P=(5-x)(x+5)+(x-3)^2+5x

=25-x^2+x^2-6x+9+5x

=-x+34

a, \(\left(x-8\right)^2-9^2=\left(x-8-9\right)\left(x-8+9\right)=\left(x-17\right)\left(x+1\right)\)

b, \(16-x^2+4xy-4y^2=4^2-\left(x^2-4xy+4y^2\right)=4^2-\left(x-2y\right)^2\)

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

a 

( x - 8 ) ^ 2 - 81 

= ( x - 8 ) ^ 2 - 9 ^ 2 

= ( x - 8 - 9 ) ( x - 8 + 9 ) 

= ( x - 17 ) ( x + 1 ) 

b 

16 - x^2 + 4xy - 4y^2 

= 4^2 - ( x^2 - 4xy + 4y^2 ) 

= 4^2 - ( x - 2y ) ^ 2 

= [ 4 - ( x - 2y ) ] [ 4 + ( x - 2y ) ] 

= ( 4 - x + 2y ) ( 4 + x - 2y ) 

Ta có x + y + z = 0

=> x + y = -z

y + z = -x

x + z = -y

Khi đó Q = \(\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)=\frac{x+y}{y}.\frac{y+z}{z}.\frac{z+x}{x}=\frac{-z}{y}.\frac{-x}{y}.\frac{.-y}{x}=-1\)

Vậy khi x + y + z = 0 thì Q = -1

ta có 

\(Q=\frac{x+y}{y}.\frac{y+z}{z}.\frac{x+z}{x}=\frac{-z}{y}.\frac{-x}{z}.\frac{-y}{x}=-1\)

Áp dụng bất đẳng thức Bunhiacopxki, ta có:
VT.[a(b+c)+b(c+d)+c(d+a)+d(a+b)]≥(a+b+c+d)²
Ta cần chứng minh:
(a+b+c+d)²≥2(ab+bc+cd+da+2ca+2bd)

⇔a²+b²+c²+d²≥2ca+2bd⇔(a−c)²+(b−d)²≥0
Bất đẳng thức đã được chứng minh

dấu = xảy ra khia a=c, b=d

Bc nhé = 35

1. A = x² + 2y² + 2xy - 4y - 3

= ( x² + 2xy + y² ) + ( y² - 4y + 4 ) - 7

= ( x + y )² + ( y - 2 )² - 7

Vì ( x + y )²\(\ge\)0\(\forall\)x;y ; ( y - 2 )²\(\ge\)0\(\forall\)y

=> A = ( x + y )² + ( y - 2 )² - 7 \(\ge\)- 7

Dấu "=" xảy ra <=> \(\orbr{\begin{cases}\left(x+y\right)^2=0\\\left(y-2\right)^2=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=-2\\y=2\end{cases}}\)

Vậy minA = - 7 <=> \(\orbr{\begin{cases}x=-2\\y=2\end{cases}}\)

2. B = 3x² + 4y² + 4xy - 4x - 2

= ( x² + 4xy + 4y² ) + ( 2x² - 4x + 2 ) - 4

= ( x + 2y )² + 2 ( x - 1 )² - 4\(\ge\)- 4

Dấu "=" xảy ra <=> \(\orbr{\begin{cases}\left(x+2y\right)^2=0\\2\left(x-1\right)^2=0\end{cases}}\)<=> \(\orbr{\begin{cases}y=-\frac{x}{2}\\x=1\end{cases}}\)<=> \(\orbr{\begin{cases}y=-\frac{1}{2}\\x=1\end{cases}}\)

Vậy minB = - 4 <=> \(\orbr{\begin{cases}y=-\frac{1}{2}\\x=1\end{cases}}\)

C = 4x² + 2y² - 4xy - 6y + 5

= ( 4x² - 4xy + y² ) + ( y² - 6y + 9 ) - 4

= ( 2x - y )² + ( y - 3 )² - 4\(\ge\)- 4

Dấu "=" xảy ra <=> \(\orbr{\begin{cases}\left(2x-y\right)^2=0\\\left(y-3\right)^2=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=\frac{y}{2}\\y=3\end{cases}}\)<=> \(\orbr{\begin{cases}x=\frac{3}{2}\\y=3\end{cases}}\)

Vậy minC = - 4 <=> \(\orbr{\begin{cases}x=\frac{3}{2}\\y=3\end{cases}}\)