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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

Đặt \(\sqrt{x^2+4x+5}=t\Rightarrow t\in\left[\sqrt{5};\sqrt{17}\right]\)

\(\Rightarrow y=f\left(t\right)=t^2-2t+7\)

\(-\dfrac{b}{2a}=1\notin\left[\sqrt{5};\sqrt{17}\right]\)

\(f\left(\sqrt{5}\right)=10+4\sqrt{5}\) ; \(f\left(\sqrt{17}\right)=22+4\sqrt{17}\)

\(\Rightarrow y_{min}=10+4\sqrt{5}\) ; \(y_{max}=22+4\sqrt{17}\)

a) \(A=5+\sqrt{-4x^2-4x}\) 

\(A==5+\sqrt{-4x\left(x+1\right)}\)

Có: \(-4x\left(x+1\right)\le0\)

\(\Rightarrow\sqrt{-4x\left(x+1\right)}=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

Vậy: \(Max_A=5\) tại \(\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

b) \(B=\sqrt{x-2}+\sqrt{4-x}\)

ĐKXĐ: \(\hept{\begin{cases}x\ge2\\x\le4\end{cases}}\Rightarrow x\in\left\{2;3;4\right\}\)

Thay \(x=2\Rightarrow\sqrt{2-2}+\sqrt{4-2}=\sqrt{2}\)

Thay \(x=3\Rightarrow\sqrt{3-1}+\sqrt{4-3}=2\)

Thay \(x=4\Rightarrow\sqrt{4-2}+\sqrt{4-4}=\sqrt{2}\)

Vậy: \(Max_B=2\) tại \(x=3\)

Bài 2:

a)\(A=\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}+\sqrt{x^2-6x+9}\)

\(=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}+\sqrt{\left(x-3\right)^2}\)

\(=\left|x-1\right|+\left|x-2\right|+\left|x-3\right|\)

\(\ge x-1+0+3-x=2\)

Dấu = khi \(\hept{\begin{cases}x-1\ge0\\x-2=0\\x-3\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge1\\x=2\\x\le3\end{cases}}\Leftrightarrow x=2\)

Vậy MinA=2 khi x=2

bÀI 2 PHẦN b bạn nhân 2 ngoặc 1 r` đặt ẩn là t =>min...

\(E=x+\sqrt{-x^2-2x+3}\)

\(\Leftrightarrow2\sqrt{2}E=2\sqrt{2}x+2\sqrt{2}\sqrt{-x^2-2x+3}\)

\(=\left(-\left(-x^2-2x+3\right)+2\sqrt{2}\sqrt{-x^2-2x+3}-2\right)+\left(-x^2+2.x.\left(\sqrt{2}-1\right)-\left(3-2\sqrt{2}\right)\right)+8-2\sqrt{2}\)

\(=-\left(\sqrt{-x^2-2x+3}-\sqrt{2}\right)^2-\left(x-\left(\sqrt{2}-1\right)\right)^2+8-2\sqrt{2}\le8-2\sqrt{2}\)

\(\Rightarrow E\le2\sqrt{2}-1\)

Dấu = xảy ra khi \(x=\sqrt{2}-1\)

\(C=3\sqrt{x}+2\sqrt{1-4x}\)

\(\Leftrightarrow2C=6\sqrt{x}+4\sqrt{1-4x}\)

\(=\left(-10x+6\sqrt{x}-\frac{9}{10}\right)+\left(-\frac{5}{2}\left(1-4x\right)+4\sqrt{1-4x}-\frac{8}{5}\right)+5\)

\(=-10\left(x-2.\sqrt{x}.\frac{3}{10}+\frac{9}{100}\right)-\frac{5}{2}\left(\sqrt{1-4x}-2.\left(1-4x\right).\frac{4}{5}+\frac{16}{25}\right)+5\)

\(=-10\left(\sqrt{x}-\frac{3}{10}\right)^2-\frac{5}{2}\left(\sqrt{1-4x}-\frac{4}{5}\right)^2+5\le5\)

\(\Rightarrow C\le\frac{5}{2}\)

Dấu = xảy ra khi \(x=\frac{9}{100}\)

a)\(y=\sqrt{-x^2+2x-1+2}=\sqrt{-\left(x^2-2x+1\right)+2}\)

\(=\sqrt{-\left(x-1\right)^2+2}\)

Dễ thấy: \(-\left(x-1\right)^2\le0\Rightarrow-\left(x-1\right)^2+2\le2\)

\(\Rightarrow y=\sqrt{-\left(x-1\right)^2+2}\le\sqrt{2}\)

Đẳng thức xảy ra khi \(x=1\)

b)\(y=2-\sqrt{4x^2-4x+1}\)

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

Dễ thấy: \(\sqrt{\left(2x-1\right)^2}\ge0\Rightarrow-\sqrt{\left(2x-1\right)^2}\le0\)

\(y=2-\sqrt{4x^2-4x+1}\le2\)

Đẳng thức xảy ra khi \(x=\frac{1}{2}\)

Akai Haruma Bonking