\(1,\\ \left\{{}\begin{matrix}\dfrac{3}{\sqrt{x-4}}+\dfrac{4}{y+2}=7\\\dfrac{5}{\sqrt{x-4}}-\dfrac{1}{y+2}=4\end{matrix}\right.\left(x>4;y\ne-2\right)\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{\sqrt{x-4}}+\dfrac{4}{y+2}=7\left(1\right)\\\dfrac{20}{\sqrt{x-4}}-\dfrac{4}{y+2}=16\left(2\right)\end{matrix}\right.\\ \left(1\right)+\left(2\right)=\dfrac{23}{\sqrt{x-4}}=23\\ \Leftrightarrow\sqrt{x-4}=1\Leftrightarrow x-4=1\Leftrightarrow x=5\left(tm\right)\\ \Leftrightarrow\dfrac{3}{\sqrt{5-4}}+\dfrac{4}{y+2}=7\\ \Leftrightarrow3+\dfrac{4}{y+2}=7\\ \Leftrightarrow y+2=1\Leftrightarrow y=-1\left(tm\right)\)
Vậy hệ pt có nghiệm \(\left(x;y\right)=\left(5;-1\right)\)
\(2,\\ a,m=1\Leftrightarrow x^2-4x+3=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
\(b,\) Áp dụng hệ thức Viét:
\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{2\left(m+1\right)}{1}=2\left(m+1\right)\\x_1x_2=m^2+2m\end{matrix}\right.\)
\(x_1^2+x_2^2\\ =\left(x_1+x_2\right)^2-2x_1x_2\\ =\left[2\left(m+1\right)\right]^2-2\left(m^2+2m\right)\\ =4m^2+8m+4-2m^2-4m\\ =2m^2+4m+4\\ =2\left(m^2+2m+1\right)+2=2\left(m+1\right)^2+2\ge2\)
Dấu \("="\Leftrightarrow m=-1\)