37: ĐKXĐ: x>3 hoặc x<-1
Ta có: \(\left(x-3\right)\left(x+1\right)+4\left(x-3\right)\cdot\sqrt{\frac{x+1}{x-3}}=-3\)
=>\(\left(x-3\right)\left(x+1\right)+4\cdot\sqrt{\left(x-3\right)\left(x+1\right)}+3=0\)
=>\(\left(\sqrt{\left(x-3\right)\left(x+1\right)}+3\right)\left(\sqrt{\left(x-3\right)\left(x+1\right)}+1\right)=0\)
mà \(\sqrt{\left(x-3\right)\left(x+1\right)}+3\ge3>0\forall x\) thỏa mãn ĐKXĐ
và \(\sqrt{\left(x-3\right)\left(x+1\right)}+1\ge1>0\forall x\) thỏa mãn ĐKXĐ
nên x∈∅
19:
ĐKXĐ: \(\begin{cases}x<>0\\ x^2+x-5<>0\end{cases}\Rightarrow\begin{cases}x<>0\\ x^2+x+\frac14-\frac{21}{4}<>0\end{cases}\)
=>\(\begin{cases}x<>0\\ \left(x+\frac12\right)^2<>\frac{21}{4}\end{cases}\Rightarrow\begin{cases}x<>0\\ x+\frac12<>\frac{\sqrt{21}}{2}\\ x+\frac12<>-\frac{\sqrt{21}}{2}\end{cases}\)
=>\(x\notin\left\lbrace0;\frac{\sqrt{21}-1}{2};\frac{-\sqrt{21}-1}{2}\right\rbrace\)
\(\frac{x^2+x-5}{x}+\frac{3x}{x^2+x-5}+4=0\)
=>\(\frac{\left(x^2+x-5\right)^2+3x^2+4x\left(x^2+x-5\right)}{x\left(x^2+x-5\right)}=0\)
=>\(\left(x^2+x-5\right)^2+4x\left(x^2+x-5\right)+3x^2=0\)
=>\(\left(x^2+x-5+x\right)\left(x^2+x-5+3x\right)=0\)
=>\(\left(x^2+2x-5\right)\left(x^2+4x-5\right)=0\)
TH1: \(x^2+2x-5=0\)
=>\(x^2+2x+1-6=0\)
=>\(\left(x+1\right)^2=6\)
=>\(x+1=\pm\sqrt6\)
=>\(\left[\begin{array}{l}x=\sqrt6-1\left(nhận\right)\\ x=-\sqrt6-1\left(nhận\right)\end{array}\right.\)
TH2: \(x^2+4x-5=0\)
=>(x+5)(x-1)=0
=>\(\left[\begin{array}{l}x+5=0\\ x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-5\left(nhận\right)\\ x=1\left(nhận\right)\end{array}\right.\)
