Решите срочно, номер 618!!! плиз очень надо

Question 1

Question 2

$1)\quad \sqrt{x+\sqrt{6x-9}}+\sqrt{x-\sqrt{6x-9}}=\sqrt6\; ,\\\\(x+\sqrt{6x-9)}+2\sqrt{x^2-(6x-9)}+(x-\sqrt{6x-9})=6\\\\2x+2\sqrt{x^2-6x+9}=6\; |:2\\\\x+\sqrt{(x-3)^2}=3\\\\|x-3|=3-x\; \; verno\; pri\; x-3\leq 0,x\leq3\\\\ODZ:\; a)\; 6x-9 \geq 0\; \; \to \; \; x \geq \frac{3}{2}\; ;\\\\b)\; x+\sqrt{6x-9} \geq 0\; \to \; \; \sqrt{6x-9} \geq -x\; \to \; \; \left \{ {{-x \leq 0} \atop {6x-9 \geq 0}} \right. \to$

$\left \{ {{x \geq 0} \atop {x \geq \frac{3}{2}}} \right. \; \to \; \; x \geq \frac{3}{2}\quad ili\\\\ \left \{ {{-x \geq 0} \atop {6x-9 \geq x^2}} \right.\; \left \{ {{x \leq 0} \atop {x^2-6x+9 \geq 0}} \right.\; \left \{ {{x \leq 0} \atop {(x-3)^2 \geq 0}} \right.\; \to \; x \leq 0\\\\c)\; x-\sqrt{6x-9} \geq 0\; \to \; \; \sqrt{6x-9} \leq x\; \; \to \; \left \{ {{x \geq \frac{3}{2}} \atop {6x-9 \leq x^2}} \right. \\\\ \left \{ {{x \geq \frac{3}{2}} \atop {(x-3)^2 \leq 0}} \right. \; \; net\; reshenij$

$x\geq \frac{3}{2}$

Ответ: х=3/2.
$2)\quad \sqrt{x+\sqrt{x+11}}+\sqrt{x-\sqrt{x+11}}=4\\\\x+\sqrt{x+11}+2\sqrt{x^2-(x+11)}+x-\sqrt{x+11}=16\\\\2x+2\sqrt{x^2-x-11}=16\\\\\sqrt{x^2-x-11}=8-x\\\\x^2-x-11=(8-x)^2\\\\x^2-x-11=x^2-16x+64\\\\15x=75\\\\x=5\\\\Proverka:\; \; \sqrt{5+\sqrt{16}}+\sqrt{5-\sqrt{16}}=\sqrt{5+4}+\sqrt{5-4}=\\\\=\sqrt9+\sqrt1=3+1=4\\\\Otvet:\; \; x=5.$

Question 3

спасибо огромное при огромное!