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 \[4\]
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