Sai rồi                               

Ta có : \(2x^2-6x+15\)

\(=2\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}\right)+\frac{21}{2}\)

\(=2\left(x-\frac{3}{2}\right)^2+\frac{21}{2}>0\)

Nhi Lô kcj đâu bạn tick mình đúng nha !

( x - 2 )( x - 4 ) + 3

<=> x² - 6x + 8 + 3

<=> ( x² - 6x + 9 ) + 2

<=> ( x - 3 )² + 2 ≥ 2 > 0 ∀ x ( đpcm )

Đề sai! có vô số số thực x mà \(x^2-4\) không phải là số dương, cụ thể là: \(x^2-4\le0,\forall x\in\left[-2;2\right]\)

\(a.A=x^2-2x+5=x^2-2x+1+4=\left(x-1\right)^2+4>0\text{∀}x\)

\(b.B=x^2-2x+9y^2-6y+3=x^2-2x+1+9y^2-6y+1+1=\left(x-1\right)^2+\left(3y-1\right)^2+1>0\text{∀}x,y\)

a. \(A=x^2-2x+5=x^2-2x+1+4=\left(x-1\right)^2+4>0\forall x\)

b. \(B=x^2-2x+9y^2-6y+3=\left(x^2-2x+1\right)+\left(9y^2-6y+1\right)+1=\left(x-1\right)^2+\left(3y-1\right)^2+1>0\forall x;y\)

a ) Ta có : \(2x-x^2-3=-\left(x^2-2x+3\right)=-\left(x^2-2x+1+2\right)=-\left[\left(x-1\right)^2+2\right]=-\left(x-1\right)^2-2\)

Do \(-\left(x-1\right)^2\le0\forall x\)

\(\Rightarrow-\left(x-1\right)^2-2\le-2< 0\forall x\)

\(\left(đpcm\right)\)

b ) Đề thiếu

:D

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

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