Question

cho đa thức: M= x^2 - x+1

a) chứng minh: M>0 với mọi x

b)tìm x để M đạt giá trị nhỏ nhất.

Answer 1

a) \(M=x^2-x+1\)

\(\Leftrightarrow M=x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{2}\right)^2+1\)

\(\Leftrightarrow M=\left[x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]-\left(\dfrac{1}{2}\right)^2+1\)

\(\Leftrightarrow M=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

Vì \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)

Nên \(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\)

Vậy \(M=x^2-x+1>0\forall x\)

b) \(M=x^2-x+1\)

\(\Leftrightarrow M=x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{2}\right)^2+1\)

\(\Leftrightarrow M=\left[x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]-\left(\dfrac{1}{2}\right)^2+1\)

\(\Leftrightarrow M=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

Vì \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)

Nên \(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)

Vậy GTNN của \(M=\dfrac{3}{4}\) khi \(x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)

Answer 2

M= x² - x+1

a) M = x²-x\(+\dfrac{1}{4}+\dfrac{3}{4}\)

= \(\left(x^2-x+\dfrac{1}{4}\right)^{ }+\dfrac{3}{4}\)

=\(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

d0 \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)

=> \(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

=>M \(\ge\dfrac{3}{4}\Rightarrow M\ge0\left(đpcm\right)\)

b)vì \(M\ge\dfrac{3}{4}\left(a\right)\)

GTNN M =\(\dfrac{3}{4}khi\) x-\(\dfrac{1}{2}=0\)

=> x=\(\dfrac{1}{2}\)