1 + 2xy - x mũ 2 - y mũ 2
a mũ 2 + b mũ 2 - c mũ 2 - d mũ 2 - 2ab + 2cd
a mũ 3 b mũ 3 - 1
x mũ 2 ( y - z) + y mũ 2 ( z - x ) + z mũ 2 ( x - y)
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DD
0
15 tháng 3 2022
a, \(=-91x-y+5z\)
b, \(=4x^2+x^2y-5y^2-\dfrac{5}{3}x^3+6xy^2+x^2y\)
\(=4x^2+2x^2y-5y^2-\dfrac{5}{3}x^3+6xy^2\)
5 tháng 10 2020
a) ( 5x - y )( 25x2 + 5xy + y2 ) = ( 5x )3 - y3 = 125x3 - y3
b) ( x - 3 )( x2 + 3x + 9 ) - ( 54 + x3 ) = x3 - 33 - 54 - x3 = -27 - 54 = -81
c) ( 2x + y )( 4x2 - 2xy + y2 ) - ( 2x - y )( 4x2 + 2xy + y2 ) = ( 2x )3 + y3 - [ ( 2x )3 - y3 ]= 8x3 + y3 - 8x3 + y3 = 2y3
d) ( x + y )2 + ( x - y )2 + ( x + y )( x - y ) - 3x2 = x2 + 2xy + y2 + x2 - 2xy + y2 + x2 - y2 - 3x2 = y2
e) ( x - 3 )3 - ( x - 3 )( x2 + 3x + 9 ) + 6( x + 1 )2
= x3 - 9x2 + 27x - 27 - ( x3 - 33 ) + 6( x2 + 2x + 1 )
= x3 - 9x2 + 27x - 27 - x3 + 27 + 6x2 + 12x + 6
= -3x2 + 39x + 6
= -3( x2 - 13x - 2 )
f) ( x + y )( x2 - xy + y2 ) + ( x - y )( x2 + xy + y2 ) - 2x3
= x3 + y3 + x3 - y3 - 2x3
= 0
g) x2 + 2x( y + 1 ) + y2 + 2y + 1
= x2 + 2x( y + 1 ) + ( y2 + 2y + 1 )
= x2 + 2x( y + 1 ) + ( y + 1 )2
= ( x + y + 1 )2
= [ ( x + y ) + 1 ]2
= ( x + y )2 + 2( x + y ) + 1
= x2 + 2xy + y2 + 2x + 2y + 1
22 tháng 9 2019
\(A=\frac{x^2+y^2-z^2+2xy}{x^2-y^2+z^2+2xz}\)
\(=\frac{\left(x^2+2xy+y^2\right)-z^2}{\left(x^2+2xz+z^2\right)-y^2}\)
\(=\frac{\left(x+y\right)^2-z^2}{\left(x+z\right)^2-y^2}\)
\(=\frac{\left(x+y+z\right)\left(x+y+z\right)}{\left(x+y+z\right)\left(x-y+z\right)}\)
\(=\frac{x+y-z}{x-y+z}\)
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 !!!
T
22 tháng 9 2019
\(A=\frac{x^2+y^2-z^2+2xy}{x^2-y^2+z^2+2xz}=\frac{\left(x^2+2xy+y^2\right)-z^2}{\left(x^2+2xz+z^2\right)-y^2}=\frac{\left(x+y\right)^2-z^2}{\left(x+z\right)^2-y^2}\)
\(=\frac{\left(x+y+z\right)\left(x+y-z\right)}{\left(x+y+z\right)\left(x-y+z\right)}=\frac{x+y-z}{x-y+z}\)
Thay \(\hept{\begin{cases}x=0\\y=2009\\z=2010\end{cases}}\) vào biểu thức :
\(\Rightarrow A=\frac{0+2009-2010}{0-2009+2010}=-1\)
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 )