Question

Que=> Solve inequalitie $\sqrt{x} - 3 \leqslant \frac{2}{ \sqrt{x} - 2}$​

Answer 1

Answer:

x € (-∞,1] U (4,16]

Answer 2

SOLUTION:

$\sf \: ( \sqrt{x} -3) - \frac{2}{ \sqrt{x} - 2 } \leqslant 0 \\ \\ \sf \frac{ ( \sqrt{x} - 3)( \sqrt{x} - 2) - 2}{ \sqrt{x} - 2} \: \leqslant 0 \\ \\ \sf \frac{x - 5 \sqrt{x} + 4 }{ \sqrt{x} - 2 } \leqslant 0 \\ \\ \sf \sqrt{x} ( \sqrt{x} - 1) - 4( \sqrt{x} - 1) \leqslant 0 \\ \\ \sf \frac{( \sqrt{x} - 1)( \sqrt{x} - 4)}{ \sqrt{x} - 2} \: = 0 \\ \\ \sf \: if \: \sqrt{x} - 1 = 0 \\ \sf \: \sqrt{x} = 1 \\ \sf \: x = {1}^{2} = 1 \ \ \\ \\ \sf \: if \: \sqrt{x} - 4 = 0 \\ \sf \: x = 16 \\ \\ \\ \sf \: if \: \sqrt{x} - 2 = 0 \\ \sf \: \sqrt{x} = 2 \\ \sf \: x = {2}^{2} = 4$