Question

Tìm cặp số tự nhiên a, b sao cho

a, 1/a=1/b+b/3

b, a/4-1/b=3/4

Answer 1

a, \(\dfrac{1}{a}=\dfrac{1}{b}+\dfrac{b}{3}\)

<=> \(\dfrac{1}{a}-\dfrac{1}{b}=\dfrac{b}{3}\)

<=> \(\dfrac{3b}{3ab}-\dfrac{3a}{3ab}=\dfrac{ab}{3ab}\)

<=> 3b - 3a = ab
<=> 3b - ab = 3a
<=> b(3 - a) = 3a
<=> 9 - b(3 - a) = 9 - 3a
<=> 9 - b(3 - a) = 3(3 - a)
<=> 9 = b(3 - a) + 3(3 - a)
<=> 9 = (b + 3)(3 - a)
Do a, b \(\in N\Rightarrow3-a;b+3\in N\)
Mà (3 - a)(b + 3) = 9
=> 3 - a; b + 3 \(\inƯ\left(9\right)=\left\{1;3;9\right\}\)
Nếu \(\left\{{}\begin{matrix}3-a=1\\b+3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=6\end{matrix}\right.\) (thỏa mãn)

Nếu \(\left\{{}\begin{matrix}3-a=3\\b+3=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0\\b=0\end{matrix}\right.\) (thỏa mãn)

Nếu \(\left\{{}\begin{matrix}3-a=9\\b+3=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-6\\b=-2\end{matrix}\right.\) (không thỏa mãn)

Vậy các cặp (a; b) thỏa mãn là (2; 6); (0; 0)
@Quỳnh Chi

Answer 2

b, \(\dfrac{a}{4}-\dfrac{1}{b}=\dfrac{3}{4}\)

<=> \(\dfrac{a}{4}-\dfrac{3}{4}=\dfrac{1}{b}\)

<=> \(\dfrac{ab}{4b}-\dfrac{3b}{4b}=\dfrac{4}{4b}\)

<=> ab - 3b = 4
<=> (a - 3)b = 4
Do a; b \(\in N\Rightarrow a-3\in N\)
Mà (a - 3)b = 4
=> a - 3; b \(\inƯ\left(4\right)=\left\{1;2;4\right\}\)
Nếu \(\left\{{}\begin{matrix}a-3=1\\b=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=4\\b=4\end{matrix}\right.\) (thỏa mãn)

Nếu \(\left\{{}\begin{matrix}a-3=2\\b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=5\\b=2\end{matrix}\right.\) (thỏa mãn)

Nếu \(\left\{{}\begin{matrix}a-3=4\\b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=7\\b=1\end{matrix}\right.\) (thỏa mãn)

Vậy các cặp (a; b) thỏa mãn là (4; 4); (5; 2); (7; 1)
@Quỳnh Chi