bài nó dàiiiiiiii , khôg hiểu chỗ nèo hỏi lại mình hen
\(\dfrac{2x}{3x^2-x+2}-\dfrac{7x}{3x^2+5x+2}=1\)
\(\Leftrightarrow\left(\dfrac{2x}{3x^2-x+2}-\dfrac{7x}{\left(3x+2\right)\left(x+1\right)}\right)=1\)
\(\Leftrightarrow\dfrac{2x\left(3x+2\right)\left(x+1\right)-\left(7x.\left(3x^2-x+2\right)\right)}{\left(3x^2-x+2\right).\left(3x+2\right)\left(x+1\right)}=\dfrac{-15x^3+17x^2-10x}{\left(3x^2-x+2\right)\left(3x+2\right)\left(x+1\right)}\)
\(\Leftrightarrow\dfrac{-15x^3+17^2-10x }{\left(3x^2-x+2\right)\left(3x+2\right)\left(x+1\right)}-1=0\)
rồi quy đồng tùm lum từa lưa nữa được như này:
\(\Leftrightarrow\dfrac{-9x^4-27x^3+10x^2-18x-4}{\left(3x^2-x+2\right)\left(3x+2\right)\left(x+1\right)}=0\)
\(\Leftrightarrow-9x^4-27x^3+10x^2-18x-4=0\)
\(\Leftrightarrow x^2+\dfrac{5}{3}.x+\dfrac{25}{26}=0\)
\(\Leftrightarrow x+\left(\dfrac{5}{6}\right)^2=\dfrac{1}{36}\)
Sử dụng công thức bậc 2 hen:
\(\Leftrightarrow x=\dfrac{-5\pm\sqrt{1}}{6}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_1=\dfrac{-5+\sqrt{1}}{6}\\x_2=\dfrac{-5-\sqrt{1}}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=-\dfrac{2}{3}\\x_2=-1\end{matrix}\right.\)