Answer:
We have, \[\frac{2x}{x-3}+\frac{1}{2x+3}+\frac{3x+9}{(x-3)(2x+3)}=0,x\ne 3,-3/2\] \[2x(2x+3)+(x-3)+(3x+9)=0\] \[4{{x}^{2}}+6x+x-3+3x+9=0\] \[4{{x}^{2}}+10x+6=0\] \[2{{x}^{2}}+5x+3=0\] \[2{{x}^{2}}+2x+3x+3=0\] \[2x(x+1)+3(x+1)=0\] \[(2x+3)(x+1)=0\] \[x=-1,\frac{-3}{2}\] \[\therefore x=-1\] [\[\because \] Given \[x\ne -3/2\]]
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