a) \(\frac{x-1}{3}=\frac{x+3}{5}\)
\(\Rightarrow5\left(x-1\right)=3\left(x+3\right)\)
\(\Rightarrow5x-5=3x+9\)
\(\Rightarrow5x-3x=9+5\)
\(\Rightarrow2x=14\)
\(\Rightarrow x=7\)
Vậy x = 7
b) \(\frac{x+1}{1}=\frac{1}{x+1}\)
\(\Rightarrow\left(x+1\right)^2=1\)
\(\Rightarrow x+1=\pm1\)
+) \(x+1=1\Rightarrow x=0\)
+) \(x+1=-1\Rightarrow x=-2\)
Vậy x = 0 hoặc x = -2
=> (x - 1).5 = (x + 3).3
=> 5x - 5 = 3x + 9
=> 5x - 3x = 9 + 5
=> 2x = 14
=> x = 14 : 2
=> x = 7
Vậy x = 7
\(\frac{x+1}{1}=\frac{1}{x+1}\)=> (x + 1)2 = 1
\(\Rightarrow\left[\begin{array}{nghiempt}x+1=1\\x+1=-1\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=-2\end{array}\right.\)
Vậy \(\left[\begin{array}{nghiempt}x=0\\x=-2\end{array}\right.\)
