\(\left(2x-1\right)^2+\left(2-x\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x-1+2-x\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\x+1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-1\end{cases}}\)
Vậy tập nghiệm của ptr là : \(S=\left\{\frac{1}{2};-1\right\}\)
\(\left(2x-1\right)^2+\left(2-x\right)\left(2x-1\right)=0.\)
\(\Leftrightarrow4x^2-4x+1+4x-2-2x^2+x=0\)
\(\Leftrightarrow2x^2+x-1=0\)
\(\Leftrightarrow2x^2+2x-x-1=0\)
\(\Leftrightarrow\left(2x^2+2x\right)-\left(x+1\right)=0\)
\(\Leftrightarrow2x\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)-\left(x+1\right)=0\)
TH1: 2x - 1 = 0
=> x = 1/2
Th2: x + 1 = 0
=> x = -1
\(\Rightarrow x\in\left\{\frac{1}{2};-1\right\}\)
\(\left(2x-1\right)^2+\left(2-x\right).\left(2x-1\right)=0\)
\(\Rightarrow\left(2x-1\right).\left(2x-1\right)+\left(2-x\right).\left(2x-1\right)=0\)
\(\Rightarrow\left(2x-1\right).\left[\left(2x-1\right)+\left(2-x\right)\right]=0\)
\(\Rightarrow\left(2x-1\right).\left(2x-1+2+x\right)=0\)
\(\Rightarrow\left(2x-1\right).\left[x.\left(2+1\right)+1\right]=0\)
\(\Rightarrow\left(2x-1\right).x.3+\left(2x-1\right)\)
\(\Rightarrow2x.x.3-x.3+\left(2x-1\right)=0\)
\(\Rightarrow x.\left(2+3\right)-x.3+\left(2x-1\right)=0\)
\(\Rightarrow x.5-x.3+\left(2x-1\right)=0\)
\(\Rightarrow x\left(5-3\right)+\left(2x-1\right)=0\)
\(\Rightarrow x.2+\left(2x-1\right)=0\)
\(\Rightarrow x.2+2x-1=0\)
\(\Rightarrow x.\left(2+2\right)-1=0\)
\(\Rightarrow x.4-1=0\)
\(\Rightarrow x.4=1\)
\(\Rightarrow x=1\div4\)
\(\Rightarrow x=0,25\)
Bài giải
\(\left(2x-1\right)^2+\left(2-x\right)\left(2x-1\right)=0\)
\(\left(2x-1\right)\left[\left(2x-1\right)+\left(2-x\right)\right]=0\)
\(\left(2x-1\right)\left[2x-1+2-x\right]=0\)
\(\left(2x-1\right)\left[x+1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}2x-1=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x=1\\x=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-1\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{\frac{1}{2}\text{ ; }-1\right\}\)
Bài giải
\(\left(2x-1\right)^2+\left(2-x\right)\left(2x-1\right)=0\)
\(\left(2x-1\right)\left[\left(2x-1\right)+\left(2-x\right)\right]=0\)
\(\left(2x-1\right)\left[2x-1+2-x\right]=0\)
\(\left(2x-1\right)\left[x+1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}2x-1=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x=1\\x=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-1\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{\frac{1}{2}\text{ ; }-1\right\}\)
\(\left(2x-1\right)^2+\left(2-x\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left[\left(2x-1\right)+\left(2-x\right)\right]=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x-1+2-x\right)=0\)\(\Leftrightarrow\left(2x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-1\end{cases}}\)
Vậy phương trình có tập nghiệm \(S=\left\{-1;\frac{1}{2}\right\}\)
