\(\Leftrightarrow\left|3x-2\right|>x+1\)

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

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

f(1)=-1

=>m+3=-1

hay m=-4

\(\dfrac{x-3}{2+x}< 1\\ \Leftrightarrow\dfrac{x-3}{2+x}-1< 0\\ \Leftrightarrow\dfrac{x-3}{2+x}-\dfrac{2+x}{2+x}< 0\\ \Leftrightarrow\dfrac{x-3-2-x}{2+x}< 0\\ \Leftrightarrow\dfrac{-5}{2+x}< 0\)

Vì \(-5< 0\)

\(\Rightarrow2+x>0\\ \Rightarrow x>-2\)

Vậy \(x>-2\)

\(B=2m^6+3m^3n^3+n^6+n^3\)

\(=2m^6+2m^3n^3+m^3n^3+n^6+n^3\)

\(=2m^3\left(m^3+n^3\right)+n^3\left(m^3+n^3\right)+n^3\)

\(=2m^3+2n^3\)

=2

Đặt \(A=\dfrac{6}{5}+\dfrac{6}{10}+...+\dfrac{6}{640}\)

=>\(2A=\dfrac{12}{5}+\dfrac{6}{5}+...+\dfrac{6}{320}\)

=>\(2A-A=\dfrac{12}{5}-\dfrac{6}{640}=\dfrac{153}{64}\)

=> \(A=\dfrac{153}{64}\)

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