Question

(2x² + 1)(3x²+4) ≥ (2 + 5x²)(2x² + 1);​

Answer 1

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$\displaystytle\bf \\(2x^2 + 1) (3x^2 + 4) \geq (2 + 5x^2)(2x^2 + 1)\\\\6x^4 + 8x^2+3x^2+4 \geq 4x^2+2+10x^4+5x^2\\\\6x^4+11x^2+4 \geq 10x^4+9x^2\\\\11x^2-9x^2\geq 10x^4-6x^4\\\\2x^2 \geq 4x^4\\\\\\-2x^4+x^2 \geq 0\\\\2x^4-x^2 \geq 0\\\\x^2(2x^2-1)\geq 0\\\\x^2\left(\sqrt{2}x+1\right)\left(\sqrt{2}x-1\right)\le 0\\\\-\frac{1}{\sqrt{2}}\leq x\leq \frac{1}{\sqrt{2}}$

ответ:

$\boxed { \displaystyle \bf -\frac{1}{\sqrt{2}}\leq x\leq \frac{1}{\sqrt{2}}}$

Answer 2

Объяснение:

$(2x^2 + 1)(3x^2+4) \geq (2 + 5x^2)(2x^2 + 1)\\\\(2x^2 + 1)(3x^2+4) - (2 + 5x^2)(2x^2 + 1)\geq 0\\\\(2x^2 + 1)*((3x^2+4 - (2 -+5x^2))\geq 0\\\\(2x^2 + 1)*((3x^2+4 - 2 - 5x^2)\geq 0\\\\(2x^2+1)*(2-2x^2)\geq 0\\\\2*(2x^2+1)*(1-x^2)\geq 0\ |:2\\\\(2x^2+1)*(1-x^2)\geq 0\\\\2x^2+1 > 0\ \ \ \ \ \ \Rightarrow\\\\1-x^2\geq 0\ |*(-1)\\\\x^2-1\leq 0\\\\(x+1)*(x-1)\leq 0$

-∞__+__-1__-__1__+__+∞          ⇒

x∈[-1;1].

Ответ: x∈[-1;1].