Question

Giúp mình bài này với

Chung minh đẳng thức:

(x+y+z)^2 - x^2 - y^2 - z^2 = 2(xy +yz +xz)

Answer 1

\(\left(x+y+z\right)^2-x^2-y^2-z^2\)

\(=x^2+y^2+z^2+2xy+2yz+2zx-x^2-y^2-z^2\)

\(=2xy+2yz+2zx\)

Answer 2

\(\left(x+y+z\right)^2-x^2-y^2-z^2=2\left(xy+yz+xz\right)\)

\(VT=\left(x+y+z\right)^2-x^2-y^2-z^2\)

\(VT=x^2+y^2+z^2+2xy+2yz+2xz-x^2-y^2-z^2\)

\(VT=2xy+2yz+2xz\)

\(VT=2\left(xy+yz+xz\right)\)

\(VT=VP\left(đpcm\right)\)

* VT: vế trái
  VP: vế phải

Answer 3

(x+y+z)²-x²-y²-z²   (1)

Ta co : 2(xy+yz+xz)

=2xy+2yz+2xz

=2xy+2yz+2xz+x²+y²+z²-x²-y²-z²

=(x+y+z)²-x²-y²-z²

Tu (1) suy ra dpcm