Question

giải chi tiết giúp mình nha,cảm ơn

Answer 1

a)\(MinA=1\Leftrightarrow x=\dfrac{1}{4}\)

b)\(MaxP=1\Leftrightarrow x=\dfrac{1}{4}\)

Answer 2

a) \(A=x-\sqrt{x}+\dfrac{5}{4}=\left(x-\sqrt{x}+\dfrac{1}{4}\right)+1=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+1\ge1\)

\(minA=1\Leftrightarrow x=\dfrac{1}{4}\)

b) \(P=\dfrac{1}{x-\sqrt{x}+\dfrac{5}{4}}=\dfrac{1}{\left(x-\sqrt{x}+\dfrac{1}{4}\right)+1}=\dfrac{1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+1}\le\dfrac{1}{1}=1\)

\(maxP=1\Leftrightarrow x=\dfrac{1}{4}\)

Answer 3

a, \(A=x-\sqrt{x}+\dfrac{5}{4}\)

\(=x-\sqrt{x}+\dfrac{1}{4}+1\)

\(=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+1\ge1\)

\(\Rightarrow minA=1\Leftrightarrow x=\dfrac{1}{4}\)

Answer 4

b, \(P=\dfrac{1}{x-\sqrt{x}+\dfrac{5}{4}}=\dfrac{1}{A}\le\dfrac{1}{1}=1\)

\(\Rightarrow maxP=1\Leftrightarrow x=\dfrac{1}{4}\)