Nhân liên hợp là ra -.-
a, Có: \(\left(\sqrt{a^2+2019}+a\right)\left(\sqrt{a^2+2019}-a\right)=a^2+2019-a^2=2019\)
Mà \(\left(\sqrt{a^2+2019}+a\right)\left(\sqrt{b^2+2019}+b\right)=2019\)
\(\Rightarrow\sqrt{a^2+2019}-a=\sqrt{b^2+2019}+b\)(1)
b,Tương tự câu a sẽ c/m được \(\sqrt{a^2+2019}+a=\sqrt{b^2+2019}-b\)(2)
Lấy (1) trừ (2) theo từng vế được
\(\sqrt{a^2+2019}-a-\sqrt{a^2+2019}-a=\sqrt{b^2+2019}+b-\sqrt{b^2+2019}+b\) \(\Leftrightarrow-2a=2b\)
\(\Leftrightarrow-a=b\)
\(\Rightarrow-a^{2019}=b^{2019}\)
Ta có: \(P=a^{2019}+b^{2019}+2019\)
\(=a^{2019}-a^{2019}+2019\)
\(=2019\)
a)Theo giả thiết thì \(VT=\frac{\left(\sqrt{a^2+2019}+a\right)\left(\sqrt{a^2+2019}-a\right)}{\sqrt{a^2+2019}-a}.\frac{\left(\sqrt{b^2+2019}+b\right)\left(\sqrt{b^2+2019}-b\right)}{\sqrt{b^2+2019}-b}=2019\)
\(\Leftrightarrow\frac{2019}{\sqrt{a^2+2019}-a}.\frac{2019}{\sqrt{b^2+2019}-b}=2019\)
\(\Rightarrow\frac{1}{\sqrt{a^2+2019}-a}.\frac{1}{\sqrt{b^2+2019}-b}=1\) (chia hai vế cho 2019)
Suy ra \(\sqrt{a^2+2019}-a=\sqrt{b^2+2019}-b\)?!? (lạ nhỉ,hay là tui làm sai gì đó chăng?)
