Soluções para a tarefa
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Pela equação, devemos ter
, pois 
é um radicando (está "dentro" de uma raiz quadrada);
![x^{2}\sqrt{x}\leq\alpha\\ \\ x^{2}\cdot x^{1/2}\leq\alpha\\ \\ x^{2+1/2}\leq\alpha\\ \\ x^{\left(4+1 \right )/2}\leq\alpha\\ \\ x^{5/2}\leq\alpha\\ \\ x\leq\alpha^{2/5}\\ \\ \boxed{x\leq \sqrt[5]{\alpha^{2}}}](https://tex.z-dn.net/?f=x%5E%7B2%7D%5Csqrt%7Bx%7D%5Cleq%5Calpha%5C%5C+%5C%5C+x%5E%7B2%7D%5Ccdot+x%5E%7B1%2F2%7D%5Cleq%5Calpha%5C%5C+%5C%5C+x%5E%7B2%2B1%2F2%7D%5Cleq%5Calpha%5C%5C+%5C%5C+x%5E%7B%5Cleft%284%2B1+%5Cright+%29%2F2%7D%5Cleq%5Calpha%5C%5C+%5C%5C+x%5E%7B5%2F2%7D%5Cleq%5Calpha%5C%5C+%5C%5C+x%5Cleq%5Calpha%5E%7B2%2F5%7D%5C%5C+%5C%5C+%5Cboxed%7Bx%5Cleq+%5Csqrt%5B5%5D%7B%5Calpha%5E%7B2%7D%7D%7D "x^{2}\sqrt{x}\leq\alpha\\ \\ x^{2}\cdot x^{1/2}\leq\alpha\\ \\ x^{2+1/2}\leq\alpha\\ \\ x^{\left(4+1 \right )/2}\leq\alpha\\ \\ x^{5/2}\leq\alpha\\ \\ x\leq\alpha^{2/5}\\ \\ \boxed{x\leq \sqrt[5]{\alpha^{2}}}")
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