Question

1/(x-1)-1 ≥ 2-1/(x-2). then solve x​

Answer 1

Step-by-step explanation:

GIVEN:-

$\frac{1}{x - 1} - 1 \geqslant 2 - \frac{1}{ x - 2}$

MAKE SURE THAT IT MUST BE GREATER THAN OR EQUAL TO 0

$\frac{1}{x - 1} - 1 - 2 + \frac{1}{ x - 2} \geqslant 0$

SOLVING IT WE GET AS FOLLOWS

$\frac{1}{x - 1} + \frac{1}{x - 2} - 3 \geqslant 0$

$\frac{x - 2}{(x - 1)(x - 2)} + \frac{x - 1}{(x - 2)(x - 1)} - \frac{3(x - 2)(x - 1)}{(x - 2)(x - 1)} \geqslant 0$

$\frac{x - 2 + x - 1 - 3(x - 1)(x - 2)}{(x - 1)( x - 2)} \geqslant 0$

$\frac{2x - 3 - 3(x ^{2} - 3x + 2) }{ {x}^{2} - 3x + 2 } \geqslant 0$

$\frac{2x - 3 - 3 {x}^{2} + 9x - 2 }{ {x}^{2} - 3x + 2} \geqslant 0$

Taking minus common from numerator the sign reverses and denominator comes to numerator

$(3 {x}^{2} - 11x +9)(x - 1)(x - 2) \leqslant 0$

so x belongs to [1,2]