Question

$(2^{x^{2} } -32)*\sqrt{3-x} =0$
$\sqrt{7x+1} -\sqrt{6-x} =\sqrt{15+2x}$

решение

Answer 1

$1)(2^{x^{2}} -32)*\sqrt{3-x} =0\\\\ODZ:3-x\geq 0\Rightarrow x \leq 3\\\\\left \{ {{2^{x^{2}}-32=0 } \atop {\sqrt{3-x}=0 }} \right.\\\\\left \{ {{2^{x^{2}} =2^{5} } \atop {3-x=0}} \right.\\\\\left \{ {{x^{2}=5 } \atop {x=3}} \right.\\\\\boxed{x_{1}=-\sqrt{5};x_{2}=\sqrt{5};x_{3}=3}$

$2)\sqrt{7x+1}-\sqrt{6-x}=\sqrt{15+2x}\\\\ODZ :x\in[-\frac{1}{7};6]\\\\(\sqrt{7x+1}-\sqrt{6-x})^{2} =(\sqrt{15+2x})^{2} \\\\7x+1-2\sqrt{(7x+1)(6-x)}+6-x=15+2x\\\\2\sqrt{(7x+1)(6-x)}=4x-8\\\\(\sqrt{(7x+1)(6-x)})^{2}=(2x-4)^{2}\\\\(7x+1)(6-x)=4x^{2}-16x+16\\\\42x-7x^{2} +6-x=4x^{2}-16x+16\\\\11x^{2}-57x+10=0\\\\D=(-57)^{2}-4*11*10=3249-440=2809=53^{2}\\\\x_{1}=\frac{57-53}{22}=\frac{4}{22}=\frac{2}{11}\\\\x_{2}=\frac{57+53}{22}=\frac{110}{22}=5$

$Otvet:\boxed{\frac{2}{11};5}$