Question

Tìm số nguyên z, biết (x+y+z)² - 2(x+y+z)(x+y) + (x+y)² = 100.

Giup với ạ

Answer 1

\(\left(x+y+z\right)^2-2\left(x+y+z\right).\left(x+y\right)+\left(x+y\right)^2=100\)

\(\left(x+y+z-x-y\right)^2=100\)

\(z^2=100\)

\(\Rightarrow z=\pm10\)