Question

Amongora'oi porãva ohechauka ojo porãva ñe'endi
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Answer 1

Respuesta:

Explicación:

$\leq \sqrt[n]{x} \\ \geq \neq \neq \sqrt{x} \sqrt[n]{x} \frac{x}{y} \frac{x}{y} \leq \leq \geq \geq \geq \geq \geq \geq \geq \geq \geq \sqrt{x} \sqrt{x} \neq \pi \alpha \alpha \sqrt[n]{x} \sqrt[n]{x} x^{2} \leq \int\limits^a_b {x} \, dx \lim_{n \to \infty} a_n \alpha \beta x_{123}$