Question

tìm GTLN(NN) của biểu thức :

a) A = \(\left(x+\dfrac{2}{3}\right)^2+\dfrac{1}{2}\) ( x thuộc Q )

b) B = \(\dfrac{2}{\left(x-\dfrac{1}{2}\right)^2+2}\) ( x thuộc Q )

Answer 1

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

\(\left(x+\dfrac{2}{3}\right)^2\ge0\forall x\in R\)

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

Dấu "=" xảy ra khi:

\(\left(x+\dfrac{2}{3}\right)^2=0\Rightarrow x=-\dfrac{2}{3}\)

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

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

\(\left(x-\dfrac{1}{2}\right)^2+2\ge2\)

\(B=\dfrac{2}{\left(x-\dfrac{1}{2}\right)^2+2}\le1\)

Dấu "=" xảy ra khi:

\(\left(x-\dfrac{1}{2}\right)^2=0\Rightarrow x=\dfrac{1}{2}\)