\(M=\dfrac{2x+y}{xy}+\dfrac{3}{2x+y}=\dfrac{2x+y}{2}+\dfrac{3}{2x+y}=\dfrac{3\left(2x+y\right)}{16}+\dfrac{3}{2x+y}+\dfrac{5}{16}\left(2x+y\right)\ge2\sqrt{\dfrac{3}{16}.3}+\dfrac{5}{16}.2\sqrt{2xy}=\dfrac{3}{2}+\dfrac{5}{4}=\dfrac{11}{4}\).

Đẳng thức xảy ra khi x = 1; y = 2.

\(M=\dfrac{2x+y}{xy}+\dfrac{3}{2x+y}=\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\)

\(M=\dfrac{3\left(2x+y\right)}{16}+\dfrac{3}{2x+y}+\dfrac{5\left(2x+y\right)}{16}\ge2\sqrt{\dfrac{9\left(2x+y\right)}{16\left(2x+y\right)}}+\dfrac{5}{16}.2\sqrt{2xy}=\dfrac{11}{4}\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(1;2\right)\)

Ta có: \(M=\dfrac{2x+y}{xy}+\dfrac{3}{2x+y}=\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\)

\(=\left(\dfrac{3}{8}\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\right)+\dfrac{5}{8}\dfrac{2x+y}{2}\)

Có: \(\dfrac{3}{8}\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\ge2\sqrt{\dfrac{3}{8}\dfrac{2x+y}{2}\dfrac{3}{2x+y}}=\dfrac{3}{2}\)

Dấu '=' xảy ra <=> \(\dfrac{3}{8}\dfrac{2x+y}{2}=\dfrac{3}{2x+y}\)

Có: \(\dfrac{5}{8}\dfrac{2x+y}{2}\ge\dfrac{5}{8}\sqrt{2xy}=\dfrac{5}{4}\)

Dấu '=' xảy ra <=> 2x=y và xy=2

\(\Rightarrow M\ge\dfrac{3}{2}+\dfrac{5}{4}=\dfrac{11}{4}\)

Dấu '=' xảy ra <=> x=1, y=2

Vậy GTNN của M là 11/4 <=> x=1;y=2

Ta có:

\(M=\dfrac{2x+y}{xx}+\dfrac{3}{2x+y}=\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\)

\(=\left(\dfrac{3}{8}\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\right)+\dfrac{5}{8}\dfrac{2x+y}{2}\)

Có: \(\dfrac{3}{8}\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\ge2\sqrt{\dfrac{3}{8}\dfrac{2x+y}{2}\dfrac{3}{2x+y}}=\dfrac{3}{2}\)

Dấu '=' xảy ra \(\Leftrightarrow\dfrac{3}{8}\dfrac{2x+y}{2}=\dfrac{3}{2x+y}\)

Có: \(\dfrac{5}{8}\dfrac{2x+y}{2}\ge\dfrac{5}{8}\sqrt{2xy}=\dfrac{5}{4}\)

Dấu '=' xảy ra \(\Leftrightarrow2x=y,xy=2\)

\(\Rightarrow M\ge\dfrac{3}{2}+\dfrac{5}{4}=\dfrac{11}{4}\)

Dấu '=' xảy ra \(\Leftrightarrow x=1,y=2\)

Vậy GTNN của M là \(\dfrac{11}{4}\Leftrightarrow x=1,y=2\)

Sao lúc thì $x,y,z$ lúc thì $a,b$ vậy bạn? Bạn coi lại đề.

We have : \(A=x+y+\dfrac{1}{2x}+\dfrac{2}{y}=\dfrac{x+y}{2}+\left(\dfrac{y}{2}+\dfrac{2}{y}\right)+\left(\dfrac{1}{2x}+\dfrac{x}{2}\right)\)

\(Applying\) C-S we have : \(\dfrac{y}{2}+\dfrac{2}{y}\ge2;\dfrac{1}{2x}+\dfrac{x}{2}\ge1\)

x + y \(\ge3\)  \(\Rightarrow\dfrac{x+y}{2}\ge\dfrac{3}{2}\)

So : \(A\ge\dfrac{3}{2}+2+1=\dfrac{9}{2}\)

" = " \(\Leftrightarrow x=1;y=2\)

kẻ lười biếng nạp card, đi ô tô

By Titu's Lemma we easy have:

\(D=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)

\(\ge\frac{\left(x+y+\frac{1}{x}+\frac{1}{y}\right)^2}{2}\)

\(\ge\frac{\left(x+y+\frac{4}{x+y}\right)^2}{2}\)

\(=\frac{17}{4}\)

Mk xin b2 nha!

\(P=\frac{1}{x^2+y^2}+\frac{1}{xy}+4xy=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}+4xy\)

\(\ge\frac{\left(1+1\right)^2}{x^2+y^2+2xy}+\left(4xy+\frac{1}{4xy}\right)+\frac{1}{4xy}\)

\(\ge\frac{4}{\left(x+y\right)^2}+2\sqrt{4xy.\frac{1}{4xy}}+\frac{1}{\left(x+y\right)^2}\)

\(\ge\frac{4}{1^2}+2+\frac{1}{1^2}=4+2+1=7\)

Dấu "=" xảy ra khi: \(x=y=\frac{1}{2}\)

1) có \(2y\le\frac{\left(x+y\right)^2}{4}=\frac{1}{4}\)

\(\Rightarrow\left(\sqrt{xy}+\frac{1}{4\sqrt{xy}}\right)^2+\frac{15}{16xy}+\frac{1}{2}\ge\frac{15}{16}\cdot4+\frac{1}{2}=\frac{17}{4}\)

Dấu "=" xảy ra <=> \(x=y=\frac{1}{2}\)

\(x^{2019}-y^{2019}+2\left(x-y\right)=0\)

<=> \(\left(x-y\right)\left(x^{2018}+x^{2017}y+...+xy^{2017}+y^{2018}\right)+2\left(x-y\right)=0\)

<=> \(\left(x-y\right)\left(x^{2018}+x^{2017}y+...+xy^{2017}+y^{2018}+2\right)=0\)(1)

Có: \(x^{2018}+x^{2017}y+...+xy^{2017}+y^{2018}+2>0\)mọi x, y.

(1) <=> \(x-y=0\)

<=> x = y

Thế vào P ta có:

\(P=x^4-2x^2+2=\left(x^2-1\right)^2+1\ge1\)

"=" xảy ra <=> \(y=x=\pm1\)

Vậy min P =1 khi và chỉ khi x = y =1 hoặc x = y =-1.

\(2.\) Ba số dương a,b,c chứ?

câu 1 bn bình phương vế 2x+y đi nhé!

Áp dụng bất đẳng thức Bunyakovsky cho hai bộ số  \(\left[\left(\sqrt{2}\right)^2+1\right]\) và  \(\left(2x^2+y^2\right)\), ta được:

\(\left[\left(\sqrt{2}\right)^2+1^2\right]\left(2x^2+y^2\right)\ge\left(\sqrt{2}.\sqrt{2}x+1.y\right)^2\)

\(\Rightarrow\)  \(3\left(2x^2+y^2\right)\ge\left(2x+y\right)^2=3^2=9\)

\(\Rightarrow\)  \(2x^2+y^2\ge3\)

Dấu  \("="\)  xảy ra  \(\Leftrightarrow\)  \(\frac{\sqrt{2}}{\sqrt{2}x}=\frac{1}{y}\)  \(\Leftrightarrow\)  \(x=y=1\)