question_answer

Solve for \[x:\frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}=\frac{2}{3},x\ne 1,2,3\]

Answer:

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 \[4\]