Question

[Nguyễn acc 2]

OvO mai cần òi , giúp . K phải thi nhe

Answer 1

[Nguyễn Ngọc Huy Toàn]

Bài 5.

\(P=\left(2x+\dfrac{1}{x}\right)^2+\left(2y+\dfrac{1}{y}\right)^2\)

Áp dụng BĐT AM-GM, ta có:

\(\left(2x+\dfrac{1}{x}\right)^2+\left(2y+\dfrac{1}{y}\right)^2\ge2\sqrt{\left(2x+\dfrac{1}{x}\right)^2.\left(2y+\dfrac{1}{y}\right)^2}\)

\(\Leftrightarrow\left(2x+\dfrac{1}{x}\right)^2+\left(2y+\dfrac{1}{y}\right)^2\ge2\left(2x+\dfrac{1}{x}\right).\left(2y+\dfrac{1}{y}\right)\)

\(\Leftrightarrow\left(2x+\dfrac{1}{x}\right)^2+\left(2y+\dfrac{1}{y}\right)^2\ge2\left(4xy+\dfrac{2x}{y}+\dfrac{2y}{x}+\dfrac{1}{xy}\right)\)

\(\Leftrightarrow\left(2x+\dfrac{1}{x}\right)^2+\left(2y+\dfrac{1}{y}\right)^2\ge8xy+4\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+\dfrac{2}{xy}\)

Áp dụng BĐT AM-GM, ta có:

\(\dfrac{x}{y}+\dfrac{y}{x}\ge2\sqrt{\dfrac{xy}{xy}}=2\sqrt{1}=2\)

\(\Leftrightarrow4\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\ge8\)

\(8xy+\dfrac{2}{xy}\ge2\sqrt{\dfrac{8xy.2}{xy}}=2\sqrt{16}=8\)

\(\Rightarrow\left(2x+\dfrac{1}{x}\right)^2+\left(2y+\dfrac{1}{y}\right)^2\ge16\)

Dấu "=" xảy ra khi:

\(\dfrac{x}{y}=\dfrac{y}{x}\)

\(\Leftrightarrow x^2=y^2\)

\(\Leftrightarrow x=y=\dfrac{1}{2}\)

Vậy \(MinP=16\) khi \(x=y=\dfrac{1}{2}\)

Answer 2

[Nguyễn Ngọc Huy Toàn]

xin giải bài cúi thoai nhé:v