Answer:
We have, \[\frac{x+1}{x-1}+\frac{x-2}{x+2}=4-\frac{2x+3}{x-2};x\ne 1,-2,2\] \[\frac{(x+1)(x+2)+(x-2)(x-1)}{(x-1)(x+2)}=\frac{4(x-2)-(2x+3)}{x-2}\] \[(x-2)[{{x}^{2}}+x+2x+2+{{x}^{2}}-2x-x+2]=[4x-8-2x-3]({{x}^{2}}+x-2)\] \[(x-2)(2{{x}^{2}}+4)=(2x-11)({{x}^{2}}+x-2)\] \[2{{x}^{3}}+4x-4{{x}^{2}}-8=2{{x}^{3}}+2{{x}^{2}}-4x-11{{x}^{2}}-11x+22\] \[4x-4{{x}^{2}}-8=-9{{x}^{2}}-15x+22\] \[5{{x}^{2}}+19x-30=0\] \[5{{x}^{2}}+25x-6x-30=0\] \[5x(x+5)-6(x+5)=0\] \[(5x-6)(x+5)=0\] \[x=-5,\frac{6}{5}\] \[\therefore x=-5\] or \[x=\frac{6}{5}\]
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