Помогите пожалуйста решить под цифрами 5) 6) 7) и 8)

Question 1

Question 2

Решите задачу:

$5)\; \; \sqrt{x^2-9}=3x-11\; \; \Rightarrow \; \; \left \{ {{3x-11\geq 0} \atop {x^2-9=9x^2-66x+121}} \right. \; \; \left \{ {{x\geq 11/3} \atop {8x^2-66x+130=0}} \right. \\\\\left \{ {{ODZ:\; \; x\geq 3\frac{2}{3}} \atop {4x^2-33x+65=0}} \right. \\\\D/4=49\; ,x_1=\frac{33-7}8}=\frac{13}{4}=3,25\; \notin ODZ\;\; ;\; \; x_2=\frac{33+7}{8}=5\\\\Otvet:\; \; x=5.\\\\6)\; \; \sqrt{x^2+5x+1}=2x-1\; \; \Rightarrow \; \; \left \{ {{2x-1\geq 0} \atop {x^2+5x+1=4x^2-4x+1}} \right. \; \; \left \{ {{x\geq 0,5} \atop {3x^2-9x=0}} \right.$

$\left \{ {{x\geq 0,5} \atop {3x(x-3)=0}} \right. \; \; \left \{ {{x\geq 0,5} \atop {x_1=0\; ,\; x_2=3}} \right. \; \; \Rightarrow \; \; x=3\; \; (x_1=0\notin ODZ)\\\\Otvet:\; \; x=3\; .\\\\7)\; \; \sqrt{x^2-9}=x^2-21\; \; \Rightarrow \; \; \left \{ {{x^2-21\geq 0} \atop {x^2-9=x^4-42x^2+441}} \right. \; \; \left \{ {{(x-\sqrt{21})(x+\sqrt{21})\geq 0} \atop {x^4-43x^2+450=0}} \right. \\\\a)\; x^4-43x^2+450=0\; ,\; \; D=49\; ,\; \; x_1=\frac{43-7}{2}=18\; ,\; \; x_2=\frac{43+7}{2}=25\\\\b)\; \; (x-\sqrt{21})(x+\sqrt{21})\geq 0\; \; ,\; \; \; +++(-\sqrt{21})---(\sqrt{21})+++\\\\x\in (-\infty ,-\sqrt{21}\, ]\cup [\, \sqrt{21},+\infty )\\\\x_1,x_2\in (-\infty ,-\sqrt{21}\, ]\cup [\sqrt{21},+\infty )\\\\Otvet:\; \; x=18\; ,\; x=25\; .$

$8)\; \; \sqrt{x^2-1}=\sqrt3\; \; \Rightarrow \; \; \left \{ {{x^2-1\geq 0} \atop {x^2-1=3}} \right. \; \; \left \{ {{x\in (-\infty ,-1\, ]\cup [\, 1,+\infty )} \atop {x^2=4}} \right. \\\\\left \{ {{x\in (-\infty ,-1\, ]\cup [\, 1,+\infty )} \atop {x_1=-2\; ,\; x_2=2}} \right. \\\\x_1\; ,\; x_2\in ODZ\\\\Otvet:\; \; x_1=-2\; ,\; x_2=2\; .$

Question 3

спасибо огромное