Question

40x-16x^2-25>=0 плиз очень надо

Answer 1

$\displaystyle 40x-16x^2-25\geq 0\\-(16x^2-40x+25)\geq 0\\-(4x-5)^2\geq 0\\(4x-5)^2\leq 0\\(4x-5)^2=0\\\\x=\frac{5}{4}=1\frac{1}{4}=1,25$