cho x+y+z=0. cm \(2\left(x^5+y^5+z^5\right)=5xyz\left(x^2+y^2+z^2\right)\)
Cho x+y+z=0. CM:
\(2\left(x^5+y^5+z^5\right)=5xyz\left(x^2+y^2+z^2\right)\)
cho x+y+z=0. CMR:
\(2.\left(x^5+y^5+z^5\right)=5xyz.\left(x^2+y^2+z^2\right)\)
Cho :\(x+y+z=0\) . CMR: \(2\left(x^5+y^5+z^5\right)=5xyz\left(x^2+y^2+z^2\right)\)
Cho \(x+y+z=0\). Chứng minh rằng :
\(5\left(x^3+y^3+z^3\right)\left(x^2+y^2+z^2\right)=6\left(x^5+y^5+z^5\right)\)
1.Cho x+y+z=0. CMR:
a) \(5\left(x^3+y^3+z^3\right)\left(x^2+y^2+z^2\right)=6\left(x^5+y^5+z^5\right)\)
b) \(x^7+y^7+z^7=7xyz\left(x^2y^2+y^2z^2+z^2x^2\right)\)
c) \(10\left(x^7+y^7+z^7\right)=7\left(x^2+y^2+z^2\right)\left(x^5+y^5+z^5\right)\)
d) \(2\left(x^5+y^5+z^5\right)=5xyz\left(x^2+y^2+z^2\right)\)
2. Tìm n∈ N để biểu thức sau là số nguyên tố
a) \(A=n^3-4n^2-4n-1\)
b) \(B=n^3-6n^2+9n-2\)
c) \(C=n^{1975}+n^{1973}+1\)
\(CMR\)nếu \(x+y+z=0\)thì:
\(2\left(x^5+y^5+z^5\right)=5xyz\left(x^2+y^2+z^2\right).\)
Cho x + y + z khác 0 ; x = y + z . Chứng minh rằng :
\(\frac{\left(xy+yz+zx\right)^2-\left(x^2y^2+y^2z^2+z^2x^2\right)}{x^2+y^2+z^2}:\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2}=yz\)
Cho x,y,z>0. Chứng minh rằng:
\(\left(\frac{x}{x+y}\right)^2+\left(\frac{y}{y+z}\right)^2+\left(\frac{z}{z+x}\right)^2\ge\frac{3}{4}\)