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

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

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

Bạn ơi , có sai đề ko z ?

Ta co :

\(B=y^2-2y\left(1-y\right)+1-2y+y^2+y^2-8y+16+x^2+2x+1+2002\)

B=\(\left(y-1+y\right)^2+\left(y-4\right)^2+(x+1)^2+2002\)

Vi \(\left(2y-1\right)^2;\left(y-4\right)^2;\left(x+1\right)^2\) luon lon hon hoac bang 0 nen

ta co : minB=2002

a: \(49-y^2=6\left(x-2021\right)^2\)

=>\(49-y^2\ge0\) và \(49-y^2\) ⋮6

=>\(y^2\in\left\lbrace1;16;25;49\right\rbrace\)

TH1: \(y^2=1\)

Ta có: \(49-y^2=6\left(x-2021\right)^2\)

=>\(6\left(x-2021\right)^2=49-1=48\)

=>\(\left(x-2021\right)^2=8\)

mà x nguyên

nên x∈∅

TH2: \(y^2=16\)

Ta có: \(49-y^2=6\left(x-2021\right)^2\)

=>\(6\left(x-2021\right)^2=49-16=33\)

=>\(\left(x-2021\right)^2=5,5\)

mà x nguyên

nên x∈∅

TH3: \(y^2=25\)

Ta có: \(49-y^2=6\left(x-2021\right)^2\)

=>\(6\left(x-2021\right)^2=49-25=24\)

=>\(\left(x-2021\right)^2=4\)

=>x-2021=2 hoặc x-2021=-2

=>x=2023(nhận) hoặc x=2019(nhận)

\(y^2=25\)

=>y=5(nhận) hoặc y=-5(nhận)

TH4: \(y^2=49\)

Ta có: \(49-y^2=6\left(x-2021\right)^2\)

=>\(6\left(x-2021\right)^2=49-49=0\)

=>\(\left(x-2021\right)^2=0\)

=>x-2021=0

=>x=2021(nhận)

\(y^2=49\)

=>y=7(nhận) hoặc y=-7(nhận)

a: \(49-y^2=6\left(x-2021\right)^2\)

=>\(49-y^2\ge0\) và \(49-y^2\) ⋮6

=>\(y^2\in\left\lbrace1;16;25;49\right\rbrace\)

TH1: \(y^2=1\)

Ta có: \(49-y^2=6\left(x-2021\right)^2\)

=>\(6\left(x-2021\right)^2=49-1=48\)

=>\(\left(x-2021\right)^2=8\)

mà x nguyên

nên x∈∅

TH2: \(y^2=16\)

Ta có: \(49-y^2=6\left(x-2021\right)^2\)

=>\(6\left(x-2021\right)^2=49-16=33\)

=>\(\left(x-2021\right)^2=5,5\)

mà x nguyên

nên x∈∅

TH3: \(y^2=25\)

Ta có: \(49-y^2=6\left(x-2021\right)^2\)

=>\(6\left(x-2021\right)^2=49-25=24\)

=>\(\left(x-2021\right)^2=4\)

=>x-2021=2 hoặc x-2021=-2

=>x=2023(nhận) hoặc x=2019(nhận)

\(y^2=25\)

=>y=5(nhận) hoặc y=-5(nhận)

TH4: \(y^2=49\)

Ta có: \(49-y^2=6\left(x-2021\right)^2\)

=>\(6\left(x-2021\right)^2=49-49=0\)

=>\(\left(x-2021\right)^2=0\)

=>x-2021=0

=>x=2021(nhận)

\(y^2=49\)

=>y=7(nhận) hoặc y=-7(nhận)

a: \(49-y^2=6\left(x-2021\right)^2\)

=>\(49-y^2\ge0\) và \(49-y^2\) ⋮6

=>\(y^2\in\left\lbrace1;16;25;49\right\rbrace\)

TH1: \(y^2=1\)

Ta có: \(49-y^2=6\left(x-2021\right)^2\)

=>\(6\left(x-2021\right)^2=49-1=48\)

=>\(\left(x-2021\right)^2=8\)

mà x nguyên

nên x∈∅

TH2: \(y^2=16\)

Ta có: \(49-y^2=6\left(x-2021\right)^2\)

=>\(6\left(x-2021\right)^2=49-16=33\)

=>\(\left(x-2021\right)^2=5,5\)

mà x nguyên

nên x∈∅

TH3: \(y^2=25\)

Ta có: \(49-y^2=6\left(x-2021\right)^2\)

=>\(6\left(x-2021\right)^2=49-25=24\)

=>\(\left(x-2021\right)^2=4\)

=>x-2021=2 hoặc x-2021=-2

=>x=2023(nhận) hoặc x=2019(nhận)

\(y^2=25\)

=>y=5(nhận) hoặc y=-5(nhận)

TH4: \(y^2=49\)

Ta có: \(49-y^2=6\left(x-2021\right)^2\)

=>\(6\left(x-2021\right)^2=49-49=0\)

=>\(\left(x-2021\right)^2=0\)

=>x-2021=0

=>x=2021(nhận)

\(y^2=49\)

=>y=7(nhận) hoặc y=-7(nhận)

a: \(49-y^2=6\left(x-2021\right)^2\)

=>\(49-y^2\ge0\) và \(49-y^2\) ⋮6

=>\(y^2\in\left\lbrace1;16;25;49\right\rbrace\)

TH1: \(y^2=1\)

Ta có: \(49-y^2=6\left(x-2021\right)^2\)

=>\(6\left(x-2021\right)^2=49-1=48\)

=>\(\left(x-2021\right)^2=8\)

mà x nguyên

nên x∈∅

TH2: \(y^2=16\)

Ta có: \(49-y^2=6\left(x-2021\right)^2\)

=>\(6\left(x-2021\right)^2=49-16=33\)

=>\(\left(x-2021\right)^2=5,5\)

mà x nguyên

nên x∈∅

TH3: \(y^2=25\)

Ta có: \(49-y^2=6\left(x-2021\right)^2\)

=>\(6\left(x-2021\right)^2=49-25=24\)

=>\(\left(x-2021\right)^2=4\)

=>x-2021=2 hoặc x-2021=-2

=>x=2023(nhận) hoặc x=2019(nhận)

\(y^2=25\)

=>y=5(nhận) hoặc y=-5(nhận)

TH4: \(y^2=49\)

Ta có: \(49-y^2=6\left(x-2021\right)^2\)

=>\(6\left(x-2021\right)^2=49-49=0\)

=>\(\left(x-2021\right)^2=0\)

=>x-2021=0

=>x=2021(nhận)

\(y^2=49\)

=>y=7(nhận) hoặc y=-7(nhận)

a: \(49-y^2=6\left(x-2021\right)^2\)

=>\(49-y^2\ge0\) và \(49-y^2\) ⋮6

=>\(y^2\in\left\lbrace1;16;25;49\right\rbrace\)

TH1: \(y^2=1\)

Ta có: \(49-y^2=6\left(x-2021\right)^2\)

=>\(6\left(x-2021\right)^2=49-1=48\)

=>\(\left(x-2021\right)^2=8\)

mà x nguyên

nên x∈∅

TH2: \(y^2=16\)

Ta có: \(49-y^2=6\left(x-2021\right)^2\)

=>\(6\left(x-2021\right)^2=49-16=33\)

=>\(\left(x-2021\right)^2=5,5\)

mà x nguyên

nên x∈∅

TH3: \(y^2=25\)

Ta có: \(49-y^2=6\left(x-2021\right)^2\)

=>\(6\left(x-2021\right)^2=49-25=24\)

=>\(\left(x-2021\right)^2=4\)

=>x-2021=2 hoặc x-2021=-2

=>x=2023(nhận) hoặc x=2019(nhận)

\(y^2=25\)

=>y=5(nhận) hoặc y=-5(nhận)

TH4: \(y^2=49\)

Ta có: \(49-y^2=6\left(x-2021\right)^2\)

=>\(6\left(x-2021\right)^2=49-49=0\)

=>\(\left(x-2021\right)^2=0\)

=>x-2021=0

=>x=2021(nhận)

\(y^2=49\)

=>y=7(nhận) hoặc y=-7(nhận)