Question

90 БАЛОВ ЗА ОДНО ЗАДАНИЕ
Найдите область определения функции:

Answer 1

$\displaystyle\bf 1)\\y=\sqrt{5x-x^{2} } \\\\5x-x^{2} \geq 0\\\\x(5-x)\geq 0$

- - - - - [0] + + + + + [5] - - - - -

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$\displaystyle\bf Otvet:x\in[0 \ ; \ 5]\\\\\\2)\\y=\frac{6}{\sqrt{8+10x-3x^{2} } } \\\\\\8+10x-3x^{2}>0\\\\3x^{2} -10x-8<0\\\\3x^{2} -10x-8=0\\\\D=(-10)^{2} -4\cdot 3\cdot (-8)=100+96=196=14^{2} \\\\\\x_{1} =\frac{10-14}{6} =-\frac{2}{3} \\\\\\x_{2} =\frac{10+14}{6} =4\\\\\\3x^{2} -10x-8=3(x+\frac{2}{3} )(x-4)\\\\\\3(x+\frac{2}{3} )(x-4)<0\\\\\\(x+\frac{2}{3} )(x-4)<0$

+ + + + + (- 2/3) - - - - - (4) + + + + +

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$\displaystyle\bf Otvet:x\in(-\frac{2}{3} \ ; \ 4)$