Question

Giải phương trình:

b)4x/x²+4x+3 - 1=6(1/x+3 - 1/2x+2)

c)x²-x-12=0

e)6x+22/x+2 - 2x+7/x+3=x+4/x^2+5x+6

Answer 1

b)

\(\frac{4x}{x^2+4x+3}-1=6\cdot\left(\frac{1}{x+3}-\frac{1}{2x+2}\right)\\ \Leftrightarrow\frac{4x}{\left(x+1\right)\cdot\left(x+3\right)}-1=6\cdot\left(\frac{1}{x+3}-\frac{1}{2\cdot\left(x+1\right)}\right)\\ \Leftrightarrow4x-\left(x+1\right)\cdot\left(x+3\right)=6\cdot\left(\frac{1}{x+3}-\frac{1}{2\cdot\left(x+2\right)}\right)\cdot\left(x+1\right)\cdot\left(x+3\right)\\ \Leftrightarrow-x^2-3=\frac{6x^2}{x+3}+\frac{24x}{x+3}+\frac{18}{x+3}-\frac{3x^2}{x+1}-\frac{12x}{x+1}-\frac{9}{x++1}\\ \Leftrightarrow-x^2\cdot\left(x+3\right)\cdot\left(x+1\right)-3\cdot\left(x+3\right)\cdot\left(x+1\right)=6x^2\cdot\left(x+1\right)+24x\cdot\left(x+1\right)+18\cdot\left(x+1\right)-3x^2\cdot\left(x+3\right)-12x\cdot\left(x+3\right)-9\cdot\left(x+3\right)\\ \Leftrightarrow-x^4-4x^3-6x^2-12x-9=3x^3+9x^2-3x-9\\ \Leftrightarrow-x^4-4x^3-6x^2-12x=3x^3+9x^2-3x\\ \Leftrightarrow x^4+4x^3+6x^2+12x+3x^3+9x^2-3x=0\\ \Leftrightarrow x^4+7x^3+15x^2+9x=0\\ \Leftrightarrow x\cdot\left(x^3+7x^2+15x+9\right)=0\\ \Leftrightarrow x\cdot\left(x^2+6x+9\right)\cdot\left(x+1\right)=0\\ \Leftrightarrow x\cdot\left(x+3\right)^2\cdot\left(x+1\right)=0\)

\(\Rightarrow x=\left[{}\begin{matrix}0\\-3\\-1\end{matrix}\right.\)

Answer 2

c)

\(x^2-x-12=0\\ \Leftrightarrow x^2+3x-4x-12=0\\ \Leftrightarrow x\cdot\left(x+3\right)-4\cdot\left(x+3\right)=0\\ \Leftrightarrow\left(x-4\right)\cdot\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)

e)

\(\frac{6x+22}{x+2}-\frac{2x+7}{x+3}=\frac{x+4}{x^2+5x+6}\\ \Leftrightarrow\frac{6x^2+40x+66}{x^2+5x+6}-\frac{2x^2+11x+14}{x^2+5x+6}-\frac{x+4}{x^2+5x+6}=0\\ \Leftrightarrow6x^2+40x+66-2x^2-11x-14-x-4=0\\ \Leftrightarrow4x^2+28x+48=0\\ \Leftrightarrow4\cdot\left(x^2+7x+12\right)=0\\ \Leftrightarrow4\cdot\left(x^4+4x+3x+12\right)=0\\ \Leftrightarrow4\cdot\left[x\cdot\left(x+4\right)+3\cdot\left(x+4\right)\right]=0\\ \Leftrightarrow4\cdot\left(x+4\right)\cdot\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+4=0\\x+3=0\end{matrix}\right.\Rightarrow}\left\{{}\begin{matrix}x=-4\\x=-3\end{matrix}\right.\)