Bài 1: 

a: \(\Leftrightarrow x-1\in\left\{1;-1;3;-3\right\}\)

hay \(x\in\left\{2;0;4;-2\right\}\)

mọi người giúp với 

B

B

B

a: Ta có: \(x^2-x+1\)

\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{4}\)

b: Ta có: \(x^2+y^2-4x+y+5\)

\(=\left(x^2-4x+4\right)+\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}\)

\(=\left(x-2\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x,y\)

Dấu '=' xảy ra khi x=2 và \(y=-\dfrac{1}{2}\)

a: =>x-xy+y=0

=>x(1-y)+1-y-1=0

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

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

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

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

b: 2x-xy-2y=3

=>x(2-y)-2y+4=7

=>x(2-y)+2(2-y)=7

=>(x+2)(y-2)=-7

=>\(\left(x+2;y-2\right)\in\left\{\left(1;-7\right);\left(-7;1\right);\left(-1;7\right);\left(7;-1\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(-1;-5\right);\left(-9;3\right);\left(-3;9\right);\left(5;1\right)\right\}\)

c: =>x(4-y)+5y-20=-3

=>x(4-y)-5(4-y)=-3

=>(4-y)(x-5)=-3

=>(x-5)(y-4)=3

=>\(\left(x-5;y-4\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(-1;-3\right);\left(-3;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(6;9\right);\left(8;5\right);\left(4;1\right);\left(2;3\right)\right\}\)