2. [tex]VT=(x+\frac{x}{x-1})^3-3x.\frac{x}{x-1}(x+\frac{x}{x-1})+\frac{3x^2}{x-1}-2=(\frac{x^2}{x-1})^2-3(\frac{x^2}{x-1})^2+3.\frac{x^2}{x-1}-2=0[/tex]
Bây giờ đặt ẩn phụ là dễ rồi.
1. [tex](\frac{1}{x-1}-\frac{1}{2x}+\frac{1}{x+1})^2=\frac{1}{(x-1)^2}+\frac{1}{4x^2}+\frac{1}{(x+1)^2}-\frac{1}{x(x-1)}-\frac{1}{x(x+1)}+\frac{2}{(x-1)(x+1)}=\frac{1}{(x-1)^2}+\frac{1}{4x^2}+\frac{1}{(x+1)^2}-\frac{1}{x}.\frac{2x}{(x-1)(x+1)}+\frac{2x}{(x-1)(x+1)}=\frac{1}{(x-1)^2}+\frac{1}{4x^2}+\frac{1}{(x+1)^2}\Rightarrow (\frac{1}{x-1}-\frac{1}{2x}+\frac{1}{x+1})^2=(\frac{3x^2+1}{12})^2\Rightarrow (\frac{3x^2+1}{2x(x^2-1)})^2=(\frac{3x^2+1}{12})^2\Rightarrow 2x(x^2-1)=\pm 12[/tex]
3. Đặt [tex]a=(x-1)^3,b=x^3,c=(x+1)^3\Rightarrow a+b+c=3x^2(x+2)\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\Rightarrow (a+b+c)(ab+bc+ca)=abc\Rightarrow (a+b)(b+c)(c+a)=0[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
