1) \(D=\left|x^2+x+3\right|+\left|x^2+x-6\right|\)
\(D=\left|x^2+x+3\right|+\left|6-x^2-x\right|\)
Áp dụng bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có :
\(D\ge\left|x^2+x+3+6-x^2-x\right|=\left|9\right|=9\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x^2+x+3\right)\left(6-x^2-x\right)\ge0\Leftrightarrow-3\le x\le2\)
2) \(C=x^2+xy+y^2-3x-3y\)
\(C=\left(x^2-2x+1\right)+\left(y^2-2y+1\right)+\left(xy-x-y+1\right)-3\)
\(C=\left(x-1\right)^2+\left(y-1\right)^2+\left(x-1\right)\left(y-1\right)-3\)
\(C=\left(x-1\right)^2+2\cdot\left(x-1\right)\cdot\frac{\left(y-1\right)}{2}+\frac{\left(y-1\right)^2}{4}+\frac{3\left(y-1\right)^2}{4}-3\)
\(C=\left(x-1-\frac{y-1}{2}\right)^2+\frac{3\left(y-1\right)^2}{4}-3\ge-3\forall x;y\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-1-\frac{y-1}{2}=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
3) \(B=x^4-2x^3+3x^2-2x+1\)
\(B=x^2\left(x^2-2x+3-\frac{2}{x}+\frac{1}{x^2}\right)\)
\(B=x^2\left[\left(x^2+2+\frac{1}{x^2}\right)-2\left(x+\frac{1}{x}\right)+1\right]\)
\(B=x^2\left[\left(x+\frac{1}{x}\right)^2-2\left(x+\frac{1}{x}\right)+1\right]\)
\(B=x^2\left(x+\frac{1}{x}-1\right)^2\)
\(B=\left[x\left(x+\frac{1}{x}-1\right)\right]^2\)
\(B=\left(x^2-x+1\right)^2\)
Xét \(x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)
\(\Rightarrow B=\left(x^2-x+1\right)^2\ge\left(\frac{3}{4}\right)^2=\frac{9}{16}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=\frac{1}{2}\)
4) \(A=ax^2+bx+c\)
\(A=a\left(x^2+\frac{bx}{a}+\frac{c}{a}\right)\)
\(A=a\left(x^2+2\cdot x\cdot\frac{b}{2a}+\frac{b^2}{4a^2}+\frac{c}{a}-\frac{b^2}{4a^2}\right)\)
\(A=a\left[\left(x+\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a^2}\right]\)
\(A=a\left(x+\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a}\ge\frac{4ac-b^2}{4a}\forall x;a;b;c\)
Dấu "=" xảy ra \(\Leftrightarrow x=\frac{-b}{2a}\)
