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

\(=x^2-2xy+y^2-x^2-2xy-y^2\)

=-4xy

2: \(\left(7n-2\right)^2-\left(2n-7\right)^2\)

\(=\left(7n-2+2n-7\right)\left(7n-2-2n+7\right)\)

\(=\left(9n-9\right)\left(5n+5\right)\)

\(=9\left(n-1\right)\left(5n+5\right)⋮9\)

3: \(P=-x^2+6x+1\)

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

\(=-\left(x^2-6x+9-10\right)\)

\(=-\left(x-3\right)^2+10< =10\forall x\)

Dấu '=' xảy ra khi x-3=0

=>x=3

4: \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)

=>\(a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+b^2y^2+2abxy\)

=>\(a^2y^2-2abxy+b^2x^2=0\)

=>\(\left(ay-bx\right)^2=0\)

=>ay-bx=0

\(\dfrac{x+5}{2006}+\dfrac{x+6}{2005}+\dfrac{x+7}{2004}=-3\)

=>\(\left(\dfrac{x+5}{2006}+1\right)+\left(\dfrac{x+6}{2005}+1\right)+\left(\dfrac{x+7}{2004}+1\right)=-3+3=0\)

=>\(\dfrac{x+2011}{2006}+\dfrac{x+2011}{2005}+\dfrac{x+2011}{2004}=0\)

=>\(\left(x+2011\right)\left(\dfrac{1}{2006}+\dfrac{1}{2005}+\dfrac{1}{2004}\right)=0\)

=>x+2011=0

=>x=-2011

\(1,x^2+9-16y^2+6x\\ =\left(x^2+6x+9\right)-\left(4y\right)^2\\ =\left(x+3\right)^2-\left(4y\right)^2\\ =\left(x-4y+3\right)\left(x+4y+3\right)\\ 2,x^2-9+y^2+2xy\\ =\left(x^2+2xy+y^2\right)-9\\ =\left(x+y\right)^2-3^2\\ =\left(x+y-3\right)\left(x+y+3\right)\\ 3,x^2-4x+4-9y^2\\ =\left(x-2\right)^2-\left(3y\right)^2\\ =\left(x-3y-2\right)\left(x+3y-2\right)\\ 4,x^2-4xy+4y^2-81\\ =\left(x-2y\right)^2-9^2\\ =\left(x-2y-9\right)\left(x-2y+9\right)\\ 5,6x^2+6y^2-24+12xy\\ =\left(6x^2+12xy+6y^2\right)-24\\ =6\left[\left(x^2+2xy+y^2\right)-4\right]\\ =6\left[\left(x+y\right)^2-2^2\right]\\ =6\left(x+y-2\right)\left(x+y+2\right)\\ 6,9x^2-6x+1-25\\ =\left(3x-1\right)^2-5^2\\ =\left(3x-1-5\right)\left(3x-1+5\right)\\ =\left(3x-6\right)\left(3x+4\right)\)

7: \(x^2+4x+4-49y^2\)

\(=\left(x^2+4x+4\right)-49y^2\)

\(=\left(x+2\right)^2-49y^2\)

=(x+2+7y)(x+2-7y)

8: \(a^3+9a-ab^2-6a^2\)

\(=a\left(a^2-6a+9-b^2\right)\)

\(=a\left[\left(a-3\right)^2-b^2\right]\)

\(=a\left(a-3-b\right)\left(a-3+b\right)\)

9: \(8x^2-16x+8-32y^2\)

\(=8\left(x^2-2x+1-4y^2\right)\)

\(=8\left[\left(x-1\right)^2-\left(2y\right)^2\right]\)

=8(x-1-2y)(x-1+2y)

10: \(4x^2-4x+1-81a^2\)

\(=\left(4x^2-4x+1\right)-81a^2\)

\(=\left(2x-1\right)^2-\left(9a\right)^2\)

=(2x-1-9a)(2x-1+9a)

11: \(x^2-6xy+9y^2-121\)

\(=\left(x^2-6xy+9y^2\right)-121\)

\(=\left(x-3y\right)^2-11^2=\left(x-3y-11\right)\left(x-3y+11\right)\)

12: \(12x^2-24x+12-3y^2\)

\(=3\left(4x^2-8x+4-y^2\right)\)

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

\(1)A=x^2-7x+2\\ =\left(x^2-2\cdot x\cdot\dfrac{7}{2}+\dfrac{49}{4}\right)-\dfrac{41}{4}\\ =\left(x-\dfrac{7}{2}\right)^2-\dfrac{41}{4}\)

Ta có: `(x-7/2)^2>=0` với mọi x

`=>A=(x-7/2)^2-41/4>=-41/4` với mọi x

Dấu "=" xảy ra: `x-7/2=0<=>x=7/2` 

\(2)B=9x^2-12x+5\\ =\left(9x^2-12x+4\right)+1\\ =\left[\left(3x\right)^2-2\cdot3x\cdot2+2^2\right]+1\\ =\left(3x-2\right)^2+1\)

Ta có: `(3x-2)^2>=0` với mọi x

`=>B=(3x-2)^2+1>=1` với mọi x

Dấu "=" xảy ra: `3x-2=0<=>x=2/3` 

\(x^2-7x+2=x^2-2\cdot x\cdot\dfrac{7}{2}+\dfrac{49}{4}-\dfrac{41}{4}\)

\(=\left(x-\dfrac{7}{2}\right)^2-\dfrac{41}{4}>=-\dfrac{41}{4}\forall x\)

Dấu '=' xảy ra khi \(x-\dfrac{7}{2}=0\)

=>\(x=\dfrac{7}{2}\)

\(x^2-12x+5\)

\(=x^2-12x+36-31=\left(x-6\right)^2-31>=-31\forall x\)

Dấu '=' xảy ra khi x-6=0

=>x=6

`x^2 - 7x + 2`

`= x^2 - 2.x . 7/2 + (7/2)^2 - 41/4`

`= (x - 7/2)^2 - 41/4`

Do `(x - 7/2)^2 >= 0=>  (x - 7/2)^2 - 41/4 >= - 41/4`

Dấu = xảy ra khi: 

`x - 7/2 = 0`

`<=> x = 7/2`

Vậy ...

-----------------------

`x^2 - 12x + 5`

`= x^2 - 2.x.6 + 6^2 - 31`

`= (x-6)^2 - 31`

Do `(x-6)^2 >= 0 =>  (x-6)^2 - 31>= -31`

Dấu = có khi: 

`x - 6 = 0`

`<=> x = 6`

Vậy .... (không có max )

\(x+3y=5\Rightarrow x=5-3y\)

Ta có:

\(A=x^2+y^2+16y+2x\)

\(A=\left(5-3y\right)^2+y^2+16y+2\left(5-3y\right)\)

\(A=25-30y+9y^2+y^2+16y+10-6y\)

\(A=10y^2-20y+10+25\)

\(A=10\left(y-1\right)^2+25\ge5\forall y\)

Dấu "=" xảy ra khi \(y=1\Rightarrow x=2\)

Vậy \(A_{min}=25\) khi \(x=2\) và \(y=1\)

\(2P=6ab+2c\left(a+b\right)\)

\(2P=3\left(a^2+b^2+c^2\right)+6ab+2c\left(a+b\right)-3\left(a^2+b^2+c^2\right)\)

\(2P=3\left(a+b\right)^2+2c\left(a+b\right)+3c^2-3\left(a^2+b^2+c^2\right)\)

\(2P=\left(a+b+c\right)^2+2\left(a+b\right)^2+2c^2-3\left(a^2+b^2+c^2\right)\)

\(2P\ge-3\left(a^2+b^2+c^2\right)\ge-54\)

\(\Rightarrow P\ge-27\)

\(P_{min}=-27\) khi \(\left\{{}\begin{matrix}a^2+b^2+c^2=18\\a+b+c=0\\a+b=0\\c=0\end{matrix}\right.\) \(\Rightarrow\left(a;b;c\right)=\left(-3;3;0\right);\left(3;-3;0\right)\)

\(P=\dfrac{-\left(x^2+1\right)+2x^2-8x+8}{x^2+1}=-1+\dfrac{2\left(x-2\right)^2}{x^2+1}\ge-1\)

\(P_{min}=-1\) khi \(x-2=0\Rightarrow x=2\)

\(P=\dfrac{9\left(x^2+1\right)-8x^2-8x-2}{x^2+1}=9-\dfrac{2\left(2x+1\right)^2}{x^2+1}\le9\)

\(P_{max}=9\) khi \(2x+1=0\Rightarrow x=-\dfrac{1}{2}\)

\[
P = \frac{x^2 - 8x + 7}{x^2 + 1}
\]

\[
x^2 - 8x + 7 = (x^2 - 8x + 16) - 9 = (x-4)^2 - 9
\]

\[
P = \frac{(x-4)^2 - 9}{x^2 + 1}
\]

- Tại \( x = 0 \):

\[
P(0) = \frac{0^2 - 8 \times 0 + 7}{0^2 + 1} = \frac{7}{1} = 7
\]

- Tại \( x = 1 \):

\[
P(1) = \frac{1^2 - 8 \times 1 + 7}{1^2 + 1} = \frac{1 - 8 + 7}{2} = \frac{0}{2} = 0
\]

- Tại \( x = 2 \):

\[
P(2) = \frac{2^2 - 8 \times 2 + 7}{2^2 + 1} = \frac{4 - 16 + 7}{4 + 1} = \frac{-5}{5} = -1
\]

- Tại \( x = 4 \)

\[
P(4) = \frac{4^2 - 8 \times 4 + 7}{4^2 + 1} = \frac{16 - 32 + 7}{16 + 1} = \frac{-9}{17}
\]

- Tại \( x = -1 \):

\[
P(-1) = \frac{(-1)^2 - 8 \times (-1) + 7}{(-1)^2 + 1} = \frac{1 + 8 + 7}{1 + 1} = \frac{16}{2} = 8
\]

Dựa trên các giá trị đã tính, ta thấy rằng giá trị lớn nhất của \( P \) là \( 8 \) và giá trị nhỏ nhất là \( -1 \).

=> Max = 8

Min = -1

\(E=x^2-2xy+3y^2-2x-10y+20\\ =\left(x^2+y^2+1-2xy-2x+2y\right)+\left(2y^2-12y+72\right)-53\\ =\left(-x+y+1\right)^2+2\left(y-6\right)^2-53\)
Ta có:

`(-x+y+1)^2>=0` với mọi x,y

`2(y-6)^2>=0` với mọi y

`=>E=(-x+y+1)^2+2(y-6)^2-53>=-53` với mọi x,y

Dấu "=" xảy ra: `-x+y+1=0` và `y-6=0`

`<=>-x+7=0` và `y=6`

`<=>x=7` và `y=6`