(d₃₎//(d₄₎\(\Leftrightarrow\left\{{}\begin{matrix}m^2+6m=7\\2n+7\ne-n^2-9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}m^2+6m-7=0\\n^2+2n+16\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(m-1\right)\left(m+7\right)=0\\\left(n+1\right)^2+15\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m-1=0\\m+7=0\end{matrix}\right.\\\left(n+1\right)^2+15\ne0\left(luônđúng\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=1\\m=-7\end{matrix}\right.\)

\(\left(d3\right)\equiv\left(d4\right)\Leftrightarrow\left\{{}\begin{matrix}m^2+6m=7\\2n+7=-n^2-9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m^2+6m-7=0\\n^2+2n+16=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(m-1\right)\left(m+7\right)=0\\\left(n+1\right)^2+15=0\left(vôlí\right)\end{matrix}\right.\)

\(a.\left[bn\right]=b.\left[an\right]\)

\(\Rightarrow\frac{a}{b}=\frac{an}{bn}\)

\(\Rightarrow\frac{a}{b}=\frac{a}{b}\)

\(\Rightarrow\left(a,b\right)\in R\)

Bài 1: Theo đề bài: \(VT=\left(a-1\right)+\frac{1}{\left(a-1\right)}+1\ge2\sqrt{\left(a-1\right).\frac{1}{a-1}}+1=2+1=3^{\left(đpcm\right)}\)

Đẳng thức xảy ra khi \(\left(a-1\right)=\frac{1}{a-1}\Leftrightarrow a=2\)

Bài 2: \(BĐT\Leftrightarrow\left(a^2+2\right)^2\ge4\left(a^2+1\right)\)

\(\Leftrightarrow a^4+4a^2+4\ge4a^2+4\)

\(\Leftrightarrow a^4\ge0\) (đúng). Đẳng thức xảy ra khi a = 0

Bài 3: Hình như sai đề thì phải ạ. Nếu a = 1,5 ; b = 1 thì \(\frac{19}{10}=1,9< 3\)