![1)\; \; f(x)=\frac{\sqrt{8-2x-x^2}}{x+3}\\\\OOF:\ \; \left \{ {{8-2x-x^2\geq 0} \atop {x+3\ne 0}} \right.\; \left \{ {{x^2+2x-8\leq 0} \atop {x\ne -3}} \right.\; \left \{ {{(x+4)(x-2)\leq 0} \atop {x\ne -3}} \right. \; \left \{ {{x\in [-4,2\, ]} \atop {x\ne -3}} \right. \; \Rightarrow \\\\\underline {x\in [-4,-3)\cup (-3,2\, ]}\; \; -\; \; otvet\\\\x^2+2x-8=0\; \; \to \; \; x_1=-4\; ,\; x_2=2\; \; (teorema\; Vieta)\\\\(x+4)(x-2)\leq 0\; ,\; \; \; \; +++[-4\, ]---[\, 2\, ]+++\; \; ,\; \; x\in [-4,2\, ] 1)\; \; f(x)=\frac{\sqrt{8-2x-x^2}}{x+3}\\\\OOF:\ \; \left \{ {{8-2x-x^2\geq 0} \atop {x+3\ne 0}} \right.\; \left \{ {{x^2+2x-8\leq 0} \atop {x\ne -3}} \right.\; \left \{ {{(x+4)(x-2)\leq 0} \atop {x\ne -3}} \right. \; \left \{ {{x\in [-4,2\, ]} \atop {x\ne -3}} \right. \; \Rightarrow \\\\\underline {x\in [-4,-3)\cup (-3,2\, ]}\; \; -\; \; otvet\\\\x^2+2x-8=0\; \; \to \; \; x_1=-4\; ,\; x_2=2\; \; (teorema\; Vieta)\\\\(x+4)(x-2)\leq 0\; ,\; \; \; \; +++[-4\, ]---[\, 2\, ]+++\; \; ,\; \; x\in [-4,2\, ]](https://tex.z-dn.net/?f=1%29%5C%3B%20%5C%3B%20f%28x%29%3D%5Cfrac%7B%5Csqrt%7B8-2x-x%5E2%7D%7D%7Bx%2B3%7D%5C%5C%5C%5COOF%3A%5C%20%5C%3B%20%5Cleft%20%5C%7B%20%7B%7B8-2x-x%5E2%5Cgeq%200%7D%20%5Catop%20%7Bx%2B3%5Cne%200%7D%7D%20%5Cright.%5C%3B%20%5Cleft%20%5C%7B%20%7B%7Bx%5E2%2B2x-8%5Cleq%200%7D%20%5Catop%20%7Bx%5Cne%20-3%7D%7D%20%5Cright.%5C%3B%20%5Cleft%20%5C%7B%20%7B%7B%28x%2B4%29%28x-2%29%5Cleq%200%7D%20%5Catop%20%7Bx%5Cne%20-3%7D%7D%20%5Cright.%20%5C%3B%20%5Cleft%20%5C%7B%20%7B%7Bx%5Cin%20%5B-4%2C2%5C%2C%20%5D%7D%20%5Catop%20%7Bx%5Cne%20-3%7D%7D%20%5Cright.%20%5C%3B%20%5CRightarrow%20%5C%5C%5C%5C%5Cunderline%20%7Bx%5Cin%20%5B-4%2C-3%29%5Ccup%20%28-3%2C2%5C%2C%20%5D%7D%5C%3B%20%5C%3B%20-%5C%3B%20%5C%3B%20otvet%5C%5C%5C%5Cx%5E2%2B2x-8%3D0%5C%3B%20%5C%3B%20%5Cto%20%5C%3B%20%5C%3B%20x_1%3D-4%5C%3B%20%2C%5C%3B%20x_2%3D2%5C%3B%20%5C%3B%20%28teorema%5C%3B%20Vieta%29%5C%5C%5C%5C%28x%2B4%29%28x-2%29%5Cleq%200%5C%3B%20%2C%5C%3B%20%5C%3B%20%5C%3B%20%5C%3B%20%2B%2B%2B%5B-4%5C%2C%20%5D---%5B%5C%2C%202%5C%2C%20%5D%2B%2B%2B%5C%3B%20%5C%3B%20%2C%5C%3B%20%5C%3B%20x%5Cin%20%5B-4%2C2%5C%2C%20%5D)


0}|-|\sqrt{b}-1|=\\\\=\sqrt{b}+1-|\sqrt{b}-1|=\left \{ {{\sqrt{b}+1-(\sqrt{b}-1)=2\; ,\; esli\; b\geq 1\; ,} \atop {\sqrt{b}+1-(1-\sqrt{b})=2\sqrt{b}\; ,\; esli\; b<1\; .}} \right." alt="3)\; \; \sqrt{(\sqrt{b}-1)^2+4\sqrt{b}}-\sqrt{(\sqrt{b}+1)^2-4\sqrt{b}}=\\\\=\sqrt{b-2\sqrt{b}+1+4\sqrt{b}}}-\sqrt{b+2\sqrt{b}+1-4\sqrt{b}}=\\\\=\sqrt{b+2\sqrt{b}+1}-\sqrt{b-2\sqrt{b}+1}=\\\\=\sqrt{(\sqrt{b}+1)^2}-\sqrt{(\sqrt{b}-1)^2}=|\underbrace {\sqrt{b}+1}_{>0}|-|\sqrt{b}-1|=\\\\=\sqrt{b}+1-|\sqrt{b}-1|=\left \{ {{\sqrt{b}+1-(\sqrt{b}-1)=2\; ,\; esli\; b\geq 1\; ,} \atop {\sqrt{b}+1-(1-\sqrt{b})=2\sqrt{b}\; ,\; esli\; b<1\; .}} \right." align="absmiddle" class="latex-formula">