\(A=x^2-3x+5\)
\(A=x^2-3x+\dfrac{9}{4}+\dfrac{11}{4}\)
\(A=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\)
Dấu "=" xảy ra khi: \(x=\dfrac{3}{2}\)
\(B=\left(2x-1\right)^2+\left(x+2\right)^2\)
\(B=4x^2-4x+1+x^2+4x+4\)
\(B=5x^2+5\ge5\)
Dấu "=" xảy ra khi: \(x=0\)
\(C=x^2-2x+y^2-4y+7\)
\(C=\left(x^2-2x+1\right)+\left(y^2-4y+4\right)+2\)
\(C=\left(x-1\right)^2+\left(y-2\right)^2+2\ge2\)
Dấu "=" xảy ra khi: \(x=1;y=2\)
\(D=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(D=\left(x+6\right)\left(x-1\right)\left(x+2\right)\left(x+3\right)\)
\(D=\left(x^2-x+6x-6\right)\left(x^2+2x+3x+6\right)\)
\(D=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(D=\left(x^2+5x\right)^2-36\)
\(D=\left(x^2+5x+\dfrac{25}{4}-\dfrac{25}{4}\right)^2-36\)
\(D=\left[\left(x+\dfrac{5}{2}\right)^2-\dfrac{25}{4}\right]^2-36\ge\dfrac{625}{8}-36=\dfrac{340}{8}\)
Dấu "=" xảy ra khi: \(x=-\dfrac{5}{2}\)
