a) Ta có: (x+1)(y-2)=-2

nên x+1; y-2 là các ước của -2

Trường hợp 1:

\(\left\{{}\begin{matrix}x+1=-1\\y-2=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=4\end{matrix}\right.\)

Trường hợp 2: 

\(\left\{{}\begin{matrix}x+1=2\\y-2=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

Trường hợp 3: 

\(\left\{{}\begin{matrix}x+1=-2\\y-2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=3\end{matrix}\right.\)

Trường hợp 4: 

\(\left\{{}\begin{matrix}x+1=1\\y-2=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

Vậy: (x,y)\(\in\){(-2;4);(1;1);(-3;3);(0;0)}

b) Ta có: (x+1)(xy-1)=3

nên x+1;xy-1 là các ước của 3

Trường hợp 1: 

\(\left\{{}\begin{matrix}x+1=1\\xy-1=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\-1=3\end{matrix}\right.\Leftrightarrow loại\)

Trường hợp 2: 

\(\left\{{}\begin{matrix}x+1=3\\xy-1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\2y-1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\2y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

Trường hợp 3: 

\(\left\{{}\begin{matrix}x+1=-1\\xy-1=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\-2y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=1\end{matrix}\right.\)

Trường hợp 4: 

\(\left\{{}\begin{matrix}x+1=-3\\xy-1=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\-4y-1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=-\dfrac{1}{2}\end{matrix}\right.\left(loại\right)\)

Vậy: \(\left(x,y\right)\in\left\{\left(2;1\right);\left(-2;1\right)\right\}\)

c) Ta có: \(\left(x+y\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-x\\x=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)

Vây: (x,y)=(-1;1)

d) Ta có: \(\left|x+y\right|\cdot\left(x-y\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left|x+y\right|=0\\x-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x=y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2y=0\\x=y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

Vậy: (x,y)=(0;0)

thanks bạn

a:

ĐKXĐ: x<>0; y<>0

\(\frac{2}{x}+\frac{1}{y}=3\)

=>\(\frac{2y+x}{xy}=3\)

=>3xy=x+2y

=>3xy-x-2y=0

=>x(3y-1)-\(2y+\frac23=\frac23\)

=>\(3x\left(y-\frac13\right)-2\left(y-\frac13\right)=\frac23\)

=>\(\left(3x-2\right)\left(y-\frac13\right)=\frac23\)

=>(3x-2)(3y-1)=2

=>(3x-2;3y-1)∈{(1;2);(2;1);(-1;-2);(-2;-1)}

=>(3x;3y)∈{(3;3);(4;2);(1;-1);(0;0)}

=>(x;y)∈{(1;1);(4/3;2/3);(1/3;-1/3);(0;0)}

mà x,y nguyên

nên x=1; y=1

b: ĐKXĐ: x<>0; y<>0

\(\frac{2}{y}-\frac{1}{x}=\frac{8}{xy}+1\)

=>\(\frac{2x-y}{xy}=\frac{8+xy}{xy}\)

=>xy+8=2x-y

=>xy-2x+y+8=0

=>x(y-2)+y-2+10=0

=>(x+1)(y-2)=-10

=>(x+1;y-2)∈{(1;-10);(-10;1);(-1;10);(10;-1);(2;-5);(-5;2);(-2;5);(5;-2)}

=>(x;y)∈{(0;-8);(-11;3);(-2;12);(9;1);(1;-3);(-6;4);(-3;7);(4;0)}

mà x<>0; y<>0

nên (x;y)∈{(-11;3);(-2;12);(9;1);(1;-3);(-6;4);(-3;7)}

d: ĐKXĐ: x<>0; y<>0

\(-\frac{3}{y}-\frac{12}{xy}=1\)

=>\(\frac{-3x-12}{xy}=1\)

=>xy=-3x-12

=>xy+3x=-12

=>x(y+3)=-12

=>(x;y+3)∈{(1;-12);(-12;1);(-1;12);(12;-1);(2;-6);(-6;2);(-2;6);(6;-2);(3;-4);(-4;3);(-3;4);(4;-3)}

=>(x;y)∈{(1;-15);(-12;-2);(-1;9);(12;-4);(2;-9);(-6;-1);(-2;3);(6;-5);(3;-7);(-4;0);(-3;1);(4;-6)}

mà y<>0

nên (x;y)∈{(1;-15);(-12;-2);(-1;9);(12;-4);(2;-9);(-6;-1);(-2;3);(6;-5);(3;-7);(-3;1);(4;-6)}

e: ĐKXĐ: y<>0

\(\frac{x}{8}-\frac{1}{y}=\frac14\)

=>\(\frac{x}{8}-\frac14=\frac{1}{y}\)

=>\(\frac{x-2}{8}=\frac{1}{y}\)

=>(x-2)y=8

=>(x-2;y)∈{(1;8);(8;1);(-1;-8);(-8;-1);(2;4);(4;2);(-2;-4);(-4;-2)}

=>(x;y)∈{(3;8);(10;1);(1;-8);(-6;-1);(4;4);(6;2);(0;-4);(-2;-2)}

mà y<>0

nên (x;y)∈{(3;8);(10;1);(1;-8);(-6;-1);(4;4);(6;2);(0;-4);(-2;-2)}

1)

xy + x - 4y = 12

x + y(x - 4) = 12

y(x - 4) = 12 - x

\(y=\dfrac{-x+12}{x-4}\)

Vì \(x,y\inℕ\) nên

\(\left(-x+12\right)⋮\left(x-4\right)\)

\(\left(-x+12\right)-\left(x-4\right)⋮\left(x-4\right)\)

\(16⋮\left(x-4\right)\)

\(\left(x-4\right)\inƯ\left(16\right)\)

\(\left(x-4\right)\in\left\{1;-1;2;-2;4;-4;8;-8;16;-16\right\}\)

\(x\in\left\{5;3;6;2;8;0;12;-4;20;-12\right\}\)

\(y\in\left\{\dfrac{-5+12}{5-4};\dfrac{-3+12}{3-4};\dfrac{-6+12}{6-4};\dfrac{-2+12}{2-4};\dfrac{-8+12}{8-4};\dfrac{-0+12}{0-4};\dfrac{-12+12}{12-4};\dfrac{4+12}{-4-4};\dfrac{-20+12}{20-4};\dfrac{12+12}{-12-4}\right\}\)

\(y\in\left\{7;-9;3;-5;1;-3;0;-2;-\dfrac{1}{2};-\dfrac{7}{5}\right\}\)

\(\left(x;y\right)\in\left\{\left(5;7\right);\left(3;-9\right);\left(6;3\right);\left(2;-5\right);\left(8;1\right);\left(0;-3\right);\left(12;0\right);\left(-4;-2\right);\left(20;-\dfrac{1}{2}\right);\left(-12;-\dfrac{7}{5}\right)\right\}\)

Mà \(x,y\inℕ\) nên các giá trị cần tìm là \(\left(x;y\right)\in\left\{\left(5;7\right);\left(6;3\right);\left(8;1\right);\left(12;0\right)\right\}\)

2)

(2x + 3)(y - 2) = 15

\(\left(2x+3\right)\inƯ\left(15\right)\)

\(\left(2x+3\right)\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)

Ta lập bảng

Mà \(x,y\inℕ\) nên các giá trị cần tìm là \(\left(x;y\right)\in\left\{\left(0;7\right);\left(1;5\right);\left(6;3\right)\right\}\)

các thầy cô ơi giúp em vs ạ mai em phải nộp r ạ!!!

a,

xy - 2x + 5y = 12

=> x(y-2) + 5y - 10 = 2

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

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

Vậy (x,y) = (-4,4); (-3,3); (-6,0); (-7,1)

b,

xy = x + y

=> xy - x - y = 0

=> x(y-1) - (y-1)= 1

=> (x-1)(y-1)= 1

Vậy (x,y) = (2,2); (0,0)

c,

xy = x-y

=> xy - x + y = 0

=> x(y-1) + (y-1) = -1

=> (x+1)(y-1)= -1

=> (x,y) = ...

d,

3^x+1 = (y+1)²

Ta có:

(y+1)² chia 3 dư 0,1

Mà 3^x+1 chia hết cho 3 với x khác -1

+ Với x = -1

<=> 3⁰ = (y+1)²

<=> (y+1)² = 1

=> \(\left[{}\begin{matrix}y+1=1\\y+1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=0\\y=-2\end{matrix}\right.\)

Ta được hai cặp (x,y) = (-1;0); (-1;-2)

+ Với x khác -1

=> (y+1)² chia hết cho 3

=> y+1 chia hết cho 3

=> y chia 3 dư 2

Vậy với x khác -1 thì giá trị ương ứng của y sẽ bằng 3k+2

Vậy...............

câu c sai bảng còn câu d sai đề bài ok