Question

chung minh 8x - 4x^2 -5 <0 voi moi x

Answer 1

8x-4x²-5

= -4x²+8x-5

= -4x²+8x-4-1

= -(4x²-8x+4)-1

= -(2x-2)²-1

do -(2x-2)² ≤ 0 ∀x

=> -(2x-2)²-1≤ -1 ∀x

=> -(2x-2)² <0 ∀x

hay 8x-4x²-5<0 ∀x (đpcm)

Answer 2

Ta có:

\(8x-4x^2-5=-\left(4x^2-8x+5\right)=-\left(\left(2x\right)^2-2.2x.2+2^2+1\right)=-\left(2x-2\right)^2-1\)Vì \(-\left(2x-2\right)^2\le0\), Với mọi x nên