Ta có: \(\left(-50\right)^2-3x^2=73\)
\(\Leftrightarrow2500-3x^2=73\)
\(\Leftrightarrow3x^2=2500-73=2427\)
\(\Leftrightarrow x^2=809\)
hay \(x=\pm\sqrt{809}\)
Vậy: \(x=\pm\sqrt{809}\)
(-50)^2 -3 . x^2 = 73
250 - 3 . x^2 = 73
3 . x^2 = 177
x^2 = 177:3
x=\(\sqrt{\dfrac{177}{3}}\)
hoặc x =- \(\sqrt{\dfrac{177}{3}}\)
(-50)2 - 3.x2 = 73
\(\Rightarrow\) 2500 - 3x2 = 73
\(\Rightarrow\) 3x2 = 2427
\(\Rightarrow\) x2 = 809
\(\Rightarrow\) x2 - 809 = 0
\(\Rightarrow\) (x - \(\sqrt{809}\))(x + \(\sqrt{809}\)) = 0
\(\Rightarrow\) \(\left[{}\begin{matrix}x-\sqrt{809}=0\\x+\sqrt{809}=0\end{matrix}\right.\)
\(\Rightarrow\) \(\left[{}\begin{matrix}x=\sqrt{809}\\x=-\sqrt{809}\end{matrix}\right.\)
Vậy ...
(Nếu xét x là số nguyên thì phương trình vô nghiệm nha!)
Chúc bn học tốt!
