\(b.\)
\(\left(x-\dfrac{1}{5}\right)\left(1\dfrac{3}{5}+2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{5}=0\\1\dfrac{3}{5}+2x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Giải.
a) \(\left(2-x\right)\left(\dfrac{4}{5}-x\right)< 0\)
Với mọi số thực x , ta luôn có \(2-x>\dfrac{4}{5}-x\)
Nên \(\left(2-x\right)\left(\dfrac{4}{5}-x\right)< 0\Leftrightarrow\left\{{}\begin{matrix}2-x>0\\\dfrac{4}{5}-x< 0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< 2\\x>\dfrac{4}{5}\end{matrix}\right.\)\(\Leftrightarrow\dfrac{4}{5}< x< 2\)
Vậy ....
b) \(\left(x-\dfrac{1}{5}\right)\left(1\dfrac{3}{5}+2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{5}=0\\1\dfrac{3}{5}+2x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{-4}{5}\end{matrix}\right.\)
Vậy....
tik mik nha !!!
\(\left(2-x\right)\left(\dfrac{4}{5}-x\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2-x>0\Rightarrow x< 2\\\dfrac{4}{5}-x< 0\Rightarrow x>\dfrac{4}{5}\end{matrix}\right.\\\left\{{}\begin{matrix}2-x< 0\Rightarrow x>2\\\dfrac{4}{5}-x>0\Rightarrow x< \dfrac{4}{5}\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\dfrac{4}{5}< x< 2\)
\(\left(x-\dfrac{1}{5}\right)\left(1\dfrac{3}{5}+2x\right)=0\)
\(\Rightarrow\left(x-\dfrac{1}{5}\right)\left(\dfrac{8}{5}+2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{5}=0\\\dfrac{8}{5}+2x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
