Question

 giải giúp câu c

Answer 1

Bài 9 :

a, Ta có : \(A=x-2\sqrt{x}+1+4=\left(\sqrt{x}-1\right)^2+4\)

Thấy : \(\left(\sqrt{x}-1\right)^2\ge0\)

\(\Rightarrow A=\left(\sqrt{x}-1\right)^2+4\ge4\)

Vậy \(Min_A=4\Leftrightarrow x=1\)

b, Ta có : \(B=\dfrac{1}{x-\dfrac{1}{2}.2.\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\)

Thấy : \(\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

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

Vậy \(Max=\dfrac{4}{3}\Leftrightarrow\sqrt{x}-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{4}\)

Answer 2

c)\(C=\sqrt{1-2x+x^2}+\sqrt{x^2-6x+9}\)

\(=\sqrt{\left(1-x\right)^2}+\sqrt{\left(x-3\right)^2}\)\(=\left|1-x\right|+\left|x-3\right|\ge\left|1-x+x-3\right|=2\)

Dấu "=" xảy ra khi \(\left(1-x\right)\left(x-3\right)\ge0\Leftrightarrow1\le x\le3\)

Vậy \(C_{min}=2\)