Помогите, пожалуйста, срочно решить два примера 1. Sqrt(2x+3)>=x-6...

$\sqrt{2x+3} \geq x-6$
1)система: 2) система:
х-6<0 или х-6≥0 <br>2x+3≥0 2х+3≥х²-12х+36

$1) \left \{ {{x\ \textless \ 6} \atop {x \geq -1.5}} \right. \ \textless \ =\ \textgreater \ -1.5 \leq x\ \textless \ 6 \\ 2) \left \{ {{x \geq 6} \atop { x^{2} -14x+33 \leq 0}} \right. \left \{ {{x \geq 6} \atop {3 \leq x \leq 11}} \right. \ \textless \ =\ \textgreater \ 6 \leq x \leq 11 \\ -1.5 \leq x \leq 11$
отв: х∈[-1.5;11]

$\sqrt{x-5} -1= \sqrt{10-x} \\ODZ: \left \{ {{x-5 \geq 0} \atop {10-x \geq 0}} \right. \left \{ {{x \geq 5} \atop {x \leq 10}}\ \textless \ =\ \textgreater \ 5 \leq x \leq 10 \right. \\ (\sqrt{x-5} -1) ^{2} = \sqrt{10-x} ^{2} \\ x-5-2 \sqrt{x-5} +1=10-x \\ 2 \sqrt{x-5} =2x-14 \\ \sqrt{x-5}=x-7 \\ \left \{ {{x-7 \geq 0} \atop {x-5= x^{2} -14x+49}} \right. \\ \left \{ {{x\geq 7} \atop { x^{2} -15x+54=0}} \right. \\ \left \{ {{x \geq 7} \atop {x=6;x=9}} \right. \\ x=9$
отв:9