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 \(y=\dfrac{41}{12}\)
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 \(y=\dfrac{41}{12}\)
Tìm x,y biết: x + (-31/12)^2 = (49/12)^2 - x = y^2
x+ (-31/12)^2 = (49/12)^2- x = y^2
Tìm x,y
a, x+(–31/12)^2=(49/12)^2–x=y^2
b, x*(x-y)=3/10 và y*(x-y)=–3/20
Tìm x,y biết rằng:
x + (-31/12)2 = (49/12)2 - x = y2
Tìm x, y:
\(x+\left(\frac{-31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x=y^2\)
\(x+\frac{-31}{12}^2=\frac{49}{12}^2-x=y^2\)
Tìm x,y
Tìm x;y biết rằng:
\(x+\left(-\frac{31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x=y^2\)
Bài 2 Tìm x,y : \(x+\left(-\frac{31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x=y^2\)
Tìm x, y biết
x + [ -31/122 ) =( 49 /12) 2 -x= y2