Question

[Letuandan]

  giải giúp e bài 16b vs ạ

Answer 1

[Nguyễn Lê Phước Thịnh]

16:

a: ĐKXĐ: x>0

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

\(x+\dfrac{1}{x}>=2\cdot\sqrt{x\cdot\dfrac{1}{x}}=2\)

\(\sqrt{x}+\dfrac{1}{\sqrt{x}}>=2\cdot\sqrt{\sqrt{x}\cdot\dfrac{1}{\sqrt{x}}}=2\)

Do đó: \(x+\dfrac{1}{x}+\sqrt{x}+\dfrac{1}{\sqrt{x}}>=2+2=4\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x^2=1\\x=1\end{matrix}\right.\)

=>x=1