Question

if x + 2/x = 1 then,


(x² + x +2)/x²(1-x) -1 = ??

Answer 1

Answer:

$\boxed{\sf{0}}$

Step-by-step explanation:

Given :

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

$=\:>\frac{x^2+2}{x}=1$

$=\:>x^2+2=x$ .............................( 1 )

To find :

$\frac{x^2+x+2}{x^2(1-x)}-1\\\\\implies\frac{x^2+2+x}{x^2(1-x)}-1\\\\\textsf{From (1) we have :}\\\\\implies\frac{x+x}{x^2(1-x)}-1$

$\implies\frac{2x}{x^2(1-x)}-1\\\\\implies \frac{2}{x(1-x)}-1\\\\\implies \frac{2}{(x-x^2)}-1\\\\\implies \frac{2}{2}-1$

$\implies 1-1$

$\implies 0$

Answer 2

Given :

x+\frac{2}{x}=1x+

x

2

=1

=\:>\frac{x^2+2}{x}=1=>

x

x

2

+2

=1

=\:>x^2+2=x=>x

2

+2=x .............................( 1 )

To find :

\begin{lgathered}\frac{x^2+x+2}{x^2(1-x)}-1\\\\\implies\frac{x^2+2+x}{x^2(1-x)}-1\\\\\textsf{From (1) we have :}\\\\\implies\frac{x+x}{x^2(1-x)}-1\end{lgathered}

x

2

(1−x)

x

2

+x+2

−1

⟹

x

2

(1−x)

x

2

+2+x

−1

From (1) we have :

⟹

x

2

(1−x)

x+x

−1

\begin{lgathered}\implies\frac{2x}{x^2(1-x)}-1\\\\\implies \frac{2}{x(1-x)}-1\\\\\implies \frac{2}{(x-x^2)}-1\\\\\implies \frac{2}{2}-1\end{lgathered}

⟹

x

2

(1−x)

2x

−1

⟹

x(1−x)

2

−1

⟹

(x−x

2

)

2

−1

⟹

2

2

−1

\implies 1-1⟹1−1

\implies 0⟹0