Gọi pt đề bài là (*)
Ta có (*) <=> x - 1 = 32
<=> x = 10
\(\sqrt{x-1}=3.\left(2018+2019+2020\right)^0\)
\(\sqrt{x-1}=3\)
\(\sqrt{x-1}^2=3^2\)
\(x-1=9\)
\(x=9+1\)
\(\Rightarrow x=10\)
\(\sqrt{x-1}=3.\left(2018+2019+2020\right)^0\)
\(\sqrt{x-1}=3.1\)
\(\sqrt{x-1}=3\)
\(x-1=3^2\)
\(x-1=9\)
\(x=9+1\)
\(x=10\)
Hok tốt :p
\(\sqrt{x-1}=3.\left(2018+2019+2020\right)^0\)
Vì \(1^0=1\Rightarrow\left(2018+2019+2020\right)^0=1\)
\(\sqrt{x-1}=3.1\)
\(\sqrt{x-1}=3\)
\(\Rightarrow\sqrt{x-1}=3^2\)
\(\Rightarrow x-1=9\)
\(\Rightarrow\orbr{\begin{cases}x-1=9\\x-1=-9\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=9+1\\x=-9+1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=10\\x=-8\end{cases}}\)
Vậy \(x=10\)hoặc \(x=-8\)
Chúc bạn học tốt !!!
