Question

so sánh x.y và x+y khi biết 

x=\(1+\sqrt{2}\)

y=\(\sqrt{2}-1\)

Answer 1

\(\sqrt{2}+1>\sqrt{2}-1\)

nên x>y

Answer 2

\(x=1+\sqrt{2}\)

\(y=\sqrt{2}-1=-1+\sqrt{2}\)

Vì `1> -1`

nên \(1+\sqrt{2}>-1+\sqrt{2}\)

hay ` x >y`

\(x+y=1+\sqrt{2}+\sqrt{2}-1=\left(1-1\right)+\sqrt{2}+\sqrt{2}\)

\(=2\sqrt{2}\)