\(\sqrt{x-1}=3.\) \(\left(2018+2019+2020\right)^0\)
\(\sqrt{x-1}=3\)
\(\sqrt{x-1}^2=3^2\)
\(x-1=9\)
\(x=9+1\)
\(\Rightarrow x=10\)
Ta có công thức : \(\sqrt{x-1}^2=n^2\) thì mới phá được dấu căn bậc 2
Nên ta làm như sau :
\(\sqrt{x-1}=3.\) \(\left(2018+2019+2020\right)^0\)
\(\sqrt{x-1}=3\)
\(\sqrt{x-1}^2=3^2\)
\(x-1=9\)
\(x=9+1\)
\(\Rightarrow x=10\)
\(\sqrt{x-1}=3\left(2018+2019+2020\right)^0\)
\(\sqrt{x-1}=3\cdot1\)
\(\sqrt{x-1}=3\)
\(\left(\sqrt{x-1}\right)^2=3^2\)
\(x-1=9\)
\(x=9+1\)
\(x=10\)
\(\sqrt{x-1}=3\left(2018+2019+2020\right)^0\)
\(\sqrt{x-1}=3\cdot1\)
\(\sqrt{x-1}=3\)
\(\left(\sqrt{x-1}\right)^2=3^2\)
\(x-1=9\)
\(x=9+1\)
\(x=10\)
