Question 154

If $$x+\frac{1}{x}=1$$ then the value of $$\frac{x^2+3x+1}{x^2+7x+1}$$

Solution

Expression : $$x+\frac{1}{x}=1$$

=> $$x^2 + 1 = x$$ ------Eqn(1)

To find : $$\frac{x^2+3x+1}{x^2+7x+1}$$

= $$\frac{(x^2+1) + 3x}{(x^2+1) + 7x}$$

Using eqn(1),we get :

= $$\frac{x + 3x}{x + 7x} = \frac{4}{8}$$

= $$\frac{1}{2}$$

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