Ta có: \(x^2-x=y^2-1\)

=>\(4x^2-4x=4y^2-4\)

=>\(4x^2-4x+1=4y^2-3\)

=>\(\left(2x-1\right)^2-\left(2y\right)^2=-3\)

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

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

TH1: 2x-1-2y=1 và 2x-1+2y=-3

=>2x-1-2y+2x-1+2y=1-3

=>4x-2=-2

=>4x=0

=>x=0

2x-1-2y=1

=>0-1-2y=1

=>-1-2y=1

=>2y=-1-1=-2

=>y=-1(nhận)

TH2: 2x-1-2y=-3 và 2x-1+2y=1
=>2x-1-2y+2x-1+2y=1-3

=>4x-2=-2

=>4x=0

=>x=0

2x-1-2y=-3

=>0-1-2y=-3

=>2y+1=3

=>2y=2

=>y=1

TH3: 2x-1-2y=-1 và 2x-1+2y=3

=>2x-1-2y+2x-1+2y=-1+3

=>4x-2=2

=>4x=4

=>x=1(nhận)

2x-1-2y=-1

=>2-1-2y=-1

=>1-2y=-1

=>2y-1=1

=>2y=2

=>y=1(nhận)

TH4: 2x-1-2y=3 và 2x-1+2y=-1

=>2x-1-2y+2x-1+2y=-1+3

=>4x-2=2

=>4x=4

=>x=1(nhận)

2x-1-2y=3

=>2-1-2y=3

=>1-2y=3

=>2y=1-3=-2

=>y=-1(nhận)

https://hoc24.vn/cau-hoi/chung-minh-rang-phuong-trinh-sau-khong-co-nghiem-nguyena-x2-y21998b-x2y21994.262907021445

y² = x² - 1998

x² = 1998 + y²

y = \(\sqrt{x^2-1998}\)

x = \(\sqrt{1998+y^2}\)

y = x - \(\sqrt{1998}\)

x = y + \(\sqrt{1998}\)

\(4y^2=4x^4+4x^3+4x^2+4x+4\)

Ta có:

\(4x^4+4x^3+4x^2+4x+4=\left(2x^2+x\right)^2+\left(3x^2+4x+4\right)>\left(2x^2+x\right)^2\)

\(4x^4+4x^3+4x^2+4x+4=\left(2x^2+x+2\right)^2-5x^2\le\left(2x^2+x+2\right)^2\)

\(\Rightarrow\left(2x^2+x\right)^2< \left(2y\right)^2\le\left(2x^2+x+2\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}\left(2y\right)^2=\left(2x^2+x+1\right)^2\\\left(2y\right)^2=\left(2x^2+x+2\right)^2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}4x^4+4x^3+4x^2+4x+4=\left(2x^2+x+1\right)^2\\4x^4+4x^3+4x^2+4x+4=\left(2x^2+x+2\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-3=0\\5x^2=0\end{matrix}\right.\)

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

- Với \(x=-1\Rightarrow y^2=1\Rightarrow y=\pm1\)

- Với \(x=0\Rightarrow y^2=1\Rightarrow y=\pm1\)

- Với \(x=3\Rightarrow y^2=121\Rightarrow y=\pm11\)

a2 là a^2 hay a.2?

a^2