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\(\dfrac{x^3+y^3+z^3-3xyz}{\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2}\)
\(\Rightarrow\dfrac{\left(x+y\right)^3-3x^2y-3xy^2-3xyz+z^3}{x^2-2xy+y^2+y^2-2yz+z^2+x^2-2xz+z^2}\)
\(\Rightarrow\dfrac{\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y+z\right)}{2x^2+2y^2+2z^2-2xy-2yz-2xz}\)
\(\Rightarrow\dfrac{\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)}{2\left(x^2+y^2+z^2-xy-yz-xz\right)}\)
\(\Rightarrow\dfrac{\left(x+y+z\right)\left(x^2+2xy+z^2-xz-yz+z^2-3xy\right)}{2\left(x^2+y^2+z^2-xy-yz-xz\right)}\)
\(\Rightarrow\dfrac{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)}{2\left(x^2+y^2+z^2-xy-yz-xz\right)}\)
\(\Rightarrow\dfrac{x+y+z}{2}\)
\(\Rightarrow\dfrac{1}{2}\left(x+y+z\right)\)
Bài 1.
x = 14
=> 13 = x - 1 ; 15 = x + 1 ; 16 = x + 2 ; 29 = 2x + 1
Thế vào N(x) ta được :
x5 - ( x + 1 )x4 + ( x + 2 )x3 - ( 2x + 1 )x2 + ( x - 1 )x
= x5 - x5 - x4 + x4 + 2x3 - 2x3 - x2 + x2 - x
= -x = -14
Bài 2.
a) ( 1 - x - 2x3 + 3x2 )( 1 - x + 2x3 - 3x2 )
= [ ( 1 - x ) - ( 2x3 - 3x2 ) ][ ( 1 - x ) + ( 2x3 - 3x2 ) ]
= ( 1 - x )2 - ( 2x3 - 3x2 )2
= 1 - 2x + x2 - [ ( 2x3 )2 - 2.2x3.3x2 + ( 3x2 )2 ]
= x2 - 2x + 1 - ( 4x6 - 12x5 + 9x4 )
= x2 - 2x + 1 - 4x6 + 12x5 - 9x4
= -4x6 + 12x5 - 9x4 + x2 - 2x + 1
b) ( x - y + z )2 + ( z - y )2 + 2( x - y + z )( y - z )
= ( x - y + z )2 + ( z - y )2 - 2( x - y + z )( z - y )
= [ ( x - y + z ) - ( z - y ) ]2
= ( x - y + z - z + y )2
= x2
1) x2-4x+5+y2+2y=0
<=>x2-4x+4+y2+2y+1=0
<=>(x-2)2+(x+1)2=0
<=>x-2=0 và x+1=0
<=>x=2 và x=-1
2)2p.p2-(p3-1)+(p+3)2p2-3p5
<=>2p3-p3+1+2p3+6p2-3p5
<=>3p3+6p2-3p5+1
3)(0.2a3)2-0.01a4(4a2-100)=0,04a6-0,04a6+1
=1
4)a) x(2x+1)-x2(x+20)+(x3-x+3)=2x2+x-x3-20x2+x3-x+3
=-18x2+3(đề sai)
b) x(3x2-x+5)-(2x3+3x-16)-x(x2-x+2)=3x3-x2+5x-2x3-3x+16-x3+x2-2x
=16
Vậy x(3x2-x+5)-(2x3+3x-16)-x(x2-x+2) không phụ thuộc vào x
5)a) x(y-z)+y(z-x)+z(x-y)=xy-xz+yz-xy+xz-yz=0
b) x(y+z-yz)-y(z+x-xz)+z(y-x)=xy+xz-xyz-yz-xy+xyz+yz-xz=0
6)M+(12x4-15x2y+2xy2+7)=0
<=>M =-(12x4-15x2y+2xy2+7)
<=>M =-12x4+15x2y-2xy2-7