1.

Nếu \(m=0\), \(f\left(x\right)=2x\)

\(\Rightarrow m=0\) không thỏa mãn

Nếu \(x\ne0\)

Yêu cầu bài toán thỏa mãn khi \(\left\{{}\begin{matrix}m< 0\\\Delta'=\left(m-1\right)^2-4m^2< 0\end{matrix}\right.\)

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

2.

\(\dfrac{-x^2+2x-5}{x^2-mx+1}\le0\forall x\)

\(\Leftrightarrow\dfrac{-\left(x-1\right)^2-4}{x^2-mx+1}\le0\forall x\)

\(\Leftrightarrow x^2-mx+1>0\forall x\)

\(\Leftrightarrow\Delta=m^2-4< 0\Leftrightarrow-2< m< 2\)

Kết luận: \(-2< m< 2\)

1, BPT đúng với mọi x thuộc R khi vầ chỉ khi:

\(\left\{{}\begin{matrix}a>0\\\Delta\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a>0\\1-4a^2\le0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a>0\\a\le\frac{-1}{2};a\ge\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow a\ge\frac{1}{2}\)

2, điều kiện: \(\Delta< 0\\ \Leftrightarrow\left(m+2\right)^2+8\left(m-4\right)< 0\\ \Leftrightarrow m^2+12m-28< 0\\ \Leftrightarrow-14< m< 2\)

3, điều kiện: \(\Delta'< 0\\ \Leftrightarrow\left(2m-3\right)^2-\left(4m-3\right)< 0\\ \Leftrightarrow m^2-4m+3< 0\\ \Leftrightarrow1< m< 3\)

4, Nếu m=0 => f(x)=-2x-1<0 (loại)

Nếu m≠0 để f(x)<0 với ∀x ϵ R khi và chỉ khi:

\(\left\{{}\begin{matrix}m< 0\\\Delta'< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m< 0\\1+m< 0\end{matrix}\right.\)

\(\Rightarrow m< -1\)

Bài 1: Bài này tớ làm không đảm bảo đúng 100% nên nếu có gì sai sót mong bạn thông cảm

a) Nếu F(x) = G(x)

\(\Rightarrow ax+b-mx-n=0\)

\(\Rightarrow x\left(a-m\right)+\left(b-n\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x\left(a-m\right)=0\\b-n=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a-m=0\\b=n\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=m\\b=n\end{matrix}\right.\)

b) Nếu F(x) = G(x)

\(\Rightarrow ax^2+bx+c-mx^2-nx-p=0\)

\(\Rightarrow x^2\left(a-m\right)+x\left(b-n\right)+\left(c-p\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x^2\left(a-m\right)=0\\x\left(b-n\right)=0\\c-p=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a-m=0\\b-n=0\\c-p=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=m\\b=n\\c=p\end{matrix}\right.\)

Bài 2:

a) \(A\left(x\right)=0\)

\(\Leftrightarrow2\left(\dfrac{1}{3}x-\dfrac{1}{2}\right)-\dfrac{1}{2}\left(3-x\right)=0\)

\(\Leftrightarrow2.\dfrac{1}{3}x-2.\dfrac{1}{2}-\dfrac{1}{2}.3+\dfrac{1}{2}x=0\)

\(\Leftrightarrow\dfrac{2}{3}x-1-\dfrac{3}{2}+\dfrac{1}{2}x=0\)

\(\Leftrightarrow\dfrac{7}{6}x-\dfrac{5}{2}=0\)

\(\Leftrightarrow\dfrac{7}{6}x=\dfrac{5}{2}\)

\(\Leftrightarrow x=\dfrac{15}{7}\)

b) Nếu B (x) = 0

\(\Leftrightarrow\left(2x-5\right)\left(x^2-\dfrac{9}{16}\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-5=0\\x^2-\dfrac{9}{16}=0\\x^2+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=5\\x^2=\dfrac{9}{16}\\x^2=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{3}{4};x=-\dfrac{3}{4}\\x=1;x=-1\end{matrix}\right.\)

c) Nếu C(x) = 0

\(\Leftrightarrow x^3-2x=0\)

\(\Leftrightarrow x\left(x^2-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x^2=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{2};x=-\sqrt{2}\end{matrix}\right.\)

d) Nếu D(x) = 0

\(\Leftrightarrow9x^2+16=0\)

\(\Leftrightarrow9x^2=-16\)

\(\Leftrightarrow x^2=-\dfrac{16}{9}\)

Vậy không tồn tại x thỏa mãn

e) Nếu M(x) = 0

\(\Leftrightarrow x^2+4x+4=0\)

\(\Leftrightarrow\left(x+2\right)^2=0\)

\(\Leftrightarrow x+2=0\)

\(\Leftrightarrow x=-2\)

a: f(x) chia hết cho g(x)

=>2x^2+4x-x-2+a+2 chia hết cho x+2

=>a+2=0

=>a=-2

b: f(x) chia hết cho g(x)

=>3x^2+6x+(m-6)x+2m-12-2m+7 chia hết cho x+2

=>-2m+7=0

=>m=7/2