Question

1, Cho x+y = 9, xy = 14

a, x-y = ?

b, x^2 + y^2 = ?

2, Cho x+y+z = 0; xy+yz+xz = 0

CM: x=y=z

Answer 1

1,x+y=9;xy=14

a)

Ta có:\(x+y=9\)

=>\(\left(x-y\right)^2+4xy=81\)

=>\(\left(x-y\right)^2=81-4xy=81-4.14=25\)

=>\(x-y=-5\)hoặc \(x-y=5\)

Vậy..

b)Ta có:\(x+y=9\)

=>\(x^2+y^2=81-2xy=81-2.14=53\)

Vậy...

Answer 2

Bài2:

Ta có:

\(x+y+z=0\)

=>\(x^2+y^2+z^2+2xy+2xz+2yz=0\)

=>\(x^2+y^2+z^2=0\)

Với mọi x;y;z thì \(x^2\)>=0;\(y^2\)>=0;\(z^2\)>=0

=>\(x^2+y^2+z^2\)>=0

Để \(x^2+y^2+z^2=0\)thì

\(x^2=0\);\(y^2=0\);\(z^2=0\)

=>\(x=y=z=0\left(đpcm\right)\)