Question

Cho phương trình \(2018x^2-\left(m-2019\right)x-2020=0\)  Tìm m để phương trình có hai nghiệm \(x_1;x_2\)thõa mãn

\(\sqrt{x_1^2+2019}-x_2=\sqrt{x_2^2+2019}-x_2\)

Answer 1

srtgb6yyyyyyyy

Answer 2

\(2018x^2-\left(m-2019\right)x-2020=0\)

Ta có \(\Delta=b^2-4ac\)

             \(=\left[-\left(m-2019\right)\right]^2-4.2018.\left(-2020\right)\)

             \(=\left(m-2019\right)^2+4.2018.2020>0\)( vì \(\left(m-2019\right)^2\ge0\forall x\))

Phương trình có 2 nghiệm \(x_1,x_2\) Áp dụng hệ thức Vi-ét ta có

\(\hept{\begin{cases}x_1+x_2=\frac{m-2019}{2018}\left(1\right)\\x_1.x_2=\frac{-2020}{2018}\left(2\right)\end{cases}}\)

Ta có \(\sqrt{x_1^2+2019}-x_2=\sqrt{x_2^2+2019}-x_2\)

\(\Leftrightarrow\sqrt{x_1^2+2019}-x_2+x_2=\sqrt{x_2^2+2019}\)

\(\Leftrightarrow\sqrt{x_1^2+2019}+0=\sqrt{x_2^2+2019}\)

\(\Leftrightarrow x_1^2+2019=x_2^2+2019\)

\(\Leftrightarrow x_1^2-x_2^2=0\)

\(\Leftrightarrow\left(x_1-x_2\right).\left(x_1+x_2\right)=0\)

\(\Leftrightarrow\left(x_1-x_2\right).\frac{m-2019}{2018}=0\Rightarrow x_1-x_2=0\left(3\right)\)

Thay (3) vào (!) ta có \(\hept{\begin{cases}x_1+x_2=\frac{m-2019}{2018}\\x_1-x_2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x_1=\frac{m-2019}{2018}\\x_1-x_2=0\end{cases}}\)

                                                                                      \(\Leftrightarrow\hept{\begin{cases}x_1=\frac{m-2019}{4036}\\x_2=\frac{m-2019}{4036}\end{cases}}\)

\(\Rightarrow x_1.x_2=\frac{-2020}{2018}=\frac{-1010}{1009}\)

\(\Leftrightarrow\frac{m-2019}{4036}.\frac{m-2019}{4036}=\frac{-1010}{1009}\)

\(\Leftrightarrow\frac{\left(m-2019\right)^2}{4036^2}=\frac{-1010}{1009}\)

\(\Leftrightarrow\left(m-2019\right)^2=\frac{4036^2.\left(-1010\right)}{1009}\)

\(\Leftrightarrow\left(m-2019\right)^2=-16305440\left(VL\right)\)

Vậy không có m để thỏa mãn bài toán