\(b)4x\left(x-2014\right)-\left(x-2014\right)=0\)
\(\left(4x-1\right)\left(x-2014\right)=0\)
\(\Leftrightarrow TH1:4x-1=0\)
\(4x=1\)
\(x=\frac{1}{4}\)
\(TH2:x-2014=0\)
\(x=2014\)
Vậy \(x\in\left\{\frac{1}{4};2014\right\}\)
\(b,4x\left(x-2014\right)-x+2014=0\)
\(\Leftrightarrow\left(x-2014\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2014\\x=\frac{1}{4}\end{cases}}\)
\(c,\left(x+1\right)^2=x+1\)
\(\Leftrightarrow\left(x+1\right)x=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
\(c)\left(x+1\right)^2-\left(x+1\right)=0\)
\(\left(x-1-1\right)\left(x+1\right)=0\)
\(\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow TH1:x-2=0\)
\(x=2\)
\(TH2:x+1=0\)
\(x=-1\)
Vậy \(x\in\left\{-1;2\right\}\)
\(c,\left(x+1\right)^2=x+1\)
\(\Rightarrow\left(x+1\right)\left(x+1\right)-\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x+1-1\right)=0\)
\(\Rightarrow\left(x+1\right)\times x=0\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=0\end{cases}}}\)
~Std well~
#TT