• # question_answer Solve for $x:\frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}=\frac{2}{3},x\ne 1,2,3$

 We have, $\frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}=\frac{2}{3},x\ne 1,2,3$ $3(x-3)+3(x-1)=2(x-1)(x-2)(x-3)$ $3x-9+3x-3=2(x-1)(x-2)(x-3)$ $6x-12=2(x-1)(x-2)()x-3$ $6(x-2)=2(x-1)(x-2)(x-3)$ $3=(x-1)(x-3)$ $3={{x}^{2}}-3x-x+3$ ${{x}^{2}}-4x=0$ $x(x-4)=0$ $\therefore$                               $x=0$ or $4$

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