1) \(x^2-2x-5m=0\) (1)
Thay \(m=3\) vào pt (1), ta được:
\(x^2-2x-5\cdot3=0\)
\(\Leftrightarrow x^2-2x-15=0\)
\(\Leftrightarrow x^2+3x-5x-15=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
2) \(\Delta=4-4\cdot\left(-5m\right)=20m+4\)
Để pt có hai nghiệm phân biệt thì \(\Delta>0\Leftrightarrow m>-\dfrac{1}{5}\)
Theo hệ thức Vi-ét: \(\left\{{}\begin{matrix}x_1+x_2=2\\x_1x_2=-5m\end{matrix}\right.\)
Lại có: \(x_1x_2^2-x_1\left(5m+3x_2\right)=10115\)
\(\Leftrightarrow x_1x_2^2-5mx_1-3x_1x_2=10115\)
\(\Leftrightarrow x_1x_2^2+x_1^2x_2-3x_1x_2=10115\)
\(\Leftrightarrow x_1x_2\left(x_1+x_2\right)-3x_1x_2=10115\)
\(\Leftrightarrow-5m\cdot2-3\cdot\left(-5m\right)=10115\)
\(\Leftrightarrow5m=10115\)
\(\Leftrightarrow m=2023\)
$\text{#}Toru$
