![1)\; \; (x-1)\cdot \sqrt{x^2-x-2}\geq 0\; ,\; \; ODZ:\; x^2-x-2\geq 0\; ,\\\\x_1=-1\; ,\; x_2=2\; \; (teor.\; Vieta)\; \; \Rightarrow \; \; (x+1)(x-2)\geq 0\; \; \Rightarrow \\\\x\in (-\infty ,-1\, ]\cup [\, 2,+\infty )\\\\T.k.\; \sqrt{x^2-x-2}\geq 0\; \; pri\; \; x\in ODZ\; ,to\; (x-1)\cdot \sqrt{x^2-x-2}\geq 0\; pri\\\\x-1\geq 0\; \; \Rightarrow \; \; x\geq 1\\\\\left \{ {{x\geq 1} \atop {x\in (-\infty ,-1\, ]\cup [\, 2,+\infty )}} \right. \; \; \Rightarrow \; \; x\in [\,2,+\infty ) 1)\; \; (x-1)\cdot \sqrt{x^2-x-2}\geq 0\; ,\; \; ODZ:\; x^2-x-2\geq 0\; ,\\\\x_1=-1\; ,\; x_2=2\; \; (teor.\; Vieta)\; \; \Rightarrow \; \; (x+1)(x-2)\geq 0\; \; \Rightarrow \\\\x\in (-\infty ,-1\, ]\cup [\, 2,+\infty )\\\\T.k.\; \sqrt{x^2-x-2}\geq 0\; \; pri\; \; x\in ODZ\; ,to\; (x-1)\cdot \sqrt{x^2-x-2}\geq 0\; pri\\\\x-1\geq 0\; \; \Rightarrow \; \; x\geq 1\\\\\left \{ {{x\geq 1} \atop {x\in (-\infty ,-1\, ]\cup [\, 2,+\infty )}} \right. \; \; \Rightarrow \; \; x\in [\,2,+\infty )](https://tex.z-dn.net/?f=1%29%5C%3B%20%5C%3B%20%28x-1%29%5Ccdot%20%5Csqrt%7Bx%5E2-x-2%7D%5Cgeq%200%5C%3B%20%2C%5C%3B%20%5C%3B%20ODZ%3A%5C%3B%20x%5E2-x-2%5Cgeq%200%5C%3B%20%2C%5C%5C%5C%5Cx_1%3D-1%5C%3B%20%2C%5C%3B%20x_2%3D2%5C%3B%20%5C%3B%20%28teor.%5C%3B%20Vieta%29%5C%3B%20%5C%3B%20%5CRightarrow%20%5C%3B%20%5C%3B%20%28x%2B1%29%28x-2%29%5Cgeq%200%5C%3B%20%5C%3B%20%5CRightarrow%20%5C%5C%5C%5Cx%5Cin%20%28-%5Cinfty%20%2C-1%5C%2C%20%5D%5Ccup%20%5B%5C%2C%202%2C%2B%5Cinfty%20%29%5C%5C%5C%5CT.k.%5C%3B%20%5Csqrt%7Bx%5E2-x-2%7D%5Cgeq%200%5C%3B%20%5C%3B%20pri%5C%3B%20%5C%3B%20x%5Cin%20ODZ%5C%3B%20%2Cto%5C%3B%20%28x-1%29%5Ccdot%20%5Csqrt%7Bx%5E2-x-2%7D%5Cgeq%200%5C%3B%20pri%5C%5C%5C%5Cx-1%5Cgeq%200%5C%3B%20%5C%3B%20%5CRightarrow%20%5C%3B%20%5C%3B%20x%5Cgeq%201%5C%5C%5C%5C%5Cleft%20%5C%7B%20%7B%7Bx%5Cgeq%201%7D%20%5Catop%20%7Bx%5Cin%20%28-%5Cinfty%20%2C-1%5C%2C%20%5D%5Ccup%20%5B%5C%2C%202%2C%2B%5Cinfty%20%29%7D%7D%20%5Cright.%20%5C%3B%20%5C%3B%20%5CRightarrow%20%5C%3B%20%5C%3B%20x%5Cin%20%5B%5C%2C2%2C%2B%5Cinfty%20%29)
Так как левая часть неравенства равна 0 при х=1 , х=-1 , х=2, причём х=-1 ,х=2 входят в ОДЗ, пишем ответ:
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1\; ,\; \; ODZ:\; \left \{ {{2x+7\geq 0} \atop {x+2\ne 0}} \right. \; \; \left \{ {{x\geq -3,5} \atop {x\ne -2}} \right. \; \; \Rightarrow \\\\x\in [-3,5\, ;\, -2)\cup (-2;+\infty )\\\\\frac{\sqrt{2x+7}}{x+2}-1>0\; ,\; \; \frac{\sqrt{2x+7}-x-2}{x+2}>0\; \Rightarrow \\\\\left \{ {{\sqrt{2x+7}-x-2>0} \atop {x+2>0}} \right. \; \; ili\; \; \left \{ {{\sqrt{2x+7}-x-2<0} \atop {x+2<0}} \right. \\\\a)\; \; \left \{ {{\sqrt{2x+7}>x+2} \atop {x+2>0}} \right. \; \left \{ {{2x+7>x^2+2x+4} \atop {x>-2\; ,\; }} \right. " alt="2)\; \; \frac{\sqrt{2x+7}}{x+2}>1\; ,\; \; ODZ:\; \left \{ {{2x+7\geq 0} \atop {x+2\ne 0}} \right. \; \; \left \{ {{x\geq -3,5} \atop {x\ne -2}} \right. \; \; \Rightarrow \\\\x\in [-3,5\, ;\, -2)\cup (-2;+\infty )\\\\\frac{\sqrt{2x+7}}{x+2}-1>0\; ,\; \; \frac{\sqrt{2x+7}-x-2}{x+2}>0\; \Rightarrow \\\\\left \{ {{\sqrt{2x+7}-x-2>0} \atop {x+2>0}} \right. \; \; ili\; \; \left \{ {{\sqrt{2x+7}-x-2<0} \atop {x+2<0}} \right. \\\\a)\; \; \left \{ {{\sqrt{2x+7}>x+2} \atop {x+2>0}} \right. \; \left \{ {{2x+7>x^2+2x+4} \atop {x>-2\; ,\; }} \right. " align="absmiddle" class="latex-formula">
-2}} \right. \; \left \{ {{(x-\sqrt3)(x+\sqrt3)<0} \atop {x>-2}} \right. \; \left \{ {{x\in (-\sqrt3,\sqrt3)} \atop {x>-2}} \right. \; \to \; \; x\in (-\sqrt3,\sqrt3)\\\\b)\; \; \left \{ {{\sqrt{2x+7}-2}} \right. \; \left \{ {{(x-\sqrt3)(x+\sqrt3)<0} \atop {x>-2}} \right. \; \left \{ {{x\in (-\sqrt3,\sqrt3)} \atop {x>-2}} \right. \; \to \; \; x\in (-\sqrt3,\sqrt3)\\\\b)\; \; \left \{ {{\sqrt{2x+7}
![\left \{ {{x\in [-3,5\, ;\, -2)\cup (-2;+\infty )} \atop {x\in (-\infty ;-3,5\, ]\cup (-\sqrt3;\sqrt3)}} \right. \; \; \Rightarrow \; \; x\in \{\, -3,5\, \}\cup (-\sqrt3;\sqrt3) \left \{ {{x\in [-3,5\, ;\, -2)\cup (-2;+\infty )} \atop {x\in (-\infty ;-3,5\, ]\cup (-\sqrt3;\sqrt3)}} \right. \; \; \Rightarrow \; \; x\in \{\, -3,5\, \}\cup (-\sqrt3;\sqrt3)](https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7Bx%5Cin%20%5B-3%2C5%5C%2C%20%3B%5C%2C%20-2%29%5Ccup%20%28-2%3B%2B%5Cinfty%20%29%7D%20%5Catop%20%7Bx%5Cin%20%28-%5Cinfty%20%3B-3%2C5%5C%2C%20%5D%5Ccup%20%28-%5Csqrt3%3B%5Csqrt3%29%7D%7D%20%5Cright.%20%5C%3B%20%5C%3B%20%5CRightarrow%20%5C%3B%20%5C%3B%20x%5Cin%20%5C%7B%5C%2C%20-3%2C5%5C%2C%20%5C%7D%5Ccup%20%28-%5Csqrt3%3B%5Csqrt3%29)