Question

81x^2<=16 решите неравенство

Answer 1

$81 x^{2} \leq 16 \\ x^{2} \leq \frac{16}{81} \\ x \leq \sqrt{ \frac{16}{81} } \\ x \leq \frac{4}{9}$

Answer 2

81x^2-16<=
(9x-4)(9x+4)<=0
(x-4/9)(x+4/9)<=0
xe[-4/9;4/9]