Chương 4: BẤT ĐẲNG THỨC, BẤT PHƯƠNG TRÌNH

\(f\left(x\right)< 0\forall x\in R.\\ \Leftrightarrow\left\{{}\begin{matrix}m< 0.\\\Delta'=m^2-2m+1-4m^2< 0.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m< 0.\\\Delta'=-3m^2-2m+1< 0.\end{matrix}\right.\)

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

\(\Leftrightarrow m< -1.\\ \Leftrightarrow m\in\left(-\infty;-1\right).\)

\(f\left(x\right)=x^2-2\left(m-1\right)x+4m.\)

Để \(f\left(x\right)< 0\forall x\in R.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a< 0.\\\Delta'< 0.\end{matrix}\right.\)

Mà \(a=1>0.\)

\(\Rightarrow f\left(x\right)< 0\Leftrightarrow x\in\phi.\)

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Mk trl sai câu này r nhg ko xoá đc bn thông cảm nhé.

\(\left|x^2+2x-3\right|< x^2+3\\ \Leftrightarrow\left\{{}\begin{matrix}-x^2-3< x^2+2x-3\\x^2+2x-3< x^2+3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x^2+2x>0\\2x+6< 0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x< -1\\x>0\end{matrix}\right.\\x< -3\end{matrix}\right.\\ \Leftrightarrow x< -3\)

Đặt \(x^2+3x=a\left(a>=-\dfrac{9}{4}\right)\)

BPT sẽ trở thành \(a>=2+\sqrt{5a+14}\)

=>\(a-2>=\sqrt{5a+14}\)

=>\(\sqrt{5a+14}< =a-2\)

=>\(\left\{{}\begin{matrix}a-2>=0\\5a+14< =\left(a-2\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a>=2\\5a+14-a^2+4a-4< =0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a>=2\\-a^2+9a+10< =0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a>=2\\a^2-9a-10>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a>=2\\\left(a-10\right)\left(a+1\right)>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a>=2\\\left[{}\begin{matrix}a>=10\\a< =-1\end{matrix}\right.\end{matrix}\right.\)

=>a>=10

=>\(x^2+3x>=10\)

=>\(x^2+3x-10>=0\)

=>(x+5)(x-2)>=0

=>\(\left[{}\begin{matrix}x>=2\\x< =-5\end{matrix}\right.\)

Trường hợp 1: m=0

=>-3<0(luôn đúng)

=>Nhận

Trường hợp 2: m<>0

\(\text{Δ}=\left(2m\right)^2-4\cdot m\cdot\left(-3\right)=4m^2+12m=4m\left(m+3\right)\)

Để phương trình có nghiệm đúng thì \(\left\{{}\begin{matrix}4m\left(m+3\right)< 0\\m< 0\end{matrix}\right.\Leftrightarrow-3< m< 0\)

Vậy: -3<m<=0

Câu 25:

\(p=\dfrac{a+b+c}{2}=\dfrac{39+25+16}{2}=\dfrac{80}{2}=40.\)

Áp dụng công thức Herong: \(S=\sqrt{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}\) ta có:

\(S=\sqrt{40\left(40-39\right)\left(40-25\right)\left(40-16\right)}=120\) (đvdt).

Ta có: \(S=\dfrac{abc}{4R}.\) \(\Rightarrow120=\dfrac{39.25.16}{4R}.\\ \Leftrightarrow120=\dfrac{3900}{R}.\\ \Leftrightarrow R=32,5.\)

Do \(a=1>0\) nên \(f\left(x\right)\ge0;\forall x\) khi:

\(\Delta'=\left(m-3\right)^2-4m\le0\)

\(\Leftrightarrow m^2-10m+9\le0\)

\(\Rightarrow1\le m\le9\)