\(x\inƯ\left(63\right)\)
\(Ư\left(63\right)=\left\{1;3;7;9;21;63\right\}\)
\(x+1\inƯ\left(63\right)\Rightarrow x=1-1=0\)
\(x=3-2=1\)
\(x=7-1=6\)
\(x=9-8=1\)
\(x=21-1=20\)
\(x=63-1=62\)
\(\Rightarrow x\in\left\{0;2;6;8;20;62\right\}\)
Ta có x<63 => \(\frac{63}{x+1}=\frac{63}{62+1}=\frac{63}{63}=1\)1
Vậy x=62 (t/m)
để rút gon được thì 63 chia hết cho x+1
=> x+1 thuộc Ư(63)
mà Ư(63)=1;3;7;9;21;63
=> X+1=1;3;7;9;21;63
x=0;2;6;8;20;62
Để phân số \(\frac{63}{x+1}\)rút gọn được
\(\Rightarrow63⋮x+1\)
\(\Rightarrow x+1\inƯ\left(63\right)=\left\{1;3;7;9;21;63\right\}\)
mà \(x< 63\)
\(\Rightarrow x\in\left\{1;3;7;9;21\right\}\)
Vậy \(x\in\left\{1;3;7;9;21\right\}\)thì phân số \(\frac{63}{x+1}\)rút gọn được .
Gọi \(k\inƯC\left(63;x+1\right)\)
\(\Leftrightarrow\frac{63}{x+1}=\frac{63⋮k}{\left(x+1\right)⋮k}\)
\(\Leftrightarrow k\inƯ\left(63\right)\) \(\left(k\ne\pm1\right)\)
\(\LeftrightarrowƯ\left(63\right)\in\left\{\pm3;\pm7;\pm9;\pm21\right\}\)
Nhận xét:
\(\left(a\right)x+1=3t\) \(\left(t\in Z,t\ne0\right)\)
\(\Leftrightarrow x=3t-1\)
\(\left(b\right)x+1=7m\) \(\left(m\in Z,m\ne0\right)\)
\(\Leftrightarrow x=7m-1\)
Vậy \(x=3t-1;x=7m-1\)\(\left(t,m\in Z;t,m\ne0\right)\)
\(\left(-21\le t\le21;-9\le m\le9\right)\)