}^{2}}+x+2x+2)$ ${{x}^{2}}+6x+8+2{{x}^{2}}+10x+8=4{{x}^{2}}+12x+8$ $3{{x}^{2}}+16x+16=4{{x}^{2}}+12x+8$ ${{x}^{2}}-4x-8=0$ $\therefore x=\frac{4\pm \sqrt{16+32}}{2}$ $x=\frac{4\pm \sqrt{48}}{2}=2\pm 4\sqrt{2}$ $\therefore x=2\pm 2\sqrt{3}$