a, x[y-2]+3y-6=0-6

x[y-2]+3[y-2]=-6

[x+3][y-2]=-6

-6=-1.6=-6.1

tiếp theo tự làm nha

b, x[y-1]-5y-5=-8-5

x[y-1]-5[y-1]=-13

[x-5][y-1]=-13

-13=-1.13=-13.1

tiếp theo tự làm nha

NHỚ THEO DÕI MÌNH NHA

b: 2xy-3y+3x=7

=>y(2x-3)+3x-4,5=7-4,5

=>2y(x-1,5)+3(x-1,5)=2,5

=>(x-1,5)(2y+3)=2,5

=>(2x-3)(2y+3)=5

=>(2x-3;2y+3)∈{(1;5);(5;1);(-1;-5);(-5;-1)}

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

=>(x;y)∈{(2;1);(4;-1);(1;-4);(-1;-2)}

a: xy+4y+3x=28

=>y(x+4)+3x+12=28+12

=>(x+4)(y+3)=40

=>(x+4;y+3)∈{(1;40);(40;1);(-1;-40);(-40;-1);(2;20);(20;2);(-2;-20);(-20;-2);(4;10);(10;4);(-4;-10);(-10;-4);(5;8);(8;5);(-5;-8);(-8;-5)}

=>(x;y)∈{(-3;37);(36;-2);(-5;-43);(-44;-4);(-2;17);(16;-1);(-6;-23);(-24;-5);(0;7);(6;1);(-8;-13);(-14;-7);(1;5);(4;2);(-9;-11);(-12;-8)}

a) \(xy+x+2y=5\\ \Rightarrow y\left(x+2\right)+x+2=5+2\\ \Rightarrow\left(x+2\right)\left(y+1\right)=7\)

Ta xét bảng:

Vậy \(\left(x;y\right)\in\left\{\left(-1;6\right);\left(5;0\right);\left(-3;-8\right);\left(-9;-2\right)\right\}\)

b) \(xy-3x-y=0\\ \Rightarrow x\left(y-3\right)-y+3=3\\ \Rightarrow\left(y-3\right)\left(x-1\right)=3\)

Ta xét bảng:

Vậy \(\left(x;y\right)\in\left\{\left(2;6\right);\left(4;4\right);\left(0;0\right);\left(-2;2\right)\right\}\)

c) \(xy+2x+2y=-16\\ \Rightarrow x\left(y+2\right)+2y+4=-12\\ \Rightarrow\left(y+2\right)\left(x+2\right)=-12\)

Ta xét bảng:

Vậy \(\left(x;y\right)\in\left\{\left(-1;-14\right);\left(0;-8\right);\left(1;-6\right);\left(2;-5\right);\left(4;-4\right);\left(10;-3\right);\left(-3;10\right);\left(-4;4\right);\left(-5;2\right);\left(-6;1\right);\left(-8;0\right);\left(-14;-1\right)\right\}\)

a) \(\left\{{}\begin{matrix}2x+3y=5\\4x-5y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x+6y=10\\4x-5y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=5\\11y=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+3\cdot\dfrac{9}{11}=5\\y=\dfrac{9}{11}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+\dfrac{27}{11}=5\\y=\dfrac{9}{11}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=\dfrac{28}{11}\\y=\dfrac{9}{11}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{14}{11}\\y=\dfrac{9}{11}\end{matrix}\right.\)

Vậy: \(x=\dfrac{14}{11};y=\dfrac{9}{11}\)