Question

Ai giải giúp mình với(về hằng đẳng thức)

Tìm x và y biết

x^2+2x+y^2-6y-10=0

Answer 1

\(x^2+2x+y^2-6y-10=0\)

\(x^2+2x+1+y^2-6x+9=10\)

\(\left(x+1\right)^2+\left(y-3\right)^2=0\)

\(\left(x+1\right)^2=\left(y-3\right)^2=0\)

\(x+1=y-3=0\)

Vậy \(x=-1;y=3\)

Answer 2

\(x^2\)\(+2x+y^2\)\(-6y-10=0\)

\(x^2\)\(+2x+1+y^2\)\(-6x+9=10\)

\(\left(x+1\right)^2\)+\(\left(y-3\right)^2\)\(=0\)

\(\left(x+1\right)^2\)\(=\left(y-3\right)^2\)\(=0\)

\(x+1=y-3=0\)

Vậy: \(x=-1;y=3\)