(x-1)^2=9
=> (x-1)^2 = 3^2
=> x-1= 3
=> x = 3 + 1
=> x =4
\(\left(x-1\right)^2=9\)
\(\Leftrightarrow\left(x-1\right)^2=3^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=3\\x-1=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3+1\\x=-1+3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=4\\x=2\end{cases}}}\)
Vậy \(x=4\)hoặc \(x=2\)
\(\left(x-3\right)^3+15=7\)
\(\Leftrightarrow\left(x-3\right)^3=7-15\)
\(\Leftrightarrow\left(x-3\right)^3=-8\)
\(\Leftrightarrow\left(x-3\right)^3=\left(-2\right)^3\)
\(\Leftrightarrow x-3=-2\)
\(\Leftrightarrow x=-2+3\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
Trl:
\(\left(x-1\right)^2=9\)
\(\Rightarrow\left(x-1\right)^2=3^2\)
\(\Rightarrow x-1=3\)
\(\Rightarrow x=4\)
\(\left(x-3\right)^3+15=7\)
\(\Rightarrow\left(x-3\right)^3=7-15\)
\(\Rightarrow\left(x-3\right)^3=-8\)
\(\Rightarrow\left(x-3\right)^3=\left(-2\right)^3\)
\(\Rightarrow x-3=-2\)
\(\Rightarrow x=1\)
a)\(\left(x-1\right)^2=9\) b)\(\left(x-3\right)^3+15=7\)
\(\left(x-1\right)^2=3^2\) \(\left(x-3\right)^3=15-7\)
\(x-1=3\) \(\left(x-3\right)^3=-8\)
\(\Rightarrow x=2\) \(\left(x-3\right)^3=\left(-2\right)^3\)
\(x-3=-2\)
\(x=1\)
Cho mình sửa lại câu cuối!
\(\left(x-1\right)^2=9\)
\(\Leftrightarrow\left(x-1\right)^2=3^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=3\\x-1=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3+1\\x=-3+1\end{cases}\Leftrightarrow}}\orbr{\begin{cases}x=4\\x=-2\end{cases}}\)
Vậy \(x=4\)hoặc \(x=-2\)
Mấy bạn làm sai rồi, mũ chẵn chia 2 trưởng hợp nha!
