Tìm GTNN của biểu thức
A= 3x2 – x-1
B=(x+1).(x+2).(x2+3x+6)
C= (x-3).(x-4).(x2-7x+16)
D=(5x-1)2 + (5x-3)2
$* \ A=3x^2-x-1=3(x^2-\dfrac13x+\dfrac1{36})-\dfrac{13}{12}=3(x-\dfrac16)^2-\dfrac{13}{12}\geq \dfrac{-13}{12}$
Dấu '=' xảy ra khi $x=\dfrac16$
$* \ B=(x+1)(x+2)(x^2+3x+6)=(x^2+3x+2)(x^2+3x+6)$
Đặt $x^2+3x+4=y \ (y\geq \dfrac 74)$
$\Rightarrow B=(y-2)(y+2)=y^2-4\geq \dfrac{-15}{16}$
Dấu '=' xảy ra khi $y=\dfrac 74\Leftrightarrow x^2+3x+4=\dfrac 74\Leftrightarrow x=\dfrac {-3}2$
$* \ C=(x-3)(x-4)(x^2-7x+16)=(x^2-7x+12)(x^2-7x+16)$
Đặt $x^2-7x+14=y \ (y\geq \dfrac 74)$
$\Rightarrow C=(y-2)(y+2)=y^2-4\geq \dfrac{-15}{16}$
Dấu '=' xảy ra khi $y=\dfrac 74\Leftrightarrow x^2-7x+14=\dfrac 74\Leftrightarrow x=\dfrac 72$
$* \ D=(5x-1)^2+(5x-3)^2$
Đặt $5x-2=y\Rightarrow D=(y+1)^2+(y-1)^2=y^2+2y+1+y^2-2y+1=2y^2+2\geq 2$
Dấu '=' xảy ra khi $y=0\Leftrightarrow 5x-2=0\Leftrightarrow x=\dfrac 25$