\(49-y^2=12.\left(x^2-4002x+4004001\right)\)

\(\Leftrightarrow49-y^2=12x^2-48024x+48048012\)

\(\Leftrightarrow-y^2+12x^2+48024x=-49+48048012\)

Ý chết cha cái đề đâu rồi sao pit tính cái dj

\(x+\left(\frac{-31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x=y^2\)

Xét \(x+\left(\frac{-31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x\)

\(\Rightarrow2x=\left(\frac{49}{12}\right)^2-\left(\frac{-31}{12}\right)^2=\frac{2401}{144}+\frac{961}{144}\)

\(\Rightarrow2x=\frac{3362}{144}\)

\(\Rightarrow x=\frac{3362}{144}.\frac{1}{2}=\frac{1681}{144}\)

Ta lai xét :

\(x+\left(\frac{-31}{12}\right)^2=y^2\)

\(\Rightarrow\frac{1681}{144}+\frac{-961}{144}=y^2\)

\(\Rightarrow\frac{720}{144}=y^2\)

\(\Rightarrow y^2=5\)

\(\Rightarrow y=2,236067977\)

\(x+\left(\frac{-31}{12}\right)^2\left(\frac{49}{12}\right)^2-x=y\)

\(x+\frac{961}{144}.\frac{2401}{144}-x=y\)

\(x+\frac{2307361}{20736}-x=y\)

\(y=\frac{2307361}{20736}\)

         Thay vào \(x+\frac{961}{144}.\frac{2401}{144}-x=y\) ta được

\(x+\frac{2307361}{20736}-x=y\)

\(x-x+\frac{2307361}{20736}=\frac{2307361}{20736}\)

Vậy x thuộc N;\(y=\frac{2307361}{20736}\)

\(\Leftrightarrow\left(x-x\right)+\frac{961}{144}\cdot\frac{2401}{144}=y\Leftrightarrow y=\frac{2307361}{20736}\)

\(\Rightarrow x\in R\)

( hình như đề sai :v )

đề saI

​

X+(-31/12)^2 = (49/12)^2 -x=y

(-31/12)^2 - (49/12)^2 = -x-x = y

961/144 - 2410/144 = -2x

-10=-2x

10=2x

10:2=x

5=x

X+961/144=y^2

5+961/144=y^2

1681/144=y^2

=>y=41/144

Dấu phân số mình ký hiệu là / đó nha

\(x+\left(\frac{-31}{12}\right)^2=\left(-\frac{49}{12}\right)^2-x=y^2\)

\(\Rightarrow x+\frac{31^2}{12^2}=\frac{49^2}{12^2}-x=y^2\) (1)

\(\Rightarrow x+x=\frac{49^2}{12^2}-\frac{31^2}{12^2}\)

\(\Rightarrow2x=\frac{49^2-31^2}{12^2}\)

\(\Rightarrow2x=\frac{\left(49-31\right).\left(49+31\right)}{144}\)

\(\Rightarrow2x=\frac{18.80}{144}\)

\(\Rightarrow2x=10\)

\(\Rightarrow x=10:2=5\)

Thay \(x=5\) vào (1) ta có:

\(5+\frac{31^2}{12^2}=y^2\)

\(\Rightarrow5+\frac{961}{144}=y^2\)

\(\Rightarrow\frac{1681}{144}=y^2\)

\(\Rightarrow\left[\begin{array}{nghiempt}y=\frac{41}{12}\\y=\frac{-41}{12}\end{array}\right.\)

Vậy \(x=5;y\in\left\{\frac{41}{12};\frac{-41}{12}\right\}\)