Question

a,(x-7).(2x-8)=0

b,(x-1)^6=(x-1)^8

Answer 1

a) \(\left(x-7\right)\left(2x-8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\2x-8=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=7\\x=4\end{cases}}\)

b) \(\left(x-1\right)^6=\left(x-1\right)^8\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-1=1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)

Answer 2

\(\left(x-7\right)\left(2x-8\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-7=0\\2x-8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=4\end{cases}}}\)

Vậy \(x=7\) hoặc \(x=4\)

\(\left(x-1\right)^6=\left(x-1\right)^8\)

\(\Rightarrow\left(x-1\right)^8-\left(x-1\right)^6=0\)

\(\left(x-1\right)^6.\left[\left(x-1\right)^2-1\right]=0\)

\(\Rightarrow\left(x-1\right)^6=0\) hoặc \(\left(x-1\right)^2-1=0\)

\(\Leftrightarrow x-1=0\)                \(\left(x-1\right)^2=\left(\pm1\right)^2\)

\(\Leftrightarrow x=1\)                          \(\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}\)

Vậy \(x=1;x=2;x=0\)

Answer 3

\(\left(x-7\right).\left(2x-8\right)=0\Rightarrow\orbr{\begin{cases}x-7=0\\2x-8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=4\end{cases}}}\\ \)

\(\left(x-1\right)^6=\left(x-1\right)^8\Leftrightarrow\left(x-1\right)^8-\left(x-1\right)^6=0\Leftrightarrow\left(x-1\right)^6.\left(\left(x-1\right)^2-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^6=0\\\left(x-1\right)^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\\left(x-1\right)^2=1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2,x=0\end{cases}}}}\)

\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^6=0\\\left(x-1\right)^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x-1=0\\\left(x-1-1\right).\left(x-1+1\right)=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\\orbr{\begin{cases}x=2\\x=0\end{cases}}\end{cases}}}\)

Answer 4

a) \(\left(x-7\right)\left(2x-8\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-7=0\\2x-8=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=7\\2x=8\end{cases}}\Rightarrow\orbr{\begin{cases}x=7\\x=4\end{cases}}\)

Vậy x = 7 hoặc x = 4

b) \(\left(x-1\right)^6=\left(x-1\right)^8\)

\(\Rightarrow\left(x-1\right)^6-\left(x-1\right)^8=0\)

\(\Rightarrow\left(x-1\right)^6\left[1-\left(x-1\right)^2\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^6=0\\1-\left(x-1\right)^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}\left(x-1\right)^6=0\\\left(x-1\right)^2=1\end{cases}}\)

TH 1 : \(\left(x-1\right)^6=0\Rightarrow x-1=0\Rightarrow x=1\)

TH 2 : \(\left(x-1\right)^2=1\Rightarrow\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}\)

Vậy \(x\in\left\{1,2,0\right\}\)

_Chúc bạn học tốt_