Question

Tim min  

A= x^2+x-2

B=x^2-x

C=1/4x^2-x+7

D=1/2x^2+3x+1

E=(x-1)(x^2+x+1)-x(x-1)(x+1)+x^2

Tim max

A= -2(x-1)^2+(x+3)

B=-x^2+4x-1

C=-2x^2+x

D=(x-3)(2-x)-3(x+5)(x+7)

E=-3x^2+4x-1

AI HELP MIK DAU TIEN MIK SẼ HAU TẠ

 

Answer 1

a, \(A=x^2+2\cdot\frac{1}{2}x+\frac{1}{4}-\frac{9}{4}=\left(x+\frac{1}{2}\right)^2-\frac{9}{4}\)

=> \(A\ge-\frac{9}{4}\) dấu = xảy ra khi : \(x=\frac{-1}{2}\)

Answer 2

b, \(B=x^2-2.\frac{1}{2}.x+\frac{1}{4}-\frac{1}{4}=\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\)

=> \(B\ge-\frac{1}{4}\) dấu = <=> \(x=\frac{1}{2}\)

Answer 3

c, \(C=\frac{1}{4}.x^2-2.\frac{1}{2}x+1+6=\left(\frac{1}{2}x^{ }-1\right)^2+6\)

=> \(C\ge6\) dấu = <=> \(x=2\)

Answer 4

d, \(2D=x^2+6x+2=x^2+6x+9-11=\left(x+3\right)^2-11\)

=> \(2D\ge-11\)=> \(D\ge-\frac{11}{2}\) Dấu = <=> \(x=-3\)

Answer 5

e, \(E=x^3-1-x\left(x^2-1\right)+x^2=x^3-1-x^3+x+x^2\)

=> \(E=x^2+x-1=x^2+2.\frac{1}{2}x+\frac{1}{4}-\frac{5}{4}=\left(x+\frac{1}{2}\right)^2-\frac{5}{4}\)

=> \(E\ge-\frac{5}{4}\) Dấu = <=> \(x=-\frac{1}{2}\)