Question

[Sawada Tsuna Yoshi]

Tìm m, n \(\in Z\) biết \(\dfrac{1}{m}+\dfrac{n}{6}=\dfrac{1}{2}\)

Answer 1

[Hoang Hung Quan]

Giải:

Ta có: \(\dfrac{1}{m}+\dfrac{n}{6}=\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{1}{m}=\dfrac{1}{2}-\dfrac{n}{6}\)

\(\Leftrightarrow\dfrac{1}{m}=\dfrac{3}{6}-\dfrac{n}{6}=\dfrac{3-n}{6}\)

\(\Leftrightarrow1.6=6=m\left(3-n\right)\)

Mà \(6=1.6=2.3=\left(-1\right).\left(-6\right)=\left(-2\right).\left(-3\right)\)

Ta có bảng sau:

\(m\)	\(1\)	\(-1\)	\(6\)	\(-6\)	\(2\)	\(-2\)	\(3\)	\(-3\)
\(3-n\)	\(6\)	\(-6\)	\(1\)	\(-1\)	\(3\)	\(-3\)	\(2\)	\(-2\)
\(n\)	\(-3\)	\(9\)	\(2\)	\(4\)	\(0\)	\(6\)	\(1\)	\(5\)

Vậy...

Answer 2

[DANGBAHAI]

Ta có \(\dfrac{1}{m}+\dfrac{n}{6}=\dfrac{1}{2}\)

\(\Rightarrow\dfrac{1}{m}=\dfrac{1}{2}-\dfrac{n}{6}\)

\(\Rightarrow\dfrac{1}{m}=\dfrac{3}{6}-\dfrac{n}{6}\)

\(\Rightarrow\dfrac{1}{m}=\dfrac{3-n}{6}\)

\(\Rightarrow1\times6=\left(3-n\right)\times m\)

\(\Rightarrow6=\left(3-n\right)\times m\)

\(\Rightarrow\left(3-n\right);m\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)

\(\Rightarrow\left(3-n\right)\times m=6=(-1)\times\left(-6\right)=(-6)\times\left(-1\right)=\left(-2\right)\times\left(-3\right)=\left(-3\right)\times\left(-2\right)=1\times6=6\times1=2\times3=3\times2\)

Ta có bảng sau

3-n	-6	-3	-2	-1	1	2	3	6
m	-1	-2	-3	-6	6	3	2	1
n	9	6	5	4	2	1	0	-3

Vậy các cặp m,n thỏa mãn là

m	-1	-2	-3	-6	1	2	3	6
n	9	6	5	4	2	1	0	-3