Ta có :

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

\(\Rightarrow2x+\dfrac{31^2}{12^2}=\dfrac{49^2}{12^2}\Rightarrow2x=\dfrac{49^2-31^2}{12^2}=10\)

\(\Rightarrow x=5\)

\(\Rightarrow y^2=\left(\dfrac{49}{12}\right)^2-5=\dfrac{1681}{144}\)

\(\Rightarrow y=\dfrac{41}{12}\)

Bui dang kien giải sai rồi. Đáp án trong sách ghi là x=5,y=41/12 và -41/12

x+(-31/12)^2=(49/12)^2-x 
<=> 2x= (49/12)^2-(-31/12)^2 
<=> 2x=10 
<=>x=5 
=>y^2=10=>y=căn 10 hoặc - căn 10

x=5;y=41/12

Bài làm:

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

\(\Leftrightarrow2x=\frac{2401}{144}+\frac{961}{144}\)

\(\Leftrightarrow2x=\frac{1681}{72}\)

\(\Rightarrow x=\frac{1681}{144}\)

=> \(y^2=\frac{1681}{144}+\frac{961}{144}=\frac{2642}{144}\)

=> \(y=\pm\frac{\sqrt{2642}}{12}\)

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

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

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