Question

Simplify: $\frac{1}{x^2-3x+2}+\frac{1}{x^2-5x+6}+\frac{1}{x^2-4x+3}$

• A. $\frac{3x+6}{(x-1)(x-2)(x-3)}$
• B. $\frac{3x-6}{(x-1{)}^2(x-2{)}^2(x-3{)}^2}$
• C. $\frac{3x+6}{(x-1{)}^2(x-2{)}^2(x-3{)}^2}$
• D. $\frac{3x-6}{(x-1)(x-2)(x-3)}$

Answer 1

The correct option is D $\frac{3x-6}{(x-1)(x-2)(x-3)}$

Given $\frac{1}{x^2-3x+2}+\frac{1}{x^2-5x+6}+\frac{1}{x^2-4x+3}$

$x^2-3x+2=x^2-2x-x+2=x(x-2)-1(x-2)=(x-1)(x-2)$

$x^2-5x+6=x^2-2x-3x+6=x(x-2)-3(x-2)=(x-2)(x-3)$

$x^2-4x+3=x^2-3x-x+3=x(x-3)-1(x-3)=(x-1)(x-2)$

Now, $\frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}+\frac{1}{(x-3)(x-1)}$

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

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