Đáp án:
Giải thích các bước giải:
`x^3-3x-4+2m=0`
`⇔ 2m=-x^3+3x+4`
Đặt `f(x)=-x^3+3x+4`
`f'(x)=-3x^2+3`
`f'(x)=0⇒` \(\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.\)
Ta có BBT:
\(\begin{array}{|c|cc|}\hline \text{$x$}&\text{$-\infty$}&\text{}&\text{-1}&\text{}&\text{1}&\text{}&\text{$+\infty$}\\\hline \text{$y'$}&\text{}&\text{}-&\text{0}&\text{}+&\text{0}&\text{}-&\text{}\\\hline \text{$y$}&\text{}+\infty&\text{}&\text{}&\text{}&\text{6}&\text{}&\text{}\\&\text{}&\text{$\searrow$}&\text{}&\text{}\nearrow&\text{}&\text{}\searrow\\&\text{}&\text{}&\text{2}&\text{}&\text{}&\text{}&\text{}-\infty\\\hline \end{array}\)
Để PT có đúng 2 nghiệm thì:
`⇔ 2 < 2m < 6`
`⇔ 1 < m < 3`
Vậy `m \in (1;3)` thì PT có đúng 2 nghiệm thực
