x² - xy + y²

= x² - 2x\(\dfrac{y}{2}\)+ \(\dfrac{y^2}{4}-\dfrac{y^2}{4}+y^2\)

= ( x - \(\dfrac{y}{2}\))² + \(\dfrac{3y^2}{4}\)

Do : ( x - \(\dfrac{y}{2}\))² lớn hơn hoặc bằng 0 với mọi x

=> ( x - \(\dfrac{y}{2}\))² + \(\dfrac{3y^2}{4}\)lớn hơn hoặc bằng \(\dfrac{3y^2}{4}\) với mọi x và lớn hơn hoặc bằng 0

Dấu " =" xảy ra khi \(\dfrac{3y^2}{4}\)= 0 => y = 0

1: \(A=\dfrac{x-2\sqrt{xy}+y}{x-y}=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}=\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)

2: Thay \(x=3+2\sqrt{2}\) và \(y=3-2\sqrt{2}\) vào A, ta được:

\(A=\dfrac{\sqrt{2}+1-\sqrt{2}+1}{\sqrt{2}+1+\sqrt{2}-1}=\dfrac{2}{2\sqrt{2}}=\dfrac{\sqrt{2}}{2}\)

\(x^2+xy+y^2+1=\left(x^2+xy+\frac{1}{4}y^2\right)+\frac{3}{4}y^2+1=\left(x+\frac{1}{2}y\right)^2+\frac{3}{4}y^2+1>0\forall x;y\)

x² + xy + y²  + 1 = (x² + 2.x. \(\frac{y}{2}\) + (\(\frac{y}{2}\))² ) + \(\frac{3y^2}{4}\) + 1 = (x + \(\frac{y}{2}\))² + \(\frac{3y^2}{4}\) + 1 \(\ge\) 0 + 0 + 1 = 1> 0 với mọi x; y

Ta có:

x²+xy+y²+1=x²+xy+1/4.y²+3/4.y²+1=(x+1/2.y)²+3/4.y²+1

Mà (x+1/2.y)² \(\ge\)0

3/4.y²>=0

1>0

Suy ra (x+1/2.y)²+3/4.y²+1>0

Hay x²+xy+y²+1>0(đpcm)

≥0

3/4.y²>=0

1>0

Suy ra (x+1/2.y)²+3/4.y²+1>0

Hay x²+xy+y²+1>0(đpcm)

Ta có :x+y\(\ge2\sqrt{xy}\)

\(\Leftrightarrow\left(\sqrt{x}\right)^2+\left(\sqrt{y}\right)^2+2\sqrt{x}\sqrt{y}\ge0\)

\(\Leftrightarrow\left(\sqrt{x}-\sqrt{y}\right)^2\ge0\)(luôn đúng với mọi x,y\(\ge0\))

Dấu"+" xảy ra khi:\(\sqrt{x}=\sqrt{y}\Leftrightarrow x=y\)

Vậy với mọi x,y\(\ge0\) thì x+y\(\ge2\sqrt{xy}\)

đong 2 bạn đổi lại dấu +\(2\sqrt{xy}\) thành -\(2\sqrt{xy}\) giùm mình

a/ \(x^2-6x+10=x^2-2.x.3+3^2+1=\left(x-3\right)^2+1\)

Với mọi x ta có :

\(\left(x-3\right)^2\ge0\)

\(\Leftrightarrow\left(x-3\right)^2+1>0\)

\(\Leftrightarrow x^2-6x+10>0\)

b/ \(x^2-4x+7=x^2-2.x.2+2^2+3=\left(x-2\right)^2+3\)

Với mọi x ta có :

\(\left(x-2\right)^2\ge0\)

\(\Leftrightarrow\left(x-2\right)^2+3\ge3\)

\(\Leftrightarrow x^2-4x+7\ge3\left(đpcm\right)\)

c/ \(x^2+x+1=x^2+2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

Với mọi x ta có :

\(\left(x+\dfrac{1}{2}\right)^2\ge0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

\(\Leftrightarrow x^2+x+1>0\left(đpcm\right)\)

d/ \(x^2+y^2+4x-6y+15=\left(x^2+4x+2^2\right)+\left(y^2-6y+3^2\right)+2=\left(x+2\right)^2+\left(y-3\right)^2+2\)

Với mọi x,y ta có :

\(\left\{{}\begin{matrix}\left(x+2\right)^2\ge0\\\left(y-3\right)^2\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left(x+2\right)^2+\left(y-3\right)^2\ge0\)

\(\Leftrightarrow\left(x+2\right)^2+\left(y-3\right)^2+2\ge0\)

\(\Leftrightarrow x^2+y^2+4x-6y+15>0\left(đpcm\right)\)

2/ Ta có :

\(\left(a+b\right)^2-4ab=a^2+2ab+b^2-4ab=a^2-2ab+b^2=\left(a-b\right)^2\)

Vậy \(\left(a-b\right)^2=\left(a+b\right)^2-4ab\left(đpcm\right)\)

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

Mà \(x+y=7;xy=-3\)

\(\Leftrightarrow x^2+y^2=7^2-2.\left(-3\right)=49+6=55\)

\(A=\dfrac{x^3+y^3+4}{xy+1}\ge\dfrac{x^3+y^3+4}{\dfrac{x^2+y^2}{2}+1}=\dfrac{x^3+y^3+4}{2}=\dfrac{\dfrac{1}{2}\left(x^3+x^3+1\right)+\dfrac{1}{2}\left(y^3+y^3+1\right)+3}{2}\)

\(\ge\dfrac{\dfrac{3}{2}\left(x^2+y^2\right)+3}{2}=3\)

\(A_{min}=3\) khi \(x=y=1\)

Do \(x^2+y^2=2\Rightarrow\left\{{}\begin{matrix}x\le\sqrt{2}\\y\le\sqrt{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x^3\le\sqrt{2}x^2\\y^3\le\sqrt{2}y^2\end{matrix}\right.\)

\(\Rightarrow A\le\dfrac{\sqrt{2}\left(x^2+y^2\right)+4}{xy+1}=\dfrac{4+2\sqrt{2}}{xy+1}\le\dfrac{4+2\sqrt{2}}{1}=4+2\sqrt{2}\)

\(A_{max}=4+2\sqrt{2}\) khi \(\left(x;y\right)=\left(0;\sqrt{2}\right);\left(\sqrt{2};0\right)\)