\(\sqrt{2020-x}+\sqrt{2023-x}+\sqrt{2028-x}=6\)\(\left(x\le2020\right)\)
\(\Leftrightarrow\sqrt{2020-x}-1+\sqrt{2023-x}-2+\sqrt{2020-x}-3=0\)
\(\Leftrightarrow\frac{\left(\sqrt{2020-x}-1\right)\left(\sqrt{2020-x}+1\right)}{\sqrt{2020-x}+1}\) \(+\frac{\left(\sqrt{2023-x}-2\right)\left(\sqrt{2023-x}+2\right)}{\sqrt{2023-x}+2}\)\(+\frac{\left(\sqrt{2028-x}-3\right)\left(\sqrt{2028-x}+3\right)}{\left(\sqrt{2028-x}+3\right)}\)=0
\(\Leftrightarrow\frac{2019-x}{\sqrt{2020-x}+1}+\frac{2019-x}{\sqrt{2023-x}+2}+\frac{2019-x}{\left(\sqrt{2028-x}+3\right)}\)=0
\(\Leftrightarrow\left(2019-x\right)\left(\frac{1}{\sqrt{2020-x}+1}+\frac{1}{\sqrt{2023-x}+2}+\frac{1}{\sqrt{2028-x}+3}\right)\)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=2019\left(tm\right)\\\frac{1}{\sqrt{2020-x}+1}+\frac{1}{\sqrt{2023-x}+2}+\frac{1}{\sqrt{2028-x}+3}=0\left(2\right)\end{matrix}\right.\)
vì \(\sqrt{2020-x}\ge0\Rightarrow\frac{1}{\sqrt{2020-x}+1}>0\)
cmtt: \(\frac{1}{\sqrt[]{2023-x}+2}>0\)
\(\frac{1}{\sqrt{2028-x}+3}>0\)
=>\(\frac{1}{\sqrt{2020-x}+1}+\frac{1}{\sqrt{2023-x}+2}+\frac{1}{\sqrt{2028-x}+3}>0\)(3)
từ (2) và (3)=> vô lý
vậy x=2019 là nghiệm của phương trình
