If $$x+\frac{2}{x}=1$$ then the value of $$\frac{x^2+x+2}{x^2(1-x)}$$ is
Expression : $$x+\frac{2}{x}=1$$
=> $$x^2 + 2 = x$$
To find : $$\frac{x^2+x+2}{x^2(1-x)}$$
= $$\frac{x + x}{x^2(1-x)}$$
= $$\frac{2}{x -Â x^2}$$
= $$\frac{2}{2} = 1$$
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