Question

cho a+b+c = x+y+z = a/x + b/y + c/z = 0 chung minh ax^2 +by^2+cz^2 = 0

Answer 1

Do x + y + z = 0 nên

x = - (y + z) ; y = - (x + z) ; z = - (x + y)

=> x² = (y + z)² ; y² = (x + z)² ; z² = (x + y)²

=> ax² + by² + cz² = a(y² + 2yz + z²) + b(x² + 2xz + z²) + c(x² + 2xy + y²) = x²(b + c) + y²(a + c) + z²(a + b) + 2(ayz + bxz + cxy)              (1) 

Thay a = - (b + c) ; b = - (a + c) ; c = - (a + b) (Do a + b + c = 0 ) và ayz+bxz+cxy=0 (do a/x+b/y+c/z=0) vào (1) ta được ax² + by² + cz² = - (ax² + by² + cz²)

=> ax² + by² + cz² = 0

 

 

 

 

 

 

Answer 2

to ko hieu noi