Ta có: \(x^2-2xy-4z^2+y^2\)

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

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

\(=\left[6-\left(-4\right)-2\cdot45\right]\left[6-\left(-4\right)+2\cdot45\right]=-80\cdot100=-8000\)

x² - 2xy + y² - 4z²

= (x - y)² - (2z)²

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

Thay x = 6 ; y = -4 và z = 45 vào biểu thức ta được:

[6 - (-4) - 2 . 45] [6 - (-4) + 2 . 45]

= -80 . 100

= -8000

Trả lời:

x² - 2xy - 4z² + y² 

= ( x² - 2xy + y² ) - 4z²
= ( x - y )² - ( 2z )²

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

Thay x = 6; y = - 4; z = 45 vào biểu thức trên, ta được:

[ ( 6 - ( - 4 ) - 2.45 ] [ 6 - ( - 4 ) + 2.45 ] = ( 6 + 4 - 90 ) ( 6 + 4 + 90 ) = ( 10 - 90 ) ( 10 + 90 ) = - 80 . 100 = - 8000

x^2-2xy-4z^2+y^2 =(x^2-2xy+y^2)-(2z)^2 =(x-y)^2-(2z)^2 =(x-y-2z)(x-y+2z) Tại x=6;y=-4;z=45 bt có gái trị là (6+4-2.45).(6+4+45)=-80.100=-8000 Vậy bt có giá trị là -8000

x² - 2xy - 4z² + y² tại x = 6 ; y = -4 ; z = 45

= x² - 2xy + y² - 4z²

= ( x - y )² - ( 2z )²

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

Thay x = 6 ; y = -4 ; z = 45 vào biểu thức , ta có :

( x - y + 2z ) ( x - y - 2z )

= ( 6 + 4 + 2 . 45 ) ( 6 + 4 - 2 . 45 )

= 100 . ( -80 )

= -8000

\(x^2-2xy-4z^2+y^2\)

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

\(=\left(x-y\right)^2-\left(2z\right)^2\)

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

Thay \(x=6;y=-4;z=45\)vào biểu thức trên:

\(\left(x-y-2z\right)\left(x-y+2z\right)=\left(6+4-2.45\right)\left(6+4+2.45\right)\)

\(=-80.100=-8000\).

ĐS: -8000.

\(x^2-2xy-4z^2+y^2\)

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

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

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

Thay x=6 ; y=-4 ; z=45 vào biểu thức trên ta được:

\(\left(x-y-2z\right)\left(x-y+2z\right)\)

\(=\left(6-4-45.2\right)\left(6-4+2.45\right)\)

\(=\left(2-90\right)\left(2+90\right)\)

=\(-8096\)

A= x^2 -2xy + y^2 - (2z)^2 

  = ( x- y)^2 - (2z)^2

  = ( x-y - 2z)(x - y +2z)

= ( 6 - (-4) - 2.4,5) ( 6 - (-4) + 2.4,5)

= ( 10 - 90)( 10 + 90 )

= -80.100  

=-8000

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

\(=\left(6+4-90\right)\left(6+4+90\right)\)

\(=100\cdot\left(-80\right)=-8000\)

Ta có :\(\frac{x}{4y+z}=\frac{y}{4z+x}=\frac{z}{4x+y}=\frac{x+y+z}{4y+z+4z+x+4x+y}=\frac{x+y+z}{5\left(x+y+z\right)}=\frac{1}{5}\)

=> \(\hept{\begin{cases}\frac{x}{4y+z}=\frac{1}{5}\\\frac{y}{4z+x}=\frac{1}{5}\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{4y+z}=\frac{1}{5}\\\frac{4z+x}{y}=5\end{cases}}\)

Khi đó A = 2019 - 1/5 + 5 = 2023,8

\(\frac{x}{4y+z}=\frac{y}{4z+x}=\frac{z}{4x+y}=\frac{x+y+z}{4y+z+4z+x+4x+y}=\frac{x+y+z}{5\left(x+y+z\right)}=\frac{1}{5}\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{4y+z}=\frac{1}{5}\\\frac{y}{4z+x}=\frac{1}{5}\end{cases}\Rightarrow\hept{\begin{cases}\frac{x}{4y+z}=\frac{1}{5}\\\frac{4z+x}{y}=5\end{cases}}}\)

Khi đó \(A=2019-\frac{1}{5}+5=2013,8\)

a) \(x^2-2xy-4z^2+y^2=\left(x-y\right)^2-4z^2=\left(x-y-2z\right)\left(x-y+2z\right)=\left(6+4-2.45\right)\left(6+4+2.45\right)=-8000\)b) \(3\left(x-3\right)\left(x+7\right)+\left(x-4\right)^2+48=3\left(x^2+4x-21\right)+\left(x^2-8x+16\right)+48=4x^2+4x+1=\left(2x+1\right)^2=\left(2.0,5+1\right)^2=4\)

a: Ta có: \(x^2-2xy+y^2-4z^2\)

\(=\left(x-y\right)^2-\left(2z\right)^2\)

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

\(=\left(6+4-2\cdot45\right)\left(6+4+2\cdot45\right)\)

\(=-8000\)

b: Ta có: \(3\left(x-3\right)\left(x+7\right)+\left(x-4\right)^2+48\)

\(=3\left(x^2+4x-21\right)+\left(x-4\right)^2+48\)

\(=3x^2+12x-63+x^2-8x+16+48\)

\(=2x^2+4x+1\)

\(=2\cdot\dfrac{1}{4}+4\cdot\dfrac{1}{2}+1\)

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

b, Có : |x| + |x-2| = |x| + |2-x| >= |x+2-x| = 2

Lại co : |x-1| >= 0

=> |x|+|x-1|+|x-2| >= 2

Dấu "=" xảy ra <=> x.(2-x) >= 0 và x-1=0 <=> x=1

Vậy x=1

Tk mk nha