Áp dụng tính chắt dãy tỉ sỗ bằng nhau

Ta có : 5xy + x + y = 4

=> 5(5xy + x + y) = 5 . 4

=> 25xy + 5x + 5y = 20

=> (25xy + 5x) + 5y + 1 = 20 + 1

=> 5x(5y + 1) + (5y + 1) = 21

=> (5x + 1)(5y + 1) = 21

=> 5x + 1;5y + 1 ∈ Ư(21) ∈ {-21;-7;-3;-1;1;3;7;21}

Mà 5x + 1;5y + 1 chia 5 dư  1

=> 5x + 1;5y + 1 ∈ {1;21}

=>(5x + 1;5y + 1) ∈ {(1;21);(21;1)}

=> (x;y) ∈ {(0;4);(4;0)}

5xy - 5x + y = 5

<=> 5xy = 5 + 5x - y

<=> \(\left\{{}\begin{matrix}x=\dfrac{5+5x-y}{5y}\\y=\dfrac{5+5x-y}{5x}\end{matrix}\right.\)

\(5xy-5x+y=5\)

\(\Rightarrow5x\left(y-1\right)+\left(y-1\right)=4\)

\(\Rightarrow\left(y-1\right)\left(5x+1\right)=4\)

Do \(x,y\in Z\)

TH1: \(\left\{{}\begin{matrix}y-1=1\\5x+1=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=2\left(tm\right)\\x=-\dfrac{3}{5}\left(ktm\right)\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}y-1=4\\5x+1=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=5\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)

TH3: \(\left\{{}\begin{matrix}y-1=2\\5x+1=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=3\left(tm\right)\\x=\dfrac{1}{5}\left(ktm\right)\end{matrix}\right.\)

TH4: \(\left\{{}\begin{matrix}y-1=-2\\5x+1=-2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-1\left(tm\right)\\x=-\dfrac{3}{5}\left(ktm\right)\end{matrix}\right.\)

TH5: \(\left\{{}\begin{matrix}y-1=-1\\5x+1=-4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=0\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

TH6: \(\left\{{}\begin{matrix}y-1=-4\\5x+1=-1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=3\left(tm\right)\\x=-\dfrac{2}{5}\left(ktm\right)\end{matrix}\right.\)

Vậy \(\left(x;y\right)\in\left\{\left(0;5\right);\left(-1;0\right)\right\}\)

5xy-5x+y=5

5xy-5xy=5

X=55

Y=50

5xy-5x+y=5

 suy ra 5xy-5x+y-1=6

 suy ra 5x(y-1)+(y-1)=6

 suy ra (5x+1)(y-1)=6

x,y thuôc Z  suy ra 5x+1,y-1 thuộc Z

 suy ra 5x+1,y-1 thuộc Ư(6)

Ta có bảng sau:

\(\frac{x}{7}+\frac{y}{11}+\frac{z}{13}=\frac{947}{1001}\)

\(\Leftrightarrow143x+91y+77z=947=143.3+91.4+77.2\)

Vậy x = 3 ; y = 4 và z = 2

đấy là mò làm thế bà ra từ 80 đời rồi cháu ạ