Помогите пожалуйста!
Найдите область определения функции:
Приложения:
Ответы на вопрос
Ответил Artem112
0
а) ![y= sqrt{ frac{2-|x|}{x^2+2} }
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2-|x| geq 0
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|x |leq 2
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-2 leq x leq 2
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D(y)[-2; 2]](https://tex.z-dn.net/?f=y%3D+sqrt%7B+frac%7B2-%7Cx%7C%7D%7Bx%5E2%2B2%7D+%7D+%0A%5C%0A2-%7Cx%7C+geq+0%0A%5C%0A%7Cx+%7Cleq+2%0A%5C%0A-2+leq+x+leq+2%0A%5C%0AD%28y%29%5B-2%3B+2%5D%0A "y= sqrt{ frac{2-|x|}{x^2+2} }
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2-|x| geq 0
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|x |leq 2
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-2 leq x leq 2
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D(y)[-2; 2]")
б)![y= sqrt{(x-2)(x-3)} +sqrt{(5-x)(6-x)}
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left { {{(x-2)(x-3) geq 0} atop {(5-x)(6-x) geq 0}} right.
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left { {{x leq 2; x geq 3} atop {x leq 5; x geq 6}} right.
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D(y)=(-infty;2]cup[3;5]cup[6;+infty)](https://tex.z-dn.net/?f=y%3D+sqrt%7B%28x-2%29%28x-3%29%7D+%2Bsqrt%7B%285-x%29%286-x%29%7D+%0A%5C%0A+left+%7B+%7B%7B%28x-2%29%28x-3%29+geq+0%7D+atop+%7B%285-x%29%286-x%29+geq+0%7D%7D+right.+%0A%5C%0A+left+%7B+%7B%7Bx+leq+2%3B+x+geq+3%7D+atop+%7Bx+leq+5%3B+x+geq+6%7D%7D+right.+%0A%5C%0AD%28y%29%3D%28-infty%3B2%5Dcup%5B3%3B5%5Dcup%5B6%3B%2Binfty%29 "y= sqrt{(x-2)(x-3)} +sqrt{(5-x)(6-x)}
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left { {{(x-2)(x-3) geq 0} atop {(5-x)(6-x) geq 0}} right.
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left { {{x leq 2; x geq 3} atop {x leq 5; x geq 6}} right.
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D(y)=(-infty;2]cup[3;5]cup[6;+infty)")
б)
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