Question

Tìm a \(\in\) Q biết:

a) \(\frac{2a+8}{5}-\frac{a}{5}\in Z\)b) \(\frac{2a+9}{a+3}-\frac{5a+17}{a+3}-\frac{3a}{a+3}\in Z\)

Answer 1

a)\(\frac{2a+8}{5}-\frac{a}{5}=\frac{a+8}{5}\)

Để \(\frac{2a+8}{5}-\frac{a}{5}\in Z\) thì: \(a+8\in B\left(5\right)\)

b)\(\frac{2a+9}{a+3}-\frac{5a+17}{a+3}-\frac{3a}{a+3}=\frac{2a+9-5a-17-3a}{a+3}=\frac{-6a-8}{a+3}\)

\(=\frac{-6a-18}{a+3}+\frac{10}{a+3}=\frac{-6.\left(a+3\right)}{a+3}+\frac{10}{a+3}=-6+\frac{10}{a+3}\)

Để: \(\frac{2a+9}{a+3}-\frac{5a+17}{a+3}-\frac{3a}{a+3}\in Z\) thì:

\(a+3\inƯ\left(10\right)=\left\{1;-1;2;-2;5;-5;10;-10\right\}\)

=>a = -2;-4;-1;-5;2;-8;7;-13

Answer 2

@Đặng Minh Triều a là số hữu tỉ nha bạn 

Answer 3

Nhưng chỉ tìm đc a là nguyên thôi nhá James Walker

Answer 4

vc ngu thật