Question

\(\sqrt{x-1}=3.\left(2018+2019+2020\right)^0\)

Answer 1

\(\sqrt{x-1}=3.\)  \(\left(2018+2019+2020\right)^0\)

\(\sqrt{x-1}=3\)

\(\sqrt{x-1}^2=3^2\)

\(x-1=9\)

\(x=9+1\)

\(\Rightarrow x=10\)

Answer 2

Ta có công thức : \(\sqrt{x-1}^2=n^2\) thì mới phá được dấu căn bậc 2

Nên ta làm như sau : 

\(\sqrt{x-1}=3.\) \(\left(2018+2019+2020\right)^0\)

\(\sqrt{x-1}=3\)

\(\sqrt{x-1}^2=3^2\)

\(x-1=9\)

\(x=9+1\)

\(\Rightarrow x=10\)

Answer 3

\(\sqrt{x-1}=3\left(2018+2019+2020\right)^0\)

\(\sqrt{x-1}=3\cdot1\)

\(\sqrt{x-1}=3\)

\(\left(\sqrt{x-1}\right)^2=3^2\)

\(x-1=9\)

\(x=9+1\)

\(x=10\)

Answer 4

\(\sqrt{x-1}=3\left(2018+2019+2020\right)^0\)

\(\sqrt{x-1}=3\cdot1\)

\(\sqrt{x-1}=3\)

\(\left(\sqrt{x-1}\right)^2=3^2\)

\(x-1=9\)

\(x=9+1\)

\(x=10\)