a: (x+1)(y-2)=0

=>\(\begin{cases}x+1=0\\ y-2=0\end{cases}\Rightarrow\begin{cases}x=-1\\ y=2\end{cases}\)

b: (x-5)(y-7)=1

=>(x-5;y-7)∈{(1;1);(-1;-1)}

=>(x;y)∈{(6;8);(4;6)}

c: (x+4)(y-2)=2

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

=>(x;y)∈{(-3;4);(-2;3);(-5;0);(-6;1)}

d: (x+3)(y-6)=-4

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

=>(x;y)∈{(-2;2);(-7;7);(-4;10);(1;5);(-1;4);(-5;8)}

e: (x+7)(5-y)=-6

=>(x+7)(y-5)=6

=>(x+7;y-5)∈{(1;6);(6;1);(-1;-6);(-6;-1);(2;3);(3;2);(-2;-3);(-3;-2)}

=>(x;y)∈{(-6;11);(-1;6);(-8;-1);(-13;4);(-5;8);(-4;7);(-9;2);(-10;3)}

f: (12-y)(6-y)=-2

=>(x-12)(y-6)=2

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

=>(x;y)∈{(13;8);(14;7);(11;4);(10;5)}

giup minh voi

Chọn C. Cặp x, y thỏa mãn là (-1;-6) hoặc (0;2).

\(y=x+\dfrac{1}{x}-5\ge2\sqrt{x.\dfrac{1}{x}}-5=-3\)

Đáp án C

Bài 2: 

a: =>x=0 hoặc x+3=0

=>x=0 hoặc x=-3

b: =>x-2=0 hoặc 5-x=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

a: \(\left(x+5\right)^2>=0\forall x\)

\(\left(2y-8\right)^2>=0\forall y\)

Do đó: \(\left(x+5\right)^2+\left(2y-8\right)^2>=0\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x+5=0\\2y-8=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-5\\y=4\end{matrix}\right.\)

b: \(\left(x+3\right)\left(2y-1\right)=5\)

=>\(\left(x+3\right)\left(2y-1\right)=1\cdot5=5\cdot1=\left(-1\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-1\right)\)

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

=>\(\left(x,y\right)\in\left\{\left(-2;3\right);\left(2;1\right);\left(-4;-2\right);\left(-8;0\right)\right\}\)

Bài 2: 

a: =>x=0 hoặc x=-3

b: =>x-2=0 hoặc 5-x=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

a: =>x^2-6x+9+y^2+8y+16=0

=>(x-3)^2+(y+4)^2=0

=>x=3 và y=-4