$C=-10-8x-16x^2$
$C=-(10+8x+16x^2)$
$C=-[(4x)^2+2.4.x+1+9]$
$C=-(4x+1)^2-9$
$(4x+1)^2≥0,∀x$
$⇒-(4x+1)^2≤0,∀x$
$⇒-(4x+1)^2-9≤-9$
Vậy $C$ đạt GTLN $=-9$ khi $4x+1=0⇔x=\dfrac{-1}{4}$
$D=-3x^2+12x-13$
$=-3(x^2-4x+\dfrac{13}{3})$
$=-3(x^2-4x+2^2+\dfrac{1}{3}$
$=-3(x-2)^2-1$
Ta có $(x-2)^2≥0,∀x$
$⇒-3(x-2)^2≤0,∀x$
$⇒-3(x-2)^2-1≤-1$
Vậy $-3x^2+12x-13$ đạt GTLN $=-1$ khi $x-2=0⇔x=2$
