Common fraction for :$\frac{1}{x^2-2x+2}+\frac{2}{ x^2-2x+3}=\frac{6}{ x^2-2x+4}$

Question

Hi guys please help me with this equation: $$\frac{1}{x^2-2x+2}+\frac{2}{ x^2-2x+3}=\frac{6}{ x^2-2x+4}$$ My problem is with finding common fraction for denominators ($x^2-2x+2$ and $x^2-2x+3$and $x^2-2x+4$). I actually want to find a common fraction , then cross It for hole equation and to simplify it. Then solve for x.

Answer 1

HINT

We have

$$\frac{1}{(x-1)^2+1}+\frac{2}{ (x-1)^2+2}=\frac{6}{ (x-1)^2+3}$$

and by $y=(x-1)^2$

$$\frac{1}{y+1}+\frac{2}{ y+2}=\frac{6}{ y+3} \iff(y+2)(y+3)+2(y+1)(y+3)=6(y+1)(y+2)$$