a) \(\left(x-1\right)=\left(x-1\right)^3\)
\(\Leftrightarrow\left(x-1\right)^3-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\\left(x-1\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x\in\left\{2;0\right\}\end{cases}}\)
Vậy \(x\in\left\{0;1;2\right\}\)
b) \(x^3+x=0\)
\(\Leftrightarrow x\left(x^2+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=-1\left(L\right)\end{cases}}\)
Vậy x = 0
\(a,\) \(\left(x-1\right)=\left(x-1\right)^3\)
\(\left(x-1\right)^3-\left(x-1\right)=0\)
\(\left(x-1\right)\left[\left(x-1\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\\left(x-1\right)^2-1=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=1\\\left(x-1\right)^2=1\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\pm1\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{\pm1\right\}\)
Bài giải
\(a,\) \(\left(x-1\right)=\left(x-1\right)^3\)
\(\left(x-1\right)^3-\left(x-1\right)=0\)
\(\left(x-1\right)\left[\left(x-1\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\\left(x-1\right)^2-1=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=1\\\left(x-1\right)^2=1\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\pm1\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{\pm1\right\}\)
\(b,\)\(x^3+x=0\)
\(x\left(x^2+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^2+1=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=-1\left(\text{ loại }\right)\end{cases}}\)
Vậy x = 0
Bài giải
\(b,\)\(x^3+x=0\)
\(x\left(x^2+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^2+1=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=-1\left(\text{ loại vì }x^2\ge0\right)\end{cases}}\)
Vậy x = 0
\(a,\left(x-1\right)=\left(x-1\right)^3\)
\(\left(x-1\right)-\left(x-1\right)^3=0\)
\(\left(x-1\right)[1-\left(x-1\right)^2]=0\)
\(\orbr{\begin{cases}\left(x-1\right)=0\\1-\left(x-1\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\\left(x-1\right)^2=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x-1=\pm1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=0,x=2\end{cases}}\)
\(x\left(x^2+1\right)=0\)
\(\orbr{\begin{cases}x=0\\x^2+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=-1\end{cases}}\Leftrightarrow\orbr{x=0}\)
NHỚ ĐÓ NHA!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
a) (x - 1) = (x - 1)3
=> (x - 1) - (x - 1)3 = 0
=> (x - 1) [1 (x - 1)2 ] = 0
=>\(\orbr{\begin{cases}x-1=0\\1-\left(x-1\right)^2=0\end{cases}}\) => \(\orbr{\begin{cases}x=0+1\\\left(x-1\right)^2=1-0\end{cases}}\) => \(\orbr{\begin{cases}x=1\\\left(x-1\right)^2=1\end{cases}}\) => \(\orbr{\begin{cases}x=1\\x-1=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=1\\x=1+1\end{cases}}\) => \(\orbr{\begin{cases}x=1\\x=2\end{cases}}\)
Vậy x thuộc {1; 2}
b) x3 + x = 0
=> x(x2 + 1) = 0
=> \(\orbr{\begin{cases}x=0\\x^2+1=0\end{cases}}\) =>\(\orbr{\begin{cases}x=0\\x^2=1+0=1\end{cases}}\)=>\(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
Vậy x thuộc {1;0}