A) \[x-\frac{1}{x}\]
B) \[x+\frac{1}{x}-1\]
C) \[x+\frac{1}{x}\]
D) None of these
Correct Answer: B
Solution :
\[4{{x}^{2}}-20x+25=0\] \[\frac{3}{2}\] \[=(2{{a}^{2}}+2ab)+(2{{b}^{2}}+2ab)+(a+b)-(10a+10b+5)\] \[\frac{7}{2}\] \[\frac{7}{2}\] \[\frac{5}{2}\] Another way Let \[a+b=x\] Then, \[2{{(a+b)}^{2}}-9(a+b)-5=2{{x}^{2}}-9x-5\] \[\Rightarrow \] \[2{{x}^{2}}-10x+x-5\] \[\Rightarrow \]\[2x\,(x-5)+1\,(x-5)\] \[a+b+c=9\] Put the value of \[ab+bc+ca=26,\] \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc\] \[x:\frac{x-a}{b+c}+\frac{x-b}{c+a}+\frac{x-c}{a+b}=3\]
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