Question

[Nguyễn Minh Lan]

gpt

x+ căn (2-x^2)=4y^2+4y+3

Answer 1

[Kiệt Nguyễn]

Áp dụng BĐT Bunhiacopxki:

\(x+\sqrt{2-x^2}\le\sqrt{\left(1^2+1^2\right)\left[x^2+\left(2-x^2\right)\right]}\le\sqrt{2.2}=2\)

(Dấu "="\(\Leftrightarrow x=1\))

và \(4y^2+4y+3=\left(2y+1\right)^2+2\ge2\)

(Dấu "="\(\Leftrightarrow y=\frac{-1}{2}\))

\(\Rightarrow x+\sqrt{2-x^2}=4y^2+4y+3\Leftrightarrow\hept{\begin{cases}x=1\\y=\frac{-1}{2}\end{cases}}\)