Chương 4: BẤT ĐẲNG THỨC, BẤT PHƯƠNG TRÌNH

\(x^2-\left(m+1\right)x-2\left(1-m\right)< 0\) nghiệm đúng \(\forall x\in\) (0; 3]

\(\Rightarrow\) PT \(x^2-\left(m+1\right)x-2\left(1-m\right)=0\left(1\right)\) có 2 nghiệm phân biệt thỏa mãn \(x_1\le0< 5< x_2\)

PT (1) \(\Leftrightarrow x=2;x=m-1\)

Để thỏa mãn yêu cầu bài toán:

\(\Rightarrow\left[{}\begin{matrix}m-1>5\\m-1\le0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m>6\\m\le1\end{matrix}\right.\)

-4m ≤ -5

⇔ m ≥ 5/4

Đề sai, ví dụ với \(\left(a;b;c\right)=\left(3;3;2\right)\) thì vế trái xấp xỉ \(2.78< \dfrac{11930}{2821}\)

Lời giải:

a.

\((2-3x^2)^5=\sum\limits_{k=0}^52^k(-3x^2)^{5-k}=\sum\limits_{k=0}^52^k(-3)^{5-k}x^{10-2k}\)

b.

$10-2k=6$

$\Leftrightarrow k=2$

Hệ số gắn với $x^6$ là: \(2^2(-3)^{5-2}=-108\)

\(mx^2+2\left(m-1\right)x+\left(m-3\right)=0\left(1\right)\)

\(+TH_1:a=0\Leftrightarrow m=0\)

Thế \(m=0\) vào \(\left(1\right)\) \(\Rightarrow2.\left(-1\right)x-3=0\Rightarrow-2x-3=0\Rightarrow x=-\dfrac{3}{2}\left(ktm\right)\)

\(+TH_1:a\ne0\Leftrightarrow m\ne0\)

Theo Vi-ét: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{-2m+2}{m}\\x_1x_2=\dfrac{c}{a}=\dfrac{m-3}{m}\end{matrix}\right.\)

\(x_1< 1< x_1\Leftrightarrow\left\{{}\begin{matrix}\Delta>0\\\left(x_1-1\right)\left(x_2-1\right)< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[2\left(m-1\right)\right]^2-4m\left(m-3\right)>0\\x_1x_2-x_1-x_2+1< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4\left(m^2-2m+1\right)-4m^2+12m>0\\x_1x_2-\left(x_1+x_2\right)+1< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4m^2+8m+4-4m^2+12m>0\\\dfrac{m-3}{m}-\left(\dfrac{-2m+2}{m}\right)+1< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}20m+4>0\\\dfrac{m-3}{m}+\dfrac{2m-2}{m}+1< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m>-\dfrac{1}{5}\\m-3+2m-2+m< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m>-\dfrac{1}{5}\\4m-5< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m>-\dfrac{1}{5}\\m< \dfrac{5}{4}\end{matrix}\right.\)

\(KL:m\in\left(-\dfrac{1}{5};\dfrac{5}{4}\right)\)