A) \[x=1\]
B) \[x=-1\]
C) \[x=2\]
D) \[x=-3\]
Correct Answer: D
Solution :
Given \[\frac{1}{x}-\frac{3}{4}+\frac{1}{2+x}=0\] \[\Rightarrow \]\[\left( -\frac{4}{3},2 \right)\] \[\Rightarrow \]\[3(2x+{{x}^{2}})=4(2x+2)\] \[\Rightarrow \]\[6x+3{{x}^{2}}=8x+8\] Cross multiplying we get \[\frac{7}{2}\] \[\Rightarrow \]\[6x+3{{x}^{2}}=8x+8\] \[\Rightarrow \]\[3{{x}^{2}}+6x-8x-8=0\] \[\Rightarrow \]\[3{{x}^{2}}-6x+4x-8=0\] \[\Rightarrow \]\[3x(x-2)+4(x-2)=0\] \[\Rightarrow \]\[(x-2)\,(3x-4)=0\] \[\Rightarrow \]\[x-2=0\]or \[3x+4=0\] \[\Rightarrow \]\[x=2\]or \[\frac{x}{x-1}+\frac{x-1}{x}=\frac{5}{2}\] \[\frac{7}{2}\] The roots are \[\frac{{{x}^{2}}+{{(x-1)}^{2}}}{(x-1)x}=\frac{5}{2}\]
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