1: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{8}{x+1}+10\sqrt{y-3}=6\\\dfrac{8}{x+1}+2\sqrt{y-3}=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y-3}=1\\\dfrac{4}{x+1}+5=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y-3=1\\\dfrac{4}{x+1}=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=4\\x=-3\end{matrix}\right.\)
2:
a: Thay m=2 vào pt, ta được \(x^2-10x+23=0\)
\(\Leftrightarrow x^2-10x+25=2\)
\(\Leftrightarrow\left(x-5\right)^2=2\)
hay \(x\in\left\{\sqrt{2}+5;-\sqrt{2}+5\right\}\)
b: \(\text{Δ}=\left(2m+6\right)^2-4\left(6m+11\right)\)
\(=4m^2+24m+36-24m-44\)
\(=4m^2-8\)
Để phương trình có hai nghiệm phân biệt thì \(4m^2-8>0\)
hay \(m\in R\backslash\left[-\sqrt{2};\sqrt{2}\right]\)
