Question

ПОМОГТИЕ ПЖ СРОЧНО!!1 ОБЪЯСНИТЕ ХОТЯ БЫ 1 ПРИМЕР!!! ИЛИ 12

Answer 1

$1)\sqrt{x}=7-x\\\\\left \{ {{x=(7-x)^{2} } \atop {7-x\geq0 }} \right.\\\\\left \{ {{x=49-14x+x^{2} } \atop {-x\geq-7 }} \right.\\\\\left \{ {{x^{2}-15x+49=0 } \atop {x\leq7 }} \right.\\\\x^{2}-15x+49=0\\\\D=(-15)^{2}-4*49=225-196=29\\\\x_{1} =\frac{15+\sqrt{29}}{2}>7-neyd\\\\x_{2}=\frac{15-\sqrt{29} }{2}\\\\Otvet:\boxed{\frac{15-\sqrt{29} }{2}}$

$12)\sqrt{2x-4} -\sqrt{x+5}=1\\\\ODZ:\\\left \{ {{2x-4\geq0 } \atop {x+5\geq0 }} \right. \\\\\left \{ {{x\geq2 } \atop {x\geq-5 }} \right. \Rightarrow x\geq 2\\\\(\sqrt{2x-4} -\sqrt{x+5})^{2} =1^{2}\\\\2x-4-2\sqrt{(2x-4)(x+5)}+x+5=1\\\\2\sqrt{(2x-4)(x+5)}=3x\\\\(2\sqrt{(2x-4)(x+5)})^{2} =(3x)^{2} \\\\4(2x-4)(x+5)=9x^{2}\\\\8x^{2}+40x-16x-80-9x^{2}=0\\\\-x^{2}+24x-80=0\\\\x^{2}-24x+80=0\\\\D=(-24)^{2}-4*80=576-320=256=16^{2}\\\\x_{1}=\frac{24-16}{2}=4$

$x_{2} = \frac{24+16}{2} =20\\\\Proverka:\\1)\sqrt{2*4-4}-\sqrt{4+5}=1\\\\\sqrt{4}-\sqrt{9}\neq 1\\\\2)\sqrt{2*20-4}-\sqrt{20+5} =1\\\\\sqrt{36}-\sqrt{25}=1\\\\6-5=1\\\\1=1-verno\\\\Otvet:\boxed{20}$