Question

[Naa.Khahh]

giúp mk bài 3 (giúp dc bao nhiêu quý bấy nhiêu ạ)

Answer 1

[Lê Thị Thục Hiền]

\(A=\dfrac{\sqrt{3x-9}}{x}=\sqrt{\dfrac{3x-9}{x^2}}=\sqrt{-\dfrac{9}{x^2}+\dfrac{3}{x}}\)\(=\sqrt{\dfrac{1}{4}-\left(\dfrac{9}{x^2}+2.\dfrac{3}{x}.\dfrac{1}{2}+\dfrac{1}{4}\right)}\)

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

Dấu "=" xảy ra khi \(\dfrac{3}{x}-\dfrac{1}{2}=0\Leftrightarrow x=6\) (tm)

Vậy \(A_{max}=\dfrac{1}{2}\)

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

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

Dấu "=" xảy ra khi \(\dfrac{2}{x}-\dfrac{1}{2}=0\Leftrightarrow x=4\) (tm)

Vậy \(B_{max}=\dfrac{1}{2}\)

\(C=\dfrac{\sqrt{x-4}}{3x}=\sqrt{\dfrac{x-4}{9x^2}}=\sqrt{-\dfrac{4}{9x^2}+\dfrac{1}{9x}}=\sqrt{\dfrac{1}{144}-\left(\dfrac{4}{9x^2}-2.\dfrac{2}{3x}.\dfrac{1}{12}+\dfrac{1}{144}\right)}\)

\(=\sqrt{\dfrac{1}{144}-\left(\dfrac{2}{3x}-\dfrac{1}{12}\right)^2}\le\sqrt{\dfrac{1}{144}}=\dfrac{1}{12}\)

Dấu "=" xảy ra khi \(x=8\) (tm)

Vậy \(C_{max}=\dfrac{1}{12}\)

Answer 2

[Naa.Khahh]

iem cảm ơn !!!!!!<3 cj