x+ (-31/12)^2 = (49/12)^2- x = y^2
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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
Ta có: \(x+\left(-\dfrac{31}{12}\right)^2=\left(\dfrac{49}{12}\right)^2-x\)
\(\Leftrightarrow x+x=\dfrac{2401}{144}-\dfrac{961}{144}=10\)
hay x=5
\(\Leftrightarrow y^2=\left(\dfrac{49}{12}\right)^2-5=\dfrac{1681}{144}\)
hay \(y=\dfrac{41}{12}\)
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;\(y=\frac{2307361}{20736}\)