$\begin{array}{l}\underline{\text{Đáp án:}}\\3a,a=-3\\3b,a=3\\4a,GTNN_A=7↔x=1\\4b,GTNN_B=1↔x=2\\\underline{\text{Giải thích các bước giải:}}\\3a,x^3-3x^2+5x+2a \vdots x-2\\→x^3-2x^2-x^2+2x+3x-6+2a+6 \vdots x-2\\→x^2(x-2)-x(x-2)+3(x-2)+2a+6 \vdots x-2\\→2a+6=0\\→2a=-6\\→a=-3\\Vậy \,\, a=-3 \,\, thì \,\, x^3-3x^2+5x+2a \vdots x-2\\3b,x^2-4x+a \vdots x-1\\→x^2-x-3x+3+a-3 \vdots x-1\\→x(x-1)-3(x-1)+a-3 \vdots x-1\\→a-3=0\\→a=3\\Vậy \,\, a=4 \,\, thì \,\, x^2-4a+a \vdots x-1\\4a,A=x^2-2x+8\\A=x^2-2.x.1+1+7\\A=(x-1)^2+7\\Vì \,\, (x-1)^2 \geq 0\\→(x-1)^2+7 \geq 7\\Hay \,\, A \geq 7\\\text{Dấu = xảy ra khi}\\x-1=0\\→x=1\\Vậy \,\, GTNN_A=7↔x=1\\b,B=x^2+4x+5\\B=x^2+4x+4+1\\B=x^2+2.x.2+2^2+1\\B=(x-2)^2+1\\Vì \,\, (x-2)^2 \geq 0\\→(x-2)^2+1 \geq 1\\Hay \,\, B \geq 1\\\text{Dấu = xảy ra khi}\\x-2=0\\→x=2\\Vậy \,\, GTNN_B=1↔x=2\\\underline{\text{CHÚC BẠN HỌC TỐT}}\\\end{array}$
