a, Với x khác 0 ; -1
\(M=\dfrac{x^2+2x+3x+3-3}{x\left(x+1\right)}=\dfrac{x^2+5x}{x\left(x+1\right)}=\dfrac{x+5}{x+1}\)
b, Ta có \(x^2-1=0\Leftrightarrow x=1;x=-1\left(ktmđk\right)\)
Thay x = 1 vào M ta được \(M=\dfrac{1+5}{1+1}=\dfrac{6}{2}=3\)
c, Ta có \(\dfrac{M}{N}=\dfrac{x+5}{x+1}:\dfrac{x}{x+1}=\dfrac{x+5}{x}=1+\dfrac{5}{x}\Rightarrow x\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Kết hợp đk vậy x = 1 ; 5 ; -5
a.
\(M=\dfrac{x+2}{x+1}+\dfrac{3}{x}-\dfrac{3}{x\left(x+1\right)}\)
\(M=\dfrac{x\left(x+2\right)+3\left(x+1\right)-3}{x\left(x+1\right)}\)
\(M=\dfrac{x^2+2x+3x+3-3}{x\left(x+1\right)}\)
\(M=\dfrac{x^2+5x}{x\left(x+1\right)}\)
\(M=\dfrac{x\left(x+5\right)}{x\left(x+1\right)}\)
\(M=\dfrac{x+5}{x+1}\)
b.
\(N=\dfrac{x}{x+1}\)
\(N=\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(N=\dfrac{x^2-x}{x^2-1}\)
\(N=\dfrac{x^2-x}{0}\)
Vậy N ko có giá trị khi \(x^2-1=0\)
c.\(P=\dfrac{x+5}{x+1}:\dfrac{x}{x+1}\)
\(P=\dfrac{\left(x+5\right)\left(x+1\right)}{x\left(x+1\right)}\)
\(P=\dfrac{x+5}{x}\)
\(P=\dfrac{x}{x}+\dfrac{5}{x}\)
\(P=1+\dfrac{5}{x}\)
Để P nhận giá trị nguyên dương thì \(5⋮x\) hay \(x\in U\left(5\right)=\left\{1;5\right\}\) ( vì để P dương )
\(\Rightarrow x=\left\{1;5\right\}\)
Vậy \(x=\left\{1,5\right\}\) thì P nhận giá trị nguyên dương
