Question

Tìm giá trị nguyên của n để biểu thức sau nguyên:

A=\(\dfrac{n-2}{n+3}\)

B=\(\dfrac{2n-1}{n+1}\)

C=\(\dfrac{2n+3}{n+2}\)

Answer 1

\(A=\dfrac{n-2}{n+3}\)

\(A\) là số nguyên \(\Leftrightarrow n+3=1\)

\(\Leftrightarrow n=-2\)

\(B=\dfrac{2n-1}{n+1}\)

\(B\) là số nguyên \(\Leftrightarrow n+1=1\)

\(\Leftrightarrow n=0\)

\(C=\dfrac{2n+3}{n+2}\)

\(C\) là số nguyên \(\Leftrightarrow n+2=1\)

\(\Leftrightarrow n=-1\)

Answer 2

Ta có:A=\(\dfrac{n-2}{n+3}=\dfrac{\left(n+3\right)-5}{n+3}=1-\dfrac{5}{n+3}\)

Để A∈Z=>\(\dfrac{5}{n+3}\)∈Z

=>5⋮ n+3

=>n+3∈Ư₍₅₎=\(\left\{\pm1;\pm5\right\}\)

=>n∈\(\left\{-2;-4;2;-8\right\}\)

Ta có:B=\(\dfrac{2n-1}{n+1}=\dfrac{2\left(n+1\right)-3}{n+1}=2-\dfrac{3}{n+1}\)

Để B∈Z=>\(\dfrac{3}{n+1}\)∈Z=>3⋮n+1

=>n+1∈Ư(3)=\(\left\{\pm1;\pm3\right\}\)

=>n∈\(\left\{0;-2;2;-4\right\}\)

ta có :C=\(\dfrac{2n+3}{n+2}=\dfrac{2.\left(n+2\right)-1}{n+2}=2-\dfrac{1}{n+2}\)

Để C∈Z=>\(\dfrac{1}{n+2}\)∈Z=>1⋮n+2

=>n+2∈Ư₍₁₎=\(\pm\)1

=>n=-1;-3