Question

[Trần Văn Thành]

Tìm a € N để biểu thức sau có giá trị nguyên:

\(\frac{40\left|2a-1\right|+15}{10a-5}\)

Answer 1

[Phùng Minh Quân]

Ta có : 

\(\left|2a-1\right|=\orbr{\begin{cases}2a-1\left(a>0\right)\\1-2a\left(a=0\right)\end{cases}}\)

Đặt \(A=\frac{40\left|2a-1\right|+15}{10a-5}\) 

+) Xét \(a>0\) ta có : 

\(A=\frac{40\left|2a-1\right|+15}{10a-5}\)

\(A=\frac{40\left(2a-1\right)+15}{10a-5}\)

\(A=\frac{80a-40+15}{10a-5}\)

\(A=\frac{80a-40}{10a-5}+\frac{15}{10a-5}\)

\(A=\frac{8\left(10a-5\right)}{10a-5}+\frac{15}{10a-5}\)

\(A=8+\frac{15}{10a-5}\)

Để A nguyên thì \(\frac{15}{10a-5}\) nguyên hay  \(15⋮\left(10a-5\right)\)\(\Rightarrow\)\(\left(10a-5\right)\inƯ\left(15\right)\)

Mà \(Ư\left(15\right)=\left\{1;-1;3;-3;5;-5;15;-15\right\}\)

Suy ra : 

\(10a-5\)	\(1\)	\(-1\)	\(3\)	\(-3\)	\(5\)	\(-5\)	\(15\)	\(-15\)
\(a\)	\(\frac{3}{5}\)	\(\frac{2}{5}\)	\(\frac{4}{5}\)	\(\frac{1}{5}\)	\(1\)	\(0\)	\(2\)	\(-1\)

Mà \(a\inℕ\left(a>0\right)\) nên \(a\in\left\{-1;0;1;2\right\}\)

+) Xét \(a=0\) ta có : 

\(A=\frac{40\left|2a-1\right|+15}{10a-5}\)

\(A=\frac{40\left|2.0-1\right|+15}{10.0-5}\)

\(A=\frac{40\left|0-1\right|+15}{0-5}\)

\(A=\frac{40+15}{-5}\)

\(A=-11\) ( A nguyên ) 

Vậy \(a\in\left\{-1;0;1;2\right\}\)

Chúc bạn học tốt ~ 

Answer 2

[Nguyễn Tất Đạt]

Đặt \(A=\frac{40\left|2a-1\right|+15}{10a-5}\)

\(\left|2a-1\right|=2a-1\)

\(\Rightarrow A=\frac{40.\left(2a-1\right)+15}{10a-5}=\frac{80a-40+15}{10a-5}=\frac{80a-25}{10a-5}\)

Để biểu thức A nhận giá trị nguyên thì \(80a-25⋮10a-5\)

Ta có: \(8\left(10a-5\right)⋮10a-5\)\(\Rightarrow80a-40⋮10a-5\)

\(\Rightarrow80a-25-\left(80a-40\right)⋮10a-5\)

\(\Rightarrow15⋮10a-5\Rightarrow\)\(10a-5\)thuộc Ư(15)

\(Ư\left(15\right)=\left\{1;3;5;15;-1;-3;-5;-15\right\}\)

\(\Rightarrow10a-5\in\left\{1;3;5;15;-1;-3;-5;-15\right\}\)

\(\Rightarrow10a\in\left\{6;8;10;4;3;0;-10\right\}\Rightarrow a\in\left\{\frac{3}{5};\frac{4}{5};1;\frac{2}{5};\frac{3}{10};0;-1\right\}\)

Do \(a\in N\)nên \(a\in\left\{1;0\right\}\)