Vì x+y+z=0
=>x+y=-z =>(x+y)^5=-z^5
hay x^5+y^5+5(x^4y+xy^4+2x³y²+2x²y³+)=-z^5
<=>x^5+y^5+z^5+5xy(x³+y³+2x²y+2x²y)=0
<=>x5+y^5+z^5+5xy(x+y)(x²-xy+y²+2xy)=0
<=>x^5+y^5+z^5-5xyz(x²+xy+y²)=0
<=>x^5+y^5+z^5=5xyz(x²+xy+y²)
<=>2(x^5+y^5+z^5)=5xyz(2x²+2xy+2y²)
<=>2(x^5+y^5+z^5)=5xyz[x²+y²+(x+y)²]
<=>2(x^5+y^5+z^5)=5xyz(x³+y²+z²)
Cho x+y+z = 0. CMR :
a) 5( x3 + y3 + z3 ) (x2 + y2 + z2) = 6(x5 + y5 + z5 )
b) 2( x5 + y5 + z5 ) = 5xyz( x2 + y2 + z2 )
:Tính
(x-2y)(3xy+5y2)+xy2(x+y-z)2x2y(2y-y3)+x2y4(6x2-12x):(x-2)+x5y4z3:x4y4z3(m-n)2+(m+n)2+2(m-n)(m+n)(x-y)2+(x+y)2+2(x-y)(x+y)x2y(x-2x3)+2x5y(3x2-6x):(x-2)+x3y4z5:x2y4z52(a-b)(a+b)+(a+b)^2+(a-b)^2
1.Nếu \(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}=1\) thì gtbt \(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\)
cho cac so x,y,z tuy y
cmr
x^2+y^2+z^2/3>= (x+y+x/3)^2
CMR: x2+y2+z2=x(y+z)
1. Cho a,b,c biết: a.b.c khác 0
và ab+bc+ca=0. Tính: P=(a+b)(b+c)(c+a)/abc
2.CMR: Nếu a,b,c>0 và a,b,c khác nhau thì
A=a^3+b^3+c^3-3abc > 0
3.Cho(x+y+z)(xy+yz+zx)=xyz
Cmr:x^2017+y^2017+z^2017=(x+y+z)^2017
Cho x+y+z=3 Cmr x^2+y^2+z^2+zx+xy+yz>=6
rút gọn phân thức:
a)3x(1-x)/1(x-1)
b)6x2y2/8xy5
c)3(x-y)(x-z)2/6(x-y)(x-z)
Cho x,y,z>0 thỏa mãn x+y+z=1. CMR: x^4+y^4/x^3+y^3 + y^4+z^4/y^3+z^3 + z^4+x^4/z^3+x^3 >=1
