x=1

y=2

Ta có: \(x^2-2xy+5y^2=y+1\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+4y^2-y-1=0\)

\(\Leftrightarrow\left(x-y\right)^2+4y^2-y-1=0\)

Mà \(4y^2-4y-1=3y^2+\left(y^2-y\right)-1\)

\(=3y^2+y\left(y-1\right)-1\ge3\cdot1+0-1=2>0\)

\(\Rightarrow\left(x-y\right)^2+4y^2-y-1>0\)

=> pt vô nghiệm

Câu 4:

5x + 7y = 112

5(x+ y) = 112 - 2y

5(x + y) = 2(56 - y)

\(\begin{cases}x+y=2\\ 56-y=5\end{cases}\)

\(\begin{cases}x+y=2\\ y=56-5\end{cases}\)

\(\begin{cases}x=2-y\\ y=51\end{cases}\)

\(\begin{cases}x=2-51\\ y=51\end{cases}\)

\(\begin{cases}x=-49\\ y=51\end{cases}\)

Vậy (x ; y) = (-49; 51)

1)<=> \(\left(x^2+2x+1\right)-\left(y^2+4x+4\right)-7=0\)

=> \(\left(x+1\right)^2-\left(y+2\right)^2=7\)

\(\Rightarrow\left(x+1-y-2\right)\left(x+1+y+2\right)=7\)

\(\left(x-y-1\right)\left(x+y+3\right)=7\)

vì x;y là số nguyên dương=> \(x;y\ge1\Rightarrow x+y+3\ge1+1+3=5\)

=> \(x+y+3=7\Rightarrow x+y=4\)

\(x-y-1=1\Rightarrow x-y=2\)

cộng hai phương trình ta có:

=> \(\left(x+y\right)+\left(x-y\right)=4+2\)

=> \(2x=6\)

=> x=3

=> y=4-3=1

2)<=> \(\left(x^2+xy+2x\right)+\left(2xy+2y^2+4y\right)+\left(x+y+2\right)=17\)

=> \(x\left(x+y+2\right)+2y\left(x+y+2\right)+1\left(x+y+2\right)=17\)

=> \(\left(x+y+2\right)\left(x+2y+1\right)=17\)

\(3xy+x+15y-44=0\)

\(3y\left(x+5\right)+\left(x+5\right)-49=0\)

\(\left(x+5\right)\left(3y+1\right)=49\)

Vì x;y là số nguyên \(\Rightarrow\hept{\begin{cases}x+5\in Z\\3y+1\in Z\end{cases}}\)

Có \(\left(x+5\right)\left(3y+1\right)=49\)

\(\Rightarrow\left(x+5\right)\left(3y+1\right)\in\text{Ư}\left(49\right)=\left\{\pm1;\pm7;\pm49\right\}\)

b tự lập bảng nhé~

Câu a:

xy - 4x = 35 - 5y

x(y - 4) = -5(y - 4) + 15

x(y -4) + 5(y - 4) = 15

(y -4)(x+ 5) = 15

Lập bảng ta có:

Theo bảng trên ta có: (x; y) = (-20; 3); (-10; 1); (-8; -1); (-6; -11); (-4; 19); (-2; 9); (0; 7); (10; 5)

Vậy x; y) = (-20; 3); (-10; 1); (-8; -1); (-6; -11); (-4; 19); (-2; 9); (0; 7); (10; 5)