Câu 1:
Tại \(x=5\) thì ta có pt:
\(pt\Leftrightarrow10+4m^2=19\)
\(\Leftrightarrow4m^2=9\Leftrightarrow m^2=\dfrac{9}{4}\)
\(\Leftrightarrow m=\pm\sqrt{\dfrac{9}{4}}=\pm\dfrac{3}{2}\)
Vậy với \(m=\pm\dfrac{3}{2}\) thì pt có nghiệm là \(x=5\)
Câu 2:
\(\dfrac{x+5}{1999}+\dfrac{x+7}{1997}=\dfrac{x+9}{1995}+\dfrac{x+11}{1993}\)
\(\Leftrightarrow\dfrac{x+5}{1999}+1+\dfrac{x+7}{1997}+1=\dfrac{x+9}{1995}+1+\dfrac{x+11}{1993}+1\)
\(\Leftrightarrow\dfrac{x+2004}{1999}+\dfrac{x+2004}{1997}=\dfrac{x+2004}{1995}+\dfrac{x+2004}{1993}\)
\(\Leftrightarrow\dfrac{x+2004}{1999}+\dfrac{x+2004}{1997}-\dfrac{x+2004}{1995}-\dfrac{x+2004}{1993}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{1999}+\dfrac{1}{1997}-\dfrac{1}{1995}-\dfrac{1}{1993}\right)=0\)
\(\Rightarrow x+2004=0\). Do \(\dfrac{1}{1999}+\dfrac{1}{1997}-\dfrac{1}{1995}-\dfrac{1}{1993}\ne0\)
\(\Rightarrow x=-2014\)