Question

cho cac so thuc x va y thoa man

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

chung minh x+y=0

Answer 1

khong lay so 1 nho nha

Answer 2

\(\sqrt{x+2+2\sqrt{x+1}}+\sqrt{x+2-2\sqrt{ }x+1}=\frac{x+5}{2}\)\(\frac{x+5}{2}\)

Answer 3

Xét \(x^2+\sqrt{1+x^2}\)ta có:

\(x^2\ge0\)

nên \(1+x^2\ge1\)

\(\Rightarrow\sqrt{1+x^2}\ge\sqrt{1}=1\)

\(\Rightarrow x^2+\sqrt{1+x^2}\ge1\)

Tương tự ta có: 

\(y^2+\sqrt{1+y^2}\ge1\)

Do đó: \(\left(x^2+\sqrt{1+x^2}\right)\left(y^2+\sqrt{1+y^2}\right)\ge1\)

Dấu bằng xảy ra khi \(x=0;y=0\)

Khi đó \(x+y=0\left(ĐPCM\right)\)