Bài 3:
28: ĐKXĐ: \(\left[\begin{array}{l}x>1\\ x<-1\end{array}\right.\)
Ta có: \(x+\frac{x}{\sqrt{x^2-1}}=\frac{35}{12}\) (1)
=>x>0
=>x>1
(1)=>\(x^2+\frac{x^2}{x^2-1}+\frac{2x^2}{\sqrt{x^2-1}}=\left(\frac{35}{12}\right)^2=\frac{1225}{144}\)
=>\(\frac{x^4-x^2+x^2}{x^2-1}+\frac{2x^2}{\sqrt{x^2-1}}=\frac{1225}{144}\)
=>\(\frac{x^4}{x^2-1}+2\cdot\frac{x^2}{\sqrt{x^2-1}}\cdot1+1=\frac{1225}{144}+1=\frac{1369}{144}=\left(\frac{37}{12}\right)^2\)
=>\(\left(\frac{x^2}{\sqrt{x^2-1}}+1\right)^2=\left(\frac{37}{12}\right)^2\)
=>\(\frac{x^2}{\sqrt{x^2-1}}+1=\frac{37}{12}\)
=>\(\frac{x^2}{\sqrt{x^2-1}}=\frac{25}{12}\)
=>\(12x^2=25\sqrt{x^2-1}\)
=>\(144x^4=625\left(x^2-1\right)=625x^2-625\)
=>\(144x^4-625x^2+625=0\)
=>\(144x^4-225x^2-400x^2+625=0\)
=>\(9x^2\left(16x^2-25\right)-25\left(16x^2-25\right)=0\)
=>\(\left(9x^2-25\right)\left(16x^2-25\right)=0\)
=>\(\left[\begin{array}{l}9x^2-25=0\\ 16x^2-25=0\end{array}\right.\Rightarrow\left[\begin{array}{l}9x^2=25\\ 16x^2=25\end{array}\right.\Rightarrow\left[\begin{array}{l}x^2=\frac{25}{9}\\ x^2=\frac{25}{16}\end{array}\right.\)
=>\(\left[\begin{array}{l}x\in\left\lbrace\frac53;-\frac53\right\rbrace\\ x\in\left\lbrace\frac54;-\frac54\right\rbrace\end{array}\right.\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\lbrace\frac53;\frac54\right\rbrace\)
bài 2:
27:
ĐKXĐ: x>3 hoặc x<=-1
\(\left(x-3\right)\left(x+1\right)+4\left(x-3\right)\cdot\sqrt{\frac{x+1}{x-3}}=-3\)
=>\(\left(x-3\right)\left(x+1\right)+4\cdot\sqrt{\left(x-3\right)\left(x+1\right)}+3=0\)
=>\(\left(\sqrt{\left(x-3\right)\left(x+1\right)}+3\right)\left(\sqrt{\left(x-3\right)\left(x+1\right)}+1\right)=0\) (vô lý)
=>Phương trình vô nghiệm
