a) Ta có:

\(x^2-x+1\)

\(=x^2-2\cdot\dfrac{1}{2}\cdot x+\dfrac{1}{4}+\dfrac{3}{4}\)

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

Mà: \(\left(x-\dfrac{1}{2}\right)^2\ge0\) và \(\dfrac{3}{4}>0\) nên

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

\(\Rightarrow x^2-x+1>0\forall x\)

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 :  \(\hept{\begin{cases}x+2=0\\y-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-2\\y=3\end{cases}}\)

\(x-x^2-1=-\left(x^2-x+\frac{1}{4}\right)-\frac{3}{4}=-\left(x-\frac{1}{2}\right)^2-\frac{3}{4}< 0\) \(\forall x\)

Do x-x\(^2\le0\)\(\left(x^2\ge x\right)\)

---> \(x-x^2-1< 0\forall x\)

Chúc bạn học tốt

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

Đặt x-y=t 

ta có: \(t^2+t+1=\left(t+\frac{1}{2}\right)^2+\frac{3}{4}>0,\forall t\)

=> \(x^2+y^2-2xy+x-y+1>0,\forall x,y\)

a) \(x^2-5x+8=\left(x^2-5x+6,25\right)+1,75=\left(x-2,5\right)^2+1,75\ge1,75>0\rightarrowđpcm\)

b) \(-4x^2-4x-2=-\left(4x^2+4x+1\right)-1=-\left(2x+1\right)^2-1\le-1< 0\rightarrowđpcm\)

A =x² -5x +8 >0 với mọi x

= x²-5x+\(\dfrac{25}{4}+\dfrac{7}{4}\)

=\(\left(x-\dfrac{5}{2}\right)^2+\dfrac{7}{4}\)

do \(\left(x-\dfrac{5}{2}\right)^2\ge0\forall x\)

=> \(\left(x-\dfrac{5}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)

=> A luôn lớn hơn 0 vs mọi x

B= -4x² -4x-2 < 0 với mọi x

=-(4x²+4x+2)

=-4x²-4x-1-1

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

=-\(\left[4\left(x^2+x+\dfrac{1}{4}\right)+1\right]\)

= -\(\left[4\left(x+\dfrac{1}{2}\right)^2+1\right]\)

=-4\(\left(x+\dfrac{1}{2}\right)^2-1\)

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

=> -4 \(\left(x+\dfrac{1}{2}\right)^2\le0\)

=> \(-4\left(x+\dfrac{1}{2}\right)^2-1\le-1\)

vậy B luôn nhỏ hơn 0 vs mọi x

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

Do \(\left\{{}\begin{matrix}\left(x+1\right)^2\ge0\\y^2\ge0\end{matrix}\right.\) \(\forall x;y\)

\(\Rightarrow\left(x+1\right)^2+y^2\ge0\)

Dấu "=" vẫn xảy ra tại \(\left\{{}\begin{matrix}x=-1\\y=0\end{matrix}\right.\)

\(x^4+2ax^2+a^2+2x^2+2a+1\)

\(=\left[\left(x^2\right)^2+2\cdot x^2\cdot a+a^2\right]+\left(2a+2x^2\right)+1\)

\(=\left(x^2+a\right)^2+2\left(x^2+a\right)+1\)=> hằng đẳng thức số 1

\(=\left(x^2+a+1\right)^2\ge0\forall a;x\)