求方程组x^3-y^3-z^3=3xyz,x^2=2(y+z)的正整数解
x^3-y^3-z^3=3xyz,
x^3-(y+z)^3=3xyz-3yz(y+z),
(x-y-z)[x^2+x(y+z)+(y+z)^2-3yz]=0,
因为x^2+x(y+z)+(y+z)^2-3yz
=x^2+y^2+z^2+xy+xz-yz
=(x+y)^2/2+(x+z)^2/2+(y-z)^2/2=0无正整数解
所以x-y-z=0,x^2=2(y+z),
x^2=2x,x=2.
8-y^3-z^3=6yz,y+z=2,仅有正整数解y=z=1.
所以原方程的正整数解为x=2,y=z=1.
已知函数f(x)=[√（x^2+2x+x）+√（x^2-1）+√（x^2-2x+1）]^(-1) (√是三次根号 我打不出)
求f(1)+f(2)+f(3)+f(4)+……+f(2003)
f(x)的第1个括号中是x^2+2x+1?
利用立方差公式
2=(x+1)-(x-1)=[(x+1)^(1/3)-(x-1)^(1/3)]*[(x+1)^(2/3)+(x+1)^(1/3)*(x-1)^(1/3)+(x-1)^(2/3)]
f(x)=[(x+1)^(1/3)-(x-1)^(1/3)]/2.
f(1)+f(2)+f(3)+f(4)+……+f(2003)
=[2^(1/3)-0^(1/3)]/2+
+[3^(1/3)-1^(1/3)]/2+
+...+
+[2004^(1/3)-2002^(1/3)]/2
=[2004^(1/3)+2003^(1/3)-1]/2.
x^(1/3)就是x的立方根.