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\(16a^2b-16ab+4b=4b\left(4a^2-4a+1\right)=4b\left(2a-1\right)^2\\ 5a^3-10a=5a\left(a^2-2\right)=5a\left(a-\sqrt{2}\right)\left(a+\sqrt{2}\right)\\ 3x-3z+x^2-2xz+z^2\\ =3\left(x-z\right)+\left(x-z\right)^2=\left(x-z\right)\left(3+x-z\right)\)
\(16a^2b-16ab+4b=4b\left[\left(4a^2\right)-4a+1\right]=4b\left[\left(2a\right)^2-4a+1\right]=4b\left(2a-1\right)^2\)
\(5a^3-10a=5a\left(a^2-2\right)\)
\(3x-3z+x^2-2xz+z^2=\left(3x-3z\right)+\left(x^2-2xz+z^2\right)=3\left(x-z\right)+\left(x-z\right)^2=\left(x-z\right)\left[3+\left(x-z\right)\right]=\left(x-z\right)\left(3+x-z\right)\)
\(x^2-y^2+z^2-t^2-2xz+2yt=\)
\(=\left(x^2-2xz+z^2\right)-\left(y^2-2yt+t^2\right)=\)
\(=\left(x-z\right)^2-\left(y-t\right)^2=\)
\(=\left[\left(x-z\right)-\left(y-t\right)\right]\left[\left(x-z\right)+\left(y-t\right)\right]\)
\(x^2-y^2+z^2-t^2-2xz+2yt\)
\(=\left(x^2-2xz+z^2\right)-\left(y^2+2yt+t^2\right)\)
\(=\left(x-z\right)^2-\left(y-t\right)^2\)
\(=\left(x-z+y-t\right)\times\left(x-z-y+t\right)\)
\(A=\frac{x^2+y^2-z^2+2xy}{x^2-y^2+z^2+2xz}\)
\(=\frac{\left(x^2+2xy+y^2\right)-z^2}{\left(x^2+2xz+z^2\right)-y^2}\)
\(=\frac{\left(x+y\right)^2-z^2}{\left(x+z\right)^2-y^2}\)
\(=\frac{\left(x+y+z\right)\left(x+y+z\right)}{\left(x+y+z\right)\left(x-y+z\right)}\)
\(=\frac{x+y-z}{x-y+z}\)
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 :
\(\Rightarrow A=\frac{0+2009-2010}{0-2009+2010}=-1\)
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 )
