Question

cho a-b=\(\sqrt{1-b^2}\) +\(\sqrt{1-a^2}\)

tính \(\sqrt{a^2+b^2}\) +2019

Answer 1

ĐKXĐ:...

\(a-\sqrt{1-a^2}=b+\sqrt{1-b^2}\)

\(\Rightarrow-2a\sqrt{1-a^2}=2b\sqrt{1-b^2}\)

\(\Rightarrow-a\sqrt{1-a^2}=b\sqrt{1-b^2}\)

\(\Rightarrow a^2\left(1-a^2\right)=b^2\left(1-b^2\right)\)

\(\Rightarrow a^2-b^2-\left(a^4-b^4\right)=0\)

\(\Rightarrow a^2-b^2-\left(a^2-b^2\right)\left(a^2+b^2\right)=0\)

\(\Rightarrow\left(a^2-b^2\right)\left(1-a^2-b^2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a^2=b^2\\a^2+b^2=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}a=b\\-a=b\\a^2+b^2=1\end{matrix}\right.\)

- Nếu \(a=b\Rightarrow2\sqrt{1-a^2}=0\Rightarrow a^2=b^2=1\)

\(\Rightarrow\sqrt{a^2+b^2}+2019=2019+\sqrt{2}\)

- Nếu \(-a=b\Rightarrow2a=2\sqrt{1-a^2}\Rightarrow a=\sqrt{1-a^2}\Rightarrow a^2=\frac{1}{2}\)

\(\Rightarrow b^2=\frac{1}{2}\Rightarrow\sqrt{a^2+b^2}+2019=2020\)

- Nếu \(a^2+b^2=1\Rightarrow\sqrt{a^2+b^2}+2019=2020\)