Đáp án + Giải thích các bước giải:

\(A=x^2-4x+24\\=(x^2-4x+4)+20\\=(x-2)^2+20\\\text{Vì :} \ (x-2)^2 \geq 0 \ \forall x \\ \to (x-2)^2+20 \geq 20\\\text{Dấu = xảy ra khi :} \ (x-2)^2=0 \\ \Leftrightarrow x=2\\\text{Vậy} \ A_{min}=20 \Leftrightarrow x=2\) 

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\(B=-3x^2-2x+1\\=-3.\Big(x^2+\dfrac{2}{3}x-\dfrac{1}{3}\Big)\\-3.\Big(x^2+2 . x . \dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{3}-\dfrac{1}{9}\Big)\\=-3.\Big[\Big(x+\dfrac{1}{3})^2-\dfrac{4}{9}\Big]\\=-3.\Big(x+\dfrac{1}{3})^2+\dfrac{4}{3}\\\text{Vì :} \ \Big(x+\dfrac{1}{3}\Big)^2 \geq 0 \ \forall x \\ \to -3.\Big(x+\dfrac{1}{3}\Big)^2 \leq 0 \\ \to -3.\Big(x+\dfrac{1}{3}\Big)^2+\dfrac{4}{3} \leq \dfrac{4}{3}\\\text{Dấu = xảy ra khi :} \ \Big(x+\dfrac{1}{3}\Big)^2=0 \\ \Leftrightarrow x=-\dfrac{1}{3}\\\text{Vậy} \ B_{max}=\dfrac{4}{3} \Leftrightarrow x=-\dfrac{1}{3}\)

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\(C=2x^2+8x+5\\=2x^2+8x+8-3\\=2.(x^2+4x+4)-3\\=2.(x+2)^2-3\\\text{Vì :} \ (x+2)^2 \geq 0 \ \forall x\\\to 2.(x+2)^2-3 \geq -3\\\text{Dấu = xảy ra khi :} \ (x+2)^2=0 \\ \Leftrightarrow x=-2\\\text{Vậy} \ C_{min}=-3 \Leftrightarrow x=-2\)

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\(D=(y+3)^2+2(y-1)^2\\=y^2+6y+9+2(y^2-2y+1)\\=y^2+6y+9+2y^2-4y+2\\=3y^2+2y+11\\=3.\Big(y^2+\dfrac{2}{3}y+\dfrac{11}{3}\Big)\\=3.\Big(y^2+2 . y . \dfrac{1}{3}+\dfrac{1}{9}+\dfrac{11}{3}-\dfrac{1}{9}\Big)\\=3.\Big[\Big(y+\dfrac{1}{3}\Big)^2+\dfrac{32}{9}\Big]\\=3.\Big(y+\dfrac{1}{3}\Big)^2+\dfrac{32}{3}\\\text{Vì :} \ \Big(y+\dfrac{1}{3}\Big)^2 \geq 0 \ \forall x \\\to 3.\Big(y+\dfrac{1}{3}\Big)^2+\dfrac{32}{3} \geq \dfrac{32}{3}\\\text{Dấu = xảy ra khi :} \ \Big(y+\dfrac{1}{3}\Big)^2\\\text{Vậy} \ D_{min}=\dfrac{32}{3} \Leftrightarrow y=-\dfrac{1}{3}\)