\(\frac{ 4}{x + 1} - \frac{1}{x} = 1 \\ \\x+1
eq 0 \ \ \ and \ \ \ x
eq 0 \\ \\x
eq -1 \ \ \ and \ \ \ x
eq 0 \\ \\D=R\setminus \left \{ -1,0 \right \}\)
\(\frac{ 4x- (x+1)}{x(x + 1)} = 1 \\ \\\frac{ 4x- x-1 }{x^2+ x}=1 \\ \\\frac{ 3x -1 }{x^2 + x} = 1 \\ \\3x-1=x^2+x\\ \\ x^2+x-3x+1=0\)
\(x^2-2x+1=0 \\ \\ (x-1)^2=0 \\ \\x-1=0 \\ \\x=1\)
