\(VT=x^2+xy+\frac{y^2}{4}+\frac{3y^2}{4}+1=\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}+1\ge1>0\)

Xem lại đề nhé

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

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:

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

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

a: \(=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}=\sqrt{ab}-\sqrt{ab}=0\)

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

\(=\sqrt{x}-2\sqrt{y}+\sqrt{y}=\sqrt{x}-\sqrt{y}\)

c: \(=\sqrt{x}+2-\dfrac{x-4}{\sqrt{x}-2}\)

\(=\sqrt{x}+2-\sqrt{x}-2=0\)

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)

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 \(x+y=7;xy=-3\)

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

2.

Ta có hằng đẳng thức : \(\left(a-b\right)^2=a^2-2ab+b^2\left(1\right)\)

Lại có  \(\left(a+b\right)^2=a^2+2ab+b^2\)

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

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

Từ (1) và (2)  \(\Rightarrow\left(a-b\right)^2=\left(a+b\right)^2-4ab\)( đpcm )

3.

Ta có hằng đẳng thức  \(\left(x+y\right)^2=x^2+2xy+y^2\)

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

Thay  \(x+y=7\)và  \(xy=-3\)vào ta được :

\(x^2+y^2=7^2-2\left(-3\right)\)

\(\Leftrightarrow x^2+y^2=49+6=55\)

Vậy ...

1. 

a) Đặt  \(A=x^2-6x+10\)

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

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

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

\(\Rightarrow A\ge1>0\)

Vậy ...

b) Đặt \(B=x^2-4x+7\)

\(B=\left(x^2-4x+4\right)+3\)

\(B=\left(x-2\right)^2+3\)

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

\(\Rightarrow B\ge3\)

Vậy ...

1.

c) Đặt \(C=x^2+x+1\)

\(C=\left(x^2+x+\frac{1}{4}\right)+\frac{3}{4}\)

\(C=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)

Mà  \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow C\ge\frac{3}{4}>0\)

Vậy ...

d) Đặt \(D=x^2+y^2+4x-6y+15\)

\(D=\left(x^2+4x+4\right)+\left(y^2-6y+9\right)\)

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

Mà  \(\left(x+2\right)^2\ge0\forall x\)

      \(\left(y-3\right)^2\ge0\forall y\)

\(\Rightarrow D\ge0\)

Dấu "=" xảy ra khi :  \(\hept{\begin{cases}x+2=0\\y-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-2\\y=3\end{cases}}\)