\(A=\frac{2x+3y}{2x+y+2}\)

\(\Leftrightarrow A\left(2x+y+2\right)=2x+3y\)

\(\Leftrightarrow2A=2x\left(1-A\right)+y\left(3-A\right)\)

\(\Leftrightarrow\left(2A\right)^2=\left(2x\left(1-A\right)+y\left(3-A\right)\right)^2\le\left(4x^2+y^2\right)\left(\left(1-A\right)^2+\left(3-A\right)^2\right)\)

\(\Leftrightarrow\left(2A\right)^2\le\left(\left(1-A\right)^2+\left(3-A\right)^2\right)\)

\(\Leftrightarrow-5\le A\le1\)

Chết em ấn nhầm nút rồi

ĐKXĐ: \(x\ge1;y\ge25\)

\(D=\frac{1}{x}\sqrt{\frac{x-1}{\left(x-2\right)^2+25}}+\frac{1}{y}\sqrt{\frac{y-25}{\left(y-50\right)^2+1}}\)

Vì x>=1,y>=25 => x-1>=0,y-25>=0 

=> D >= 0

Dấu "=" xảy ra <=> x=1,y=25

Vậy MinD=0 khi x=1,y=25

Ta có: \(\left(x-2\right)^2+25\ge25;\left(y-50\right)^2+1\ge1\)

=>\(\frac{1}{x}\sqrt{\frac{x-1}{\left(x-2\right)^2+25}}\le\frac{1}{x}\sqrt{\frac{x-1}{25}};\frac{1}{y}\sqrt{\frac{y-25}{\left(y-50\right)^2+1}}\le\frac{1}{y}\sqrt{y-25}\)

=>\(D\le\frac{1}{x}\sqrt{\frac{x-1}{25}}+\frac{1}{y}\sqrt{y-25}\)

Vì x>=1 => x-1>=0. Áp dụng bđt cosi với 2 số dương x-1 và 1 ta có:

\(\sqrt{x-1}=\sqrt{\left(x-1\right).1}\le\frac{x-1+1}{2}=\frac{x}{2}\)

=>\(\frac{1}{x}\sqrt{\frac{x-1}{25}}\le\frac{1}{x}\cdot\frac{x}{2}\cdot\frac{1}{\sqrt{25}}=\frac{1}{10}\)

Vì y>=25 => y-25>=0. ÁP dụng bđt cô si cho 2 số dương 25 và y-25 ta có:

\(\sqrt{y-25}=\frac{\sqrt{25\left(y-25\right)}}{5}\le\frac{25+y-25}{2.5}=\frac{y}{10}\)

=>\(\frac{1}{y}\sqrt{y-25}=\frac{1}{y}\cdot\frac{y}{10}=\frac{1}{10}\)

Suy ra \(D\le\frac{1}{10}+\frac{1}{10}=\frac{1}{5}\)

Dấu "=" xảy ra <=> x=2,y=50

Vậy MaxD = 1/5 khi x=2,y=50

Akai Haruma Bonking