Question

Chứng minh đẳng thức :

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

Answer 1

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

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

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

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

 

Answer 2

\(\left[\left(x+y\right)+z\right]^2=\left[\left(x+y\right)^2+2.\left(x+y\right)z+z^2\right]=x^2+2xy+y^2+2xz+2yz+z^2\)\(+z^2\)

Thay vào: x^2+y^2+z^2+ 2xy+2yz+2xz - x^2 - y^2 - z^2= 2(xy+yz+xz) (đpcm)