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

=\(\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}+1\) \(\ge\)0

vậy....

b

\(x\left(x-1\right)+y\left(y-3\right)+10\)

\(=x^2-x+y^2-3+10\)

\(=\left(x^2-2\cdot\frac{1}{2}x+\frac{1}{4}\right)+\left(y^2-2\cdot\frac{3}{2}y+\frac{9}{4}\right)+\frac{15}{2}\)

\(=\left(x-\frac{1}{2}\right)^2+\left(y-\frac{3}{2}\right)^2+\frac{15}{2}\ge\frac{15}{2}\) 

x( x - 1 ) + y( y - 3 ) + 10

= x² - x + y² - 3y + 10

= x² - x + y² - 3y + 1/4 + 9/4 + 15/2

= ( x² - x + 1/4 ) + ( y² - 3y + 9/4 ) + 15/2

= ( x - 1/2 )² + ( y - 3/2 )² + 15/2 ≥ 15/2 > 0 ∀ x, y ( đpcm )

Giải:

a) \(x^2+xy+y^2+1\)

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

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

\(=\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}+1\ge1>0;\forall x\)

Vậy ...

(x-3)² +y² +1 >0

với mọi x;y

x² - 6x + 10 + y²

= x² - 6x + 9 + 1 + y²

= x² - 2.x.3 + 3² + 1 + y²

= (x - 3)² + 1 + y²

Ta có: (x - 3)² + 1 \(\ge\)1 và y² \(\ge\)0

=> (x - 3)² + 1 + y² \(\ge\)1 > 0 với mọi x, y (đpcm).

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

b/ \(x^2+5y^2+2x-4xy-10y+14\)

\(=\left(x^2-4xy+4y^2\right)+2\left(x-2y\right)+1+\left(y^2-6y+9\right)+4\)

\(=\left(x-2y\right)^2+2\left(x-2y\right)+1+\left(y-3\right)^2+4\)

\(=\left(x-2y+1\right)^2+\left(y-3\right)^2+4>0\)

a)  \(A=x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)       với mọi x

b)   \(B=x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\) với mọi x

c)  \(x^2+xy+y^2+1=\left(x+\frac{1}{2}y\right)^2+\frac{3}{4}y^2+1>0\)  với mọi x,y

d)  bạn kiểm tra lại đề câu d) nhé:

 \(x^2+4y^2+z^2-2x-6y+8z+15\)

\(=\left(x-1\right)^2+\left(2y-\frac{6}{4}\right)^2+\left(z+4\right)^2-\frac{13}{4}\)

Đề câu d đúng mà!