D = x² + 4xy + 4y² - z² + 2xt - t² 

 = (x + 2y)² - (z - t)² 

= (x + 2y - z + t)(x + 2y + z - t) 

Thay x = 10 ; y = 40 ; z = 30 ; t = 20 vào D 

\(\Rightarrow D=\left(10+40.2-30+20\right)\left(10+40.2+30-20\right)=80.100=8000\)

D = x\(^2\) + 4xy + 4y \(^2\) - z \(^2\) + 2zt - t \(^2\)

D = (x + 2y)\(^2\) - z\(^2\)+ z\(^2\) + 2zt + t\(^2\) - t\(^2\) 

D = (10 + 80)\(^2\) - 30\(^2\) + (z + t)\(^2\) - 20\(^2\) 

D = 90\(^2\) - 900 - 900 + (30 + 20)\(^2\) - 400

D = 8100 - 900 + 2500 - 400 

D =8600

HT

8000 nha

D = x\(^2\) + 2xy + y\(^2\) - z\(^2\) - 2zt - t\(^2\)

D = (x + y)\(^2\)  - z\(^2\) + z\(^2\) - 2zt + t\(^2\) - t\(^2\)

D = (89 + 11)\(^2\) +(z - t)\(^2\) - z\(^2\) - t\(^2\)

D = 100\(^2\) + (60 - 30)\(^2\) - 60\(^2\) - 30\(^2\)

D = 10 000 + 900 - 3600 - 900

D = 6400

Học tốt

D= 6400 nha

a: \(A=\dfrac{2}{3}x^3y\cdot\dfrac{3}{4}xy^2\cdot z^2\)

\(=\left(\dfrac{2}{3}\cdot\dfrac{3}{4}\right)\cdot\left(x^3\cdot x\right)\cdot\left(y\cdot y^2\right)\cdot z^2\)

\(=\dfrac{1}{2}x^4y^3z^2\)

b: \(A=\dfrac{1}{2}x^4y^3z^2\)

bậc của đa thức A là 4+3+2=9

c: \(A=\dfrac{1}{2}x^4y^3z^2\)

Hệ số là \(\dfrac{1}{2}\)

Phần biến là \(x^4;y^3;z^2\)

d: Thay x=-1;y=-2;z=-3 vào A, ta được:

\(A=\dfrac{1}{2}\cdot\left(-1\right)^4\cdot\left(-2\right)^3\cdot\left(-3\right)^2\)

\(=\dfrac{1}{2}\left(-8\right)\cdot9=-4\cdot9=-36\)

\(A=\frac{x^2+y^2-z^2+2xy}{x^2-y^2+z^2+2xz}\)

       \(=\frac{\left(x^2+2xy+y^2\right)-z^2}{\left(x^2+2xz+z^2\right)-y^2}\)

         \(=\frac{\left(x+y\right)^2-z^2}{\left(x+z\right)^2-y^2}\)

           \(=\frac{\left(x+y+z\right)\left(x+y+z\right)}{\left(x+y+z\right)\left(x-y+z\right)}\)

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

Ta thay : \(x=0;y=2009;z=2010\) ta được :

\(A=\frac{0+2009-2010}{0-2009+2010}=-\frac{1}{1}=-1\)

Chúc bạn học tốt !!!

\(A=\frac{x^2+y^2-z^2+2xy}{x^2-y^2+z^2+2xz}=\frac{\left(x^2+2xy+y^2\right)-z^2}{\left(x^2+2xz+z^2\right)-y^2}=\frac{\left(x+y\right)^2-z^2}{\left(x+z\right)^2-y^2}\)

\(=\frac{\left(x+y+z\right)\left(x+y-z\right)}{\left(x+y+z\right)\left(x-y+z\right)}=\frac{x+y-z}{x-y+z}\)

Thay \(\hept{\begin{cases}x=0\\y=2009\\z=2010\end{cases}}\) vào biểu thức :

\(\Rightarrow A=\frac{0+2009-2010}{0-2009+2010}=-1\)

Bài 1:

a) Ta có: \(2x=5y.\)

=> \(\frac{x}{y}=\frac{5}{2}\)

=> \(\frac{x}{5}=\frac{y}{2}\) và \(x.y=90.\)

Đặt \(\frac{x}{5}=\frac{y}{2}=k\Rightarrow\left\{{}\begin{matrix}x=5k\\y=2k\end{matrix}\right.\)

Có: \(x.y=90\)

=> \(5k.2k=90\)

=> \(10k^2=90\)

=> \(k^2=90:10\)

=> \(k^2=9\)

=> \(k=\pm3.\)

TH1: \(k=3\)

\(\Rightarrow\left\{{}\begin{matrix}x=3.5=15\\y=3.2=6\end{matrix}\right.\)

TH2: \(k=-3\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-3\right).5=-15\\y=\left(-3\right).2=-6\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(15;6\right),\left(-15;-6\right).\)

e) Ta có: \(\frac{x}{y}=\frac{4}{5}.\)

=> \(\frac{x}{4}=\frac{y}{5}\) và \(x.y=20.\)

Đặt \(\frac{x}{4}=\frac{y}{5}=k\Rightarrow\left\{{}\begin{matrix}x=4k\\y=5k\end{matrix}\right.\)

Có: \(x.y=20\)

=> \(4k.5k=20\)

=> \(20k^2=20\)

=> \(k^2=20:20\)

=> \(k^2=1\)

=> \(k=\pm1.\)

TH1: \(k=1\)

\(\Rightarrow\left\{{}\begin{matrix}x=1.4=4\\y=1.5=5\end{matrix}\right.\)

TH2: \(k=-1\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-1\right).4=-4\\y=\left(-1\right).5=-5\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(4;5\right),\left(-4;-5\right).\)

Chúc bạn học tốt!

Phân tích à ? -.-

a) ax - bx + ab - x²

= ( ax + ab ) - ( x² + bx )

= a( x + b ) - x( x + b )

= ( x + b )( a - x )

b) x² - 4xy + 4y² - 4

= ( x² - 4xy + 4y² ) - 4

= ( x - 2y )² - 2²

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

c) ( x² + y² - 2 )² - ( 2xy - 2 )²

= [ ( x² + y² - 2 ) - ( 2xy - 2 ) ][ ( x² + y² - 2 ) + ( 2xy - 2 ) ]

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

= ( x² - 2xy + y² )[ ( x² + 2xy + y² ) - 4 ]

= ( x - y )²[ ( x + y )² - 2² ]

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

d) ab( x² + y² ) + ( a² + b² ) ( cái này không phân tích được ((: )

1 + 2xy - x² - y²

= 1 - ( x² - 2xy + y² )

= 1² - ( x - y )²

= [ 1 - ( x - y ) ][ 1 + ( x - y ) ]

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

a² + b² - c² - d² - 2ab + 2cd

= ( a² - 2ab + b² ) - ( c² - 2cd + d² )

= ( a - b )² - ( c - d )²

= [ ( a - b ) - ( c - d ) ][ ( a - b ) + ( c - d ) ]

= ( a - b - c + d )( a - b + c - d )

a³b³ - 1

= ( ab )³ - 1³

= ( ab - 1 )[ ( ab )² + ab.1 + 1² ]

= ( ab - 1 )( a²b² + ab + 1 )

x²( y - z ) + y²( z - x ) + z²( x - y )

= z²( x - y ) + x²y - x²z + y²z + y²x

= z²( x - y ) + ( x²y - y²x ) - ( x²z - y²z )

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

= z²( x - y ) + xy( x - y ) - z( x + y )( x - y )

= ( x - y )[ z² + xy - z( x + y ) ]

= ( x - y )( z² + xy - zx - zy )

= ( x - y )[ ( z² - zx ) - ( zy - xy ) ]

= ( x - y )[ z( z - x ) - y( z - x ) ]

= ( x - y )( z - x )( z - y )