Срочно заранее спасибо

Решите задачу:

$1)\; \; 16- \sqrt{\frac{2}{3}x} =12\; ,\; \; \; ODZ:\; \; x \geq 0\\\\\sqrt{\frac{2}{3}x}=4\; ,\; \; \frac{2}{3}x=16\; ,\; \; \underline {x=24}\\\\2)\; \; 1+x=\sqrt{1+x\cdot \sqrt{x^2+12}}\; ,\; \; \; ODZ:\; \; 1+x \geq 0\; ,\; \; x \geq -1\\\\1+2x+x^2=1+x\cdot \sqrt{x^2+12}\\\\x\cdot \sqrt{x^2+12}=2x+x^2\\\\x^2(x^2+12)=4x^2+4x^3+x^4\\\\4x^3-8x^2=0\\\\4x^2(x-2)=0\\\\\underline {x_1=0\; ,\; \; x_2=2}$

$3)\; \; \sqrt{12+x}=\sqrt{7x+8}-2\; ,\; \; \; ODZ:\; \left \{ {{12+x \geq 0} \atop {7x+8 \geq 0}} \right. \; \left \{ {{x \geq -12} \atop {x \geq -\frac{8}{7}}} \right. \; \to \; x \geq -\frac{8}{7}\\\\\sqrt{12+x} +2=\sqrt{7x+8} \\\\12+x+4\sqrt{12+x}+4=7x+8\\\\4\sqrt{12+x}=6x-8\; |:2\\\\2\sqrt{12+x}=3x-4\\\\4\cdot (12+x)=9x^2-24x+16\\\\9x^2-28x-32=0\\\\D/4=196+9\cdot 32=484\\\\\underline {x_1\ \frac{14-22}{9} =-\frac{8}{9} \; ,\; \; x_2=\frac{14+22}{9} =4}$

$4)\; \; \sqrt{9-x} \leq 3\; ,$

$\left \{ {{9-x \geq 0} \atop {9-x \leq 3^2}} \right. \; \; \left \{ {{-x \geq -9} \atop {-x \leq 0}} \right. \; \; \left \{ {{x \leq 9} \atop {x \geq 0}} \right. \; \; \; \to \; \; \; 0 \leq x \leq 9\\\\\underline {x\in [\, 0,9\, ]}$