ĐKXĐ: \(x\ne\left\{0;\frac{-3\pm\sqrt{13}}{2}\right\}\)

Phương trình tương đương: \(\frac{x^2+\frac{1}{x^2}-1}{x-\frac{1}{x}+3}=\frac{1}{2}\)

Đặt \(x-\frac{1}{x}=a\Rightarrow x^2+\frac{1}{x^2}=a^2+2\)

Pt trở thành: \(\frac{a^2+1}{a+3}=\frac{1}{2}\)

\(\Leftrightarrow2a^2+2=a+3\)

\(\Leftrightarrow2a^2-a-1=0\)

\(\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x-\frac{1}{x}=1\\x-\frac{1}{x}=-\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2-x-1=0\\2x^2+x-2=0\end{matrix}\right.\) (casio)

a) \(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4}=\frac{x-4}{5}+\frac{x-5}{6}\)

\(\left(\frac{x-1}{2}+1\right)+\left(\frac{x-2}{3}+3\right)+\left(\frac{x-3}{4}+1\right)=\left(\frac{x-4}{5}+1\right)+\left(\frac{x-5}{6}+1\right)\)

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

\(\left(x-1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)\)=0

\(x-1=0\)

\(x=1\)

\(a,\frac{x}{x-3}-\frac{6}{x^2-9}=\frac{1}{x+3}\) (đkxđ: x khác 3, -3)

\(\frac{x\left(x+3\right)-6}{\left(x-3\right)\left(x+3\right)}=\frac{1}{x+3}\)

\(x\left(x+3\right)-6=x-3\)

\(x^2+2x-3=0\)

\(\left(x+3\right)\left(x-1\right)=0\)

\(\Longrightarrow\left[\begin{array}{l}x=-3\left(L\right)\\ x=1\left(N\right)\end{array}\right.\)

\(b,\frac{x^2}{x-2}+\frac{x}{1-x}=\frac{4}{x^2-3x+2}\) (đkxđ: \(x\ne1,x\ne2)\)

\(\frac{x^2}{x-2}-\frac{x}{x-1}=\frac{4}{\left(x-1\right)\left(x-2\right)}\)

\(\frac{x^2\left(x-1\right)-x\left(x-2\right)}{\left(x-1\right)\left(x-2\right)}=\frac{4}{\left(x-1\right)\left(x-2\right)}\)

\(x^2\left(x-1\right)-x\left(x-2\right)=4\)

\(x^3-x^2-x^2+2x=4\)

\(x^3-2x^2+2x-4=0\)

\(\left(x^3-2x^2\right)+\left(2x-4\right)=0\)

\(x^2\left(x-2\right)+2\left(x-2\right)=0\)

\(\left(x-2\right)\left(x^2+2\right)=0\)

vì \(x^2+2>0\forall x\) ⇒ x - 2 = 0

⇒ x = 2 (ko thoả mãn)

vậy phương trình vô nghiệm

\(x-\frac{\frac{x}{2}-\frac{3+x}{4}}{2}=3-\frac{\left(1-\frac{6-x}{3}\right).\frac{1}{2}}{2}\)

\(\Leftrightarrow2x-\frac{x}{2}+\frac{3+x}{4}=6-\frac{1}{2}+\frac{6-x}{6}\)

\(\Leftrightarrow24x-6x+9+3x=72-6+12-2x\)

\(\Leftrightarrow23x=69\)

\(\Leftrightarrow x=3\)

Vậy nghiệm của pt x=3

\(\text{GIẢI :}\)

ĐKXĐ : \(x\ne1,\text{ }x\ne-2\).

\(\frac{2}{x-1}+\frac{1}{x+2}=\frac{x^2-x}{x-1}+\left(\text{-}x\right)\)

\(\Leftrightarrow\frac{2}{x-1}+\frac{1}{x+2}=\frac{x\left(x-1\right)}{x-1}+\left(\text{-}x\right)\)

\(\Leftrightarrow\frac{2}{x-1}+\frac{1}{x+2}=x+\left(\text{-}x\right)\)

\(\Leftrightarrow\frac{2}{x-1}+\frac{1}{x+2}=0\)

\(\Leftrightarrow\frac{2\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}+\frac{x-1}{\left(x-1\right)\left(x+2\right)}=0\)

\(\Rightarrow2\left(x+2\right)+\left(x-1\right)=0\)

\(\Leftrightarrow2x+4+x-1\)

\(\Leftrightarrow3x+3=0\)

\(\Leftrightarrow3x=\text{-3}\Leftrightarrow x=\text{-1}\)

Vậy tập nghiệm của phương trình đã cho là \(S=\left\{-1\right\}\).

\(\frac{2}{x-1}+\frac{1}{x+2}=\frac{x^2-x}{x-1}+\left(-x\right)\left(đk:x\ne1;-2\right)\)

\(\frac{2\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}+\frac{\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}=\frac{x\left(x-1\right)}{x-1}-x\)

\(< =>\frac{2x+4+x-1}{\left(x-1\right)\left(x+2\right)}=x-x=0\)

\(< =>2x+4+x-1=0\)

\(< =>3x=1-4=-3\)

\(< =>x=\frac{-3}{3}=-1\left(tmđk\right)\)

Vậy nghiệm của phương trình trên là \(\left\{-1\right\}\)