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

1) \(\frac{1}{2}=\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)\(\Leftrightarrow\)\(x+y\ge8\)

\(\frac{1}{2}=\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}\)\(\Leftrightarrow\)\(xy=2\left(x+y\right)\ge16\)

\(A=\sqrt{x}+\sqrt{y}\ge2\sqrt[4]{xy}\ge2\sqrt[4]{16}=4\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=4\)

2) \(B=\sqrt{3x-5}+\sqrt{7-3x}\ge\sqrt{3x-5+7-3x}=\sqrt{2}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{7}{3}\end{cases}}\)

\(B=\sqrt{3x-5}+\sqrt{7-3x}\le\frac{3x-5+1+7-3x+1}{2}=2\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=2\)

1) 

a) \(2x^2-12x+18+2xy-6y\)

\(=2x^2-6x-6x+18+2xy-6y\)

\(=\left(2xy+2x^2-6x\right)-\left(6y+6x-18\right)\)

\(=x\left(2y+2x-6\right)-3\left(2y+2x-6\right)\)

\(=\left(x-3\right)\left(2y+2x-6\right)\)

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

b) \(x^2+4x-4y^2+8y\)

\(=x^2+4x-4y^2+8y+2xy-2xy\)

\(=\left(-4y^2+2xy+8y\right)+\left(-2xy+x^2+4x\right)\)

\(=2y\left(-2y+x+4\right)+x\left(-2y+x+4\right)\)

\(=\left(2y+x\right)\left(-2y+x+4\right)\)

2)  \(5x^3-3x^2+10x-6=0\)

\(\Leftrightarrow x^2\left(5x-3\right)+2\left(5x-3\right)=0\Leftrightarrow\left(x^2+2\right)\left(5x-3\right)=0\)

Mà \(x^2+2>0\Rightarrow5x-3=0\Rightarrow x=\frac{3}{5}\)

\(x^2+y^2-2x+4y+5=0\)

\(\Leftrightarrow x^2+y^2-2x+4y+4+1=0\)

\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2+\left(y+2\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x-1=0\\y+2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\)

3)\(P\left(x\right)=x^2+y^2-2x+6y+12\)

\(P\left(x\right)=x^2+y^2-2x+6y+1+9+2\)

\(=\left(x^2-2x+1\right)+\left(y^2+6y+9\right)+2\)

\(=\left(x-1\right)^2+\left(y+3\right)^2+2\ge2\)

Vậy \(P\left(x\right)_{min}=2\Leftrightarrow\hept{\begin{cases}x-1=0\\y+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-3\end{cases}}\)

Bài làm

a) 2x² - 12x + 18 + 2xy - 6y

= 2x² - 6x - 6x + 18 + 2xy - 6y 

= ( 2xy + 2x² - 6x ) - ( 6y + 6x - 18 )

= 2x( y + x - 3 ) - 6( y + x - 3 )

= ( 2x - 6 ) ( y + x - 3 )

# Học tốt #

Akai Haruma Bonking

Vì x>8y>0 áp dụng BĐT Cauchy cho 3 số dương

\(P=x+\dfrac{1}{y\left(x-8y\right)}=\left(x-8y\right)+8y+\dfrac{1}{y\left(x-8y\right)}\ge3\sqrt[3]{\left(x-8y\right).8y.\dfrac{1}{y\left(x-8y\right)}}=3\sqrt[3]{8}=6\)

Đẳng thức xảy ra \(\Leftrightarrow x-8y=8y=\dfrac{1}{y\left(x-8y\right)}\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=\dfrac{1}{4}\end{matrix}\right.\)

vì x>8y>0 nên x-8y>0

Ta có : P=\(x+\dfrac{1}{y\left(x-8y\right)}\)= x-8y+8y+ \(\dfrac{1}{y\left(x-8y\right)}\)

ÁP dụng BĐT côsy cho 3 số dương dạng a+b+c\(\ge\) 3\(\sqrt[3]{abc}\) ta đc:

P \(\ge\)3\(\sqrt[3]{\left(x-8y\right).8y.\dfrac{1}{y\left(x-8y\right)}}\)\(\ge\) 3.2=6

Vậy Pmin=6 khi đó dấu "=" xẫy ra khi : \(x-8y=8y=\dfrac{1}{y\left(x-8y\right)}\)

<=> \(\left\{{}\begin{matrix}x=4\\y=\dfrac{1}{4}\end{matrix}\right.\)