đặt: \(x^2-3x+4=t\Rightarrow t\ge\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{1}{t-1}+\dfrac{1}{t}-\dfrac{6}{t+1}=0\Leftrightarrow\dfrac{t\left(t+1\right)}{t\left(t^2-1\right)}+\dfrac{\left(t^2-1\right)}{t\left(t^2-1\right)}-\dfrac{6t\left(t-1\right)}{t\left(t^2-1\right)}=0\)
\(\Leftrightarrow\dfrac{t^2+t+t^2-1-6t^2+6t}{t\left(t^2-1\right)}=\dfrac{-4t^2+7t-1}{t\left(t^2-1\right)}=\dfrac{4t^2-7t+1}{t\left(1-t^2\right)}=0\)
\(\Leftrightarrow4t^2-7t+1=0\Leftrightarrow t^2-2.\dfrac{7}{8}t+\left(\dfrac{7}{8}\right)^2=\dfrac{33}{64}\)
\(\left[\begin{matrix}t=\dfrac{7-\sqrt{33}}{8}\left(loai\right)\\t=\dfrac{7+\sqrt{33}}{8}< \dfrac{7}{4}\left(loai\right)\end{matrix}\right.\) Kết luận vô nghiệm
Đặt a = x2 - 3x + 3
pt đã cho trở thành: \(\frac{1}{a}+\frac{1}{a+1}=\frac{6}{a+2}\left(đkxđ:a;a+1;a+2\ne0\right)\)
\(\Leftrightarrow\frac{a+1+a}{a\left(a+1\right)}=\frac{6}{a+2}\)\(\Leftrightarrow\frac{2a+1}{a\left(a+1\right)}=\frac{6}{a+2}\)
\(\Leftrightarrow\left(2a+1\right)\left(a+2\right)=6a\left(a+1\right)\)
\(\Leftrightarrow2a^2+a+4a+2-6a^2-6a=0\)
\(\Leftrightarrow-4a^2-a+2=0\)
\(\Leftrightarrow4a^2+a-2=0\)
\(\Leftrightarrow\left(2a\right)^2-2.2a.\frac{1}{4}+\frac{1}{16}-\frac{33}{16}=0\)
\(\Leftrightarrow\left(2a-\frac{1}{4}\right)^2=\frac{33}{16}\)
\(\Rightarrow\left[\begin{matrix}2a-\frac{1}{4}=\sqrt{\frac{33}{16}}\\2a-\frac{1}{4}=-\sqrt{\frac{33}{16}}\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}a=\frac{1+\sqrt{33}}{8}\\a=\frac{1-\sqrt{33}}{8}\end{matrix}\right.\)
Đến đây thay lần lượt các giá trị của a vào a = x2 - 3x + 3 là ra