Question

[Mr Ray]

gia su bo 3 so thuc x,y,z thoa \(x+1=y+z\) va \(xy+z^2-7z+10=0\)

 Chung minh \(x^2+y^2=-z^2+12z-19\)

Answer 1

[Mr Lazy]

\(x+1=y+z\Rightarrow x-y=z-1\Rightarrow\left(x-y\right)^2=\left(z-1\right)^2\)

\(\Leftrightarrow x^2+y^2-2xy=z^2-2z+1\)

\(\Leftrightarrow x^2+y^2-2\left(-z^2+7z-10\right)=z^2-2z+1\text{ }\left(do\text{ }xy+z^2-7z+10=0\right)\)

\(\Leftrightarrow x^2+y^2=-z^2+12z-19\text{ (đpcm)}\)

Answer 2

[Hồ Thị Uyên]

Khó                                       .