Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x+\left(\frac{-31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x=y^2\)
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+(-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 \(x=5;y\in\left\{\frac{41}{12};\frac{-41}{12}\right\}\)
\(x+\left(\frac{-31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x\)
<=>\(x+x=\frac{31^2}{12^2}+\frac{49^2}{12^2}\)
<=>\(2x=\frac{3362}{144}=\frac{1681}{72}\)
<=>\(x=\frac{1681}{144}\)
=>\(y^2=x+\left(-\frac{39}{12}\right)^2=\frac{1681}{144}+\frac{1521}{144}=\frac{1601}{72}\Rightarrow y=^+_-\sqrt{\frac{1601}{72}}\)
Vì
\(x+\left(-\frac{31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x\Rightarrow2x=\left(\frac{49}{12}\right)^2-\left(-\frac{31}{12}\right)^2=10\)
=> 2x = 10 => x = 5
Thay x = 5 vào ta có :
\(5+\left(-\frac{31}{12}\right)^2=y^2\Leftrightarrow\frac{1681}{144}=y^2=\left(\frac{41}{12}\right)^2=\left(-\frac{41}{12}\right)^2\)
=> y = 41/12 hoặc y = -41/12
Ta có:
\(x+\left(-\frac{31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x\)
\(\Rightarrow2x+\frac{961}{144}=\frac{2401}{144}\)
\(\Rightarrow2x=\frac{2401}{144}-\frac{961}{144}\)
\(\Rightarrow2x=\frac{1440}{144}\)
\(\Rightarrow2x=10\)
\(\Rightarrow x=5\)
Vậy \(x=5\)
tìm x và y nha