Question

1.Tính nhanh

A =( \(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\)) x \(\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)

2.Tìm n ∈ Z để \(\dfrac{n+3}{n-2}\)nhận giá trị nguyên 

Answer 1

1. \(\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)

\(=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right).0\)

\(=0\)

Answer 2

Bài 1: 

Ta có: \(A=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)

\(=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)\)

=0

Answer 3

Answer 4

Bài 2:

Để \(\dfrac{n+3}{n-2}\) là số nguyên thì \(n+3⋮n-2\)

\(\Leftrightarrow n-2+5⋮n-2\)

mà \(n-2⋮n-2\)

nên \(5⋮n-2\)

\(\Leftrightarrow n-2\inƯ\left(5\right)\)

\(\Leftrightarrow n-2\in\left\{1;-1;5;-5\right\}\)

hay \(n\in\left\{3;1;7;-3\right\}\)

Vậy: \(n\in\left\{3;1;7;-3\right\}\)