\(x^2-4x+9y^2+6y+10\\ =\left(x^2-4x+4\right)+\left(9y^2+6y+1\right)+5\\ =\left(x-2\right)^2+\left(3y+1\right)^2+5\ge5>0\)

1.

\(x^2\)+\(y^2\)+2y-6x+10=0

=> \(x^2\)-6x+9 +\(y^2\)+2y+1=0

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

pt vô nghiệm

4.

=> \(x^2\)+8x+16+(3y)\(^2\)-2.3.2y+4=0

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

pt vô nghiệm

3.

=> (3y)\(^2\)+2.3y+1+\(x^2\)+4x+4

=> (3y+1)\(^2\)+(x+2)\(^2\)=0

pt vô nghiệm

a) \(-2x^2+2x+1>0\)

   \(-\left(2x^2-2x-1\right)>0\)

nhân 2 vế với (-1)=> đổi dấu sao sánh

\(\Leftrightarrow2x^2-2x-1< 0\)

\(\Leftrightarrow x^2-x-\frac{1}{2}< 0\)

\(\Leftrightarrow x^2-2.\frac{1}{2}x+\left(\frac{1}{2}\right)^2-\frac{1}{4}-\frac{1}{2}< 0\)

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

ta có \(\left(x-\frac{1}{2}\right)^2\ge0\)với mọi \(x\)

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

b) \(9x^2-6x+2>0\)

<=> \(\left(3x\right)^2-2.3.x+1-1+2>0\)

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

vì \(\left(3x-1\right)^2\ge0\)với mọi \(x\)=> (1)  luôn đúng     ( bạn lí giải tương tự như trên nha)

c)\(-4x^2-4x-2< 0\)

\(\Leftrightarrow-\left(4x^2+4x+2\right)< 0\)

nhân 2 vế với (-1)=> đổi dấu so sánh 

\(4x^2+4x+2>0\)

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

lí giải tương tự như trên

=> đpcm

Câu a sai đề rồi cậu ơi

câu a sai đề rồi bn ơi

Bài 1

\(A=x^2-6x+15=x^2-2.3.x+9+6=\left(x-3\right)^2+6>0\forall x\)

\(B=4x^2+4x+7=\left(2x\right)^2+2.2.x+1+6=\left(2x+1\right)^2+6>0\forall x\)

Bài 2

\(A=-9x^2+6x-2021=-\left(9x^2-6x+2021\right)=-\left[\left(3x-1\right)^2+2020\right]=-\left(3x-1\right)^2-2020< 0\forall x\)

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

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

a) x^2 + 2x - 35 = 0

<=> (x - 5)(x + 7) = 0

<=> x = 5 hoặc x = - 7

b) 4x^2 - 12x - 27 = 0

<=> (2x - 9)(2x + 3) = 0

<=> x = 4,5 hoặc x = - 1,5

c) 9x^2 + 24x + 7 = 0

<=> (3x + 1)(3x + 7) = 0

<=> x = - 1/3 hoặc x = - 7/3

d) x^2 + y^2 - 4x + 6y + 13 = 0

<=> (x - 2)^2 + (y + 3)^2 = 0

<=> x = 2 và y = - 3

e) 25x^2 - 10x - 24 = 0

<=> (5x - 6)(5x + 4) = 0

<=> x = 1,2 hoặc x = - 0,8

a ) \(4x^2+2x+1=\left(2x\right)^2+2\cdot2x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(2x+\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)

b ) \(x^2+3x+4=\left(x^2+2\cdot\frac{3}{2}\cdot x+\frac{9}{4}\right)+\frac{7}{4}=\left(x+\frac{3}{2}\right)^2+\frac{7}{4}>0\forall x\)

c ) \(9x^2+3x+5=\left(3x\right)^2+2\cdot3x\cdot\frac{1}{2}+\frac{1}{4}+\frac{19}{4}=\left(3x+\frac{1}{2}\right)^2+\frac{19}{4}>0\forall x\)

Ta có : 4x² + 2x + 1

= (2x)² + 2.2x.\(\frac{1}{2}\) + \(\frac{1}{2}+\frac{3}{4}\)

= (2x + \(\frac{1}{2}\))² + \(\frac{3}{4}\)

Mà : (2x + \(\frac{1}{2}\))² \(\ge0\forall x\)

=> (2x + \(\frac{1}{2}\))² + \(\frac{3}{4}\) \(\ge\frac{3}{4}\forall x\)

Hay : (2x + \(\frac{1}{2}\))² + \(\frac{3}{4}\)  \(>0\forall x\)

Vậy 4x² + 2x + 1 \(>0\forall x\)

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