Question

Chứng minh đẳng thức(x+y+z)²-x²-y²-z²​= 2(xy+yz+zx)

Answer 1

Ta có:

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

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

\(=2\left(xy+yz+zx\right)\) = VP

=> đpcm

Answer 2

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

Biến đổi vế trái:

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

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

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

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

\(=2\left(xy+yz+zx\right)=\) VP