Question

solve for x...
please give solution in detail​

Answer 1

Step-by-step explanation:

We have,

$\frac{1}{x + 1} - \frac{2}{ {x}^{2} - x + 1 } < \frac{1 - 2x}{ {x}^{3} + 1 }$

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

$\implies \frac{ {x}^{2} - 3x - 1 }{(x + 1)( {x}^{2} - x + 1) } < \frac{1 - 2x}{ (x + 1)( {x}^{2} - x + 1) }$

$\implies \frac{ {x}^{2} - 3x - 1 }{(x + 1)( {x}^{2} - x + 1) } - \frac{1 - 2x}{ (x + 1)( {x}^{2} - x + 1) } < 0$

$\implies \frac{ {x}^{2} - 3x - 1 - 1 + 2x }{(x + 1)( {x}^{2} - x + 1) } < 0$

$\implies \frac{ {x}^{2} - x - 2 }{(x + 1)( {x}^{2} - x + 1) } < 0$

$\implies \frac{ {x}^{2} - 2x + x- 2 }{(x + 1)( {x}^{2} - x + 1) } < 0$

$\implies \frac{ x(x- 2) +1( x- 2 )}{(x + 1)( {x}^{2} - x + 1) } < 0$

$\implies \frac{( x + 1)(x- 2) }{(x + 1)( {x}^{2} - x + 1) } < 0$

$\implies \frac{(x- 2) }{( {x}^{2} - x + 1) } < 0$

Now,$x^2-x+1$ is always positive because its discriminant is negative and coefficient of x² is positive

So,

$x \in( - \infty ,2)$