Answer:
We have, \[\frac{a}{x-a}+\frac{b}{x-b}=\frac{2c}{x-c},x\ne a,b,c\] \[a(x-b)(x-c)+b(x-a)(x-c)=2c(x-a)(x-b)\] \[a({{x}^{2}}-bx-cx+bc)+b({{x}^{2}}-ax-cx+ac)=2c({{x}^{2}}-ax-bx+ab)\] \[a{{x}^{2}}-abx-acx+abc+b{{x}^{2}}-abx-bcx+abc=2c{{x}^{2}}-2acx-2bcx+2abc\] \[a{{x}^{2}}+b{{x}^{2}}-2abx-acx-bcx+2abc=2c{{x}^{2}}-2acx-2bcx+2abc\] \[a{{x}^{2}}+b{{x}^{2}}-2c{{x}^{2}}-2abx-acx-bcx+2acx+2bcx=0\] \[(a+b-2c){{x}^{2}}+(-2ab+ac+bc)x=0\] \[x[(a+b-2c)x+(ac+bc-2ab)]=0\] \[x=0,-\frac{(ac+bc-2ab)}{a+b-2c}\]
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