\(a,B=2x=2\cdot2,5=5\\ b,A=x^3-6x^2+12x-8+x^3-x+6x^2-18x+4x^2+x+8\\ A=2x^3+4x^2-6x\\ C=\dfrac{A}{B}\\ C=\dfrac{2x^3+4x^2-6x}{2x}=x^2+2x-3=\left(x+1\right)^2-4\ge-4\)
Dấu \("="\Leftrightarrow x=-1\)
a, Khi x=2,5 => B= 2. 2,5 =5
b, A= (x-2)3+x.(x-1).(x+1)+6x.(x-3)+4x2+x+8
A=(x-2)3+x.(x2-1)+6x2-18x+4x2+x+8
A= x3-3x2.2+3x.22+x3-x +6x -18x+4x2+x+8
A= x3-6x2+12x+x3-x +6x -18x+4x2+x+8
A=(x3+x3)+(-6x2+4x2)+(12x-x+6x-18x+x)+8
A= 2x3-2x2+8
Hoctot
c, C= (2x3-2x2+8) :2x =(2x3:2x)+(-2x2:2x)+(8:2x)
= x2-x +4
= x2-2.x.\(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{15}{4}\)
= \(\left(x-\dfrac{1}{2}\right)^2+\dfrac{15}{4}\)
Ta thấy: \(\left(x-\dfrac{1}{2}\right)^2\ge0=>C\ge\dfrac{15}{4}\)
Dấu"=" xảy ra <=> \(x-\dfrac{1}{2}=0< =>x=\dfrac{1}{2}\)
Vậy Cmin= \(\dfrac{15}{4}< =>x=\dfrac{1}{2}\)
Hoctot
