Bài làm
\(\frac{x+4}{x+3}+\frac{\left(12x+7\right)}{\left(x+3\right)\left(x+4\right)}=\frac{1-2x}{4-x}\)
\(\Leftrightarrow\frac{\left(x+4\right)^2\left(4-x\right)}{\left(x+3\right)\left(x+4\right)\left(4-x\right)}+\frac{\left(12x+7\right)\left(4-x\right)}{\left(x+3\right)\left(x+4\right)\left(4-x\right)}=\frac{\left(1-2x\right)\left(x+3\right)\left(x+4\right)}{\left(x+3\right)\left(x+4\right)\left(4-x\right)}\)
\(\Rightarrow\left(x^2+8x+16\right)\left(4-x\right)+\left(12x+7\right)\left(4-x\right)=\left(x-2x^2+3-6x\right)\left(x+4\right)\)
\(\Leftrightarrow4x^2+24x+64-x^3-8x^2-16x+48x-12x^2+28-7x=x^2-2x^3+3x-6x^2+4x-8x^2+12-24x\)
\(\Leftrightarrow\left(-x^3+2x^3\right)+\left(4x^2-8x^2-12x^2-x^2+6x^2+8x^2\right)+\left(24x-16x+48x-7x-3x-4x+24x\right)+\left(64+28-12\right)=0\)
\(\Leftrightarrow x^3-2x^2+66x+80=0\)
~ Đến đây giải tiếp ~