Question

[Hiền Bùi Ngọc]

Tìm gtln ( hoặc gtnn ):
a) \(A=2x^2-8x-10\)
b) \(B=x-x^2\)

Answer 1

[ST]

a, \(A=2x^2-8x-10=2\left(x^2-4x+4\right)-18=2\left(x-2\right)^2-18\ge-18\)

Dấu "=" xảy ra <=> x-2=0 <=> x=2

Vậy MinA = -18 khi x=2

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

Dấu "=" xảy ra <=> x-1/2=0 <=> x=1/2

Vậy MaxB = 1/4 khi x=1/2

Answer 2

[Thái Bùi Ngọc]

a) \(A=2x^2-8x-10\)
\(=2\left(x^2-4x-5\right)\)

\(=2\left(x^2-2.x.2+2^2-2^2-5\right)\)
\(=2\left[\left(x-2\right)^2-9\right]\)
\(=2\left(x-2\right)^2-18\)

Vì \(2\left(x-2\right)^2\ge0\forall x\)

Nên \(2\left(x-2\right)^2\ge-18\)

Hay \(A\ge-18\)

Vậy gtnn của A là -18 khi \(2\left(x-2\right)^2=0\)

\(x-2=0\)

\(x=2\)

b) \(B=x-x^2\)

\(=-x^2-x\)

\(=-\left(x^2-x\right)\)

\(=-\text{[}x^2-2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2\text{]}\)

\(=-\text{[}\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\text{]}\)

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

Vì \(-\left(x-\frac{1}{2}\right)^2\le0\forall x\)
Nên \(-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\forall x \)
Vậy gtln của B là \(\frac{1}{4}\)khi \(x-\frac{1}{2}=0\)

\(x=\frac{1}{2}\)