Ta có : 8 . 6 + 288 : (x - 3)2 = 50
=> 48 + 288 : (x - 3)2 = 50
=> 288 : (x - 3)2 = 50 - 48
=> 288 : (x - 3)2 = 2
=> (x - 3)2 = 288 : 2
=> (x - 3)2 = 144
=> không tìm được x thỏa mãn điều kiện trên
8.6+288:(x-3)2 = 50
<=> 48+288:(x-3)2 = 50
<=> 288:(x-3)2 = 50-48 = 2
<=> (x-3)2 = 2.288 = 576 = 242
<=> \(\orbr{\begin{cases}x-3=24\\x-3=-24\end{cases}}\)
<=> \(\orbr{\begin{cases}x=24+3=27\\x=3-24=-21\end{cases}}\)
Vậy x thuộc { 27 ; -21 }
\(8\cdot6+288:\left(x-3\right)^2=50\)
\(48+288:\left(x-3\right)^2=50\)
\(288:\left(x-3\right)^2=50-48\)
\(288:\left(x-3\right)^2=2\)
\(\left(x-3\right)^2=288:2\)
\(\left(x-3\right)^2=144\)
\(Th1:x-3=12\)\(Th2:x-3=-12\)
\(x=12+3=15\)\(x=-12+3=-9\)
Vậy \(x=15\)hoặc \(x=-9\)
\(8\times6+288:\left(x-3\right)^2=50\)
\(48+288:\left(x-3\right)^2=50\)
\(288:\left(x-3\right)^2=2\)
\(\left(x-3\right)^2=144\)
\(x^2:3^2=144\)
\(x^2=144:9\)
\(x^2=16\)
\(x=4\)
= 48 + 288 : ( x - 3 )2 = 50
288 : ( x - 3 )2 = 50 - 48
288: ( x - 3 )2= 2
(x - 3 )2= 288 : 2
(x - 3)2= 144
(x - 3)2 = 122
x - 3 = 12
x = 12 + 3 = 15