Question

Tìm x:

\(\dfrac{5}{x^2+x-6}-\dfrac{2}{x^2+4x+3}=\dfrac{-3}{2x+1}\)

Answer 1

\(\dfrac{5}{x^2+x-6}-\dfrac{2}{x^2+4x+3}=\dfrac{-3}{2x+1}\left(x\ne2;x\ne-3;x\ne-1;x\ne-\dfrac{1}{2}\right)\\ \Leftrightarrow\dfrac{5}{\left(x-2\right)\left(x+3\right)}-\dfrac{2}{\left(x+1\right)\left(x+3\right)}=\dfrac{-3}{2x+1}\\ \Leftrightarrow\dfrac{5\left(x+1\right)}{\left(x-2\right)\left(x+3\right)\left(x+1\right)}-\dfrac{2\left(x-2\right)}{\left(x-2\right)\left(x+3\right)\left(x+1\right)}=\dfrac{-3}{2x+1}\\ \Leftrightarrow\dfrac{5x+5-2x+4}{\left(x-2\right)\left(x+3\right)\left(x+1\right)}=\dfrac{-3}{2x+1}\\ \Leftrightarrow\dfrac{3x+9}{\left(x-2\right)\left(x+3\right)\left(x+1\right)}=\dfrac{-3}{2x+1}\\ \Leftrightarrow\dfrac{3\left(x+3\right)}{\left(x-2\right)\left(x+3\right)\left(x+1\right)}=\dfrac{-3}{2x+1}\\ \Leftrightarrow\dfrac{3}{\left(x-2\right)\left(x+1\right)}=\dfrac{-3}{2x+1}\\ \Leftrightarrow3\left(2x+1\right)=-3\left(x-2\right)\left(x+1\right)\\ \Leftrightarrow2x+1=-\left(x^2+x-2x-2\right)\\ \Leftrightarrow2x+1=-x^2+x+2\\ \Leftrightarrow x^2+2x-x+1-2=0\\ \Leftrightarrow x^2+x-1=0\)

\(\Delta=1^2-4\cdot1\cdot\left(-1\right)=5>0\)

\(x_1=\dfrac{-1+\sqrt{5}}{2\cdot1}=\dfrac{\sqrt{5}-1}{2}\\ x_2=\dfrac{-1-\sqrt{5}}{2\cdot1}=\dfrac{-\sqrt{5}-1}{2}\) (tm) 

Vậy: ... 

Answer 2

ĐK: \(x\notin\left\{-3;2;-1;-\dfrac{1}{2}\right\}\)

\(\dfrac{5}{x^2+x-6}-\dfrac{2}{x^2+4x+3}=\dfrac{-3}{2x+1}\)

\(\Leftrightarrow\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{2}{\left(x+1\right)\left(x+3\right)}=\dfrac{-3}{2x+1}\)

\(\Leftrightarrow\dfrac{5\left(x+1\right)-2\left(x-2\right)}{\left(x+3\right)\left(x-2\right)\left(x+1\right)}=\dfrac{-3}{2x+1}\)

\(\Leftrightarrow\dfrac{3x+9}{\left(x+3\right)\left(x-2\right)\left(x+1\right)}=\dfrac{-3}{2x+1}\)

\(\Leftrightarrow\dfrac{3}{\left(x-2\right)\left(x+1\right)}=\dfrac{-3}{2x+1}\)

\(\Rightarrow2x+1=-\left(x-2\right)\left(x+1\right)\)

\(\Leftrightarrow2x+1=-x^2+x+2\)

\(\Leftrightarrow x^2+x-1=0\)

\(\Delta=1^2-4\cdot\left(-1\right)=1+4=5>0\)

\(\rightarrow\) PT có 2 nghiệm pb 

\(x_1=\dfrac{-1+\sqrt{5}}{2}\) (T/m)

\(x_2=\dfrac{-1-\sqrt{5}}{2}\) (T/m)

Vậy \(S=\left\{\dfrac{-1+\sqrt{5}}{2};\dfrac{-1-\sqrt{5}}{2}\right\}\)