Объяснение:
1)
[tex]-2x^2-5x+3 < 0\ |*(-1)\\2x^2+5x-3 > 0\\2x^2+6x-x-3 > 0\\2x*(x+3)-(x+3) > 0\\(x+3)*(2x-1) > 0.[/tex]
-∞__+__-3__-__1/2__+__+∞
Ответ: x∈(-∞;-3)U(1/2;+∞).
2)
[tex]3x^2-4x+7 > 0\ |:3\\x^2-\frac{4x}{3}+\frac{7}{3} > 0\\ x^2-2*x*\frac{2}{3} +(\frac{2}{3})^2-( \frac{2}{3} )^2+\frac{7}{3} > 0\\ (x-\frac{2}{3})^2-\frac{4}{9}+\frac{7}{3} > 0\\ (x-\frac{2}{3})^2+\frac{7*3-4}{9} > 0\\ (x-\frac{2}{3} )^2+\frac{17}{9} > 0.\ \ \ \ \Rightarrow[/tex]
Ответ: x∈(-∞;+∞).
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Answers & Comments
Объяснение:
1)
[tex]-2x^2-5x+3 < 0\ |*(-1)\\2x^2+5x-3 > 0\\2x^2+6x-x-3 > 0\\2x*(x+3)-(x+3) > 0\\(x+3)*(2x-1) > 0.[/tex]
-∞__+__-3__-__1/2__+__+∞
Ответ: x∈(-∞;-3)U(1/2;+∞).
2)
[tex]3x^2-4x+7 > 0\ |:3\\x^2-\frac{4x}{3}+\frac{7}{3} > 0\\ x^2-2*x*\frac{2}{3} +(\frac{2}{3})^2-( \frac{2}{3} )^2+\frac{7}{3} > 0\\ (x-\frac{2}{3})^2-\frac{4}{9}+\frac{7}{3} > 0\\ (x-\frac{2}{3})^2+\frac{7*3-4}{9} > 0\\ (x-\frac{2}{3} )^2+\frac{17}{9} > 0.\ \ \ \ \Rightarrow[/tex]
Ответ: x∈(-∞;+∞).