Решите неравенство.На фотографии.. Created by claudiaschiffer. algebra-ru

Решите неравенство.На фотографии...

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$\frac{x^{2} }{3}*6\geq \frac{3x-3}{2}*6\\\\2x^{2} \geq 3(3x-3)\\\\2x^{2}-9x+9\geq 0\\\\2x^{2}-9x+9=0\\\\D=(-9)^{2}-4*2*9=81-72=9=3^{2}\\\\x_{1}=\frac{9-3}{4} =1,5\\\\x_{2}=\frac{9+3}{4}=3\\\\2x^{2} -9x+9=2(x-1,5)(x-3)\\\\(x-1,5)(x-3)\geq 0$

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______[1,5]_______[3]________

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Ответ : x ∈ (- ∞ ; 1,5] ∪ [3 ; + ∞)

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manyny06

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Ответ:

решение смотри на фотографии

Объяснение:

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