Please help in this question!! question is attached.

1.	Please help in this question!! question is attached.
Answer» ANSWER: $2 \leqslant x < 2 \sqrt{ \frac{7}{<klux>3</klux>} }$ Step-by-step explanation: $\sqrt{x + 3} > \sqrt{x - 1} + \sqrt{x - 2}$ Expression inside square root must be NON negative, so x+3 ≥ 0, x-1 ≥ 0 and x-2 ≥ 0 ⇒ x ≥ -3, x ≥ 1 and x ≥ 2 ⇒ x ≥ 2 Squaring both side of inequality, we get $x + 3 > (x - 1) + (x - 2) + 2 \sqrt{x - 1} \sqrt{x - 2}$ $6 - x > 2 \sqrt{x - 1} \sqrt{x - 2} \geqslant 0$ So 6-x ≥ 0 ⇒ x ≤ 6 Squaring the EQUATION again, we get $36 - 12x + {x}^{2} > 4(x - 1)(x - 2)$ $36 - 12x + {x}^{2} > 4( {x}^{2} - 3x + 2)$ $36 - 12x + {x}^{2} > 4 {x}^{2} - 12x + 8$ $3 {x}^{2} - (36 - 8) < 0$ ${x}^{2} - \frac{28}{3} < 0$ $(x + \sqrt{ \frac{28}{3} })( x - \sqrt{ \frac{28}{3} }) < 0$ $- 2 \sqrt{ \frac{7}{3} } < x < 2 \sqrt{ \frac{7}{3} }$ x must also satisfy the inequalities $2 \leqslant x \leqslant 6$ Hence $2 \leqslant x < 2 \sqrt{ \frac{7}{3} }$

Answer»

$2 \leqslant x < 2 \sqrt{ \frac{7}{<klux>3</klux>} }$