Fudo lm thiếu 1 trường hợp r
Ta có \(\left(2x-5\right)^2=\left|2x-5\right|\)
\(\Leftrightarrow\orbr{\begin{cases}\left(2x-5\right)^2=2x-5\\\left(2x-5^2\right)=5-2x\end{cases}}\)
TH1: \(\left(2x-5\right)^2=2x-5\)
\(\Leftrightarrow\left(2x-5\right)^2-\left(2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x-5-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-5=0\\2x-5-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=5\\2x-6=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\2x=6\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=3\end{cases}}\) (1)
TH2: \(\left(2x-5\right)^2=5-2x\)
\(\Leftrightarrow\left(2x-5\right)^2-\left(5-2x\right)=0\)
\(\Leftrightarrow\left(2x-5\right)^2+2x-5=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x-5+1\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-5=0\\2x-4=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=5\\2x=4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=2\end{cases}}\) (2)
Từ (1) và (2) \(\Leftrightarrow x\in\left\{\frac{5}{2};2;3\right\}\)
Vậy \(x\in\left\{\frac{5}{2};2;3\right\}\)
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Bài giải
\(\left(2x-5\right)^2=\left|2x-5\right|\)
Mà \(\left(2x-5\right)^2\ge0\) với mọi x nên :
\(\left(2x-5\right)^2=2x-5\)
\(\left(2x-5\right)^2-\left(2x-5\right)=0\)
\(\Rightarrow\text{ }\left(2x-5\right)\left[\left(2x-5\right)-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}2x-5=0\\2x-5-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x=5\\2x=6\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=3\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{\frac{5}{2}\text{ ; }3\right\}\)
cảm ơn các bạn nhìu nha!!!
