a, ĐKXĐ:\(2x^3-2x^2\ne0\Rightarrow2x^2\left(x-1\right)\ne0\Rightarrow\left\{{}\begin{matrix}2x^2\ne0\\x-1\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne0\\x\ne1\end{matrix}\right.\)
b, \(A=\dfrac{5x^2-5x}{2x^3-2x^2}\)
\(\Rightarrow A=\dfrac{5x\left(x-1\right)}{2x^2\left(x-1\right)}\)
\(\Rightarrow A=\dfrac{5}{2x}\)
Để A=1\(\Rightarrow\dfrac{5}{2x}=1\)
\(\Rightarrow2x=5\\ \Rightarrow x=\dfrac{5}{2}\)
a, đk \(2x^2\left(x-1\right)\ne0\Leftrightarrow x\ne0;x\ne1\)
b, \(A=\dfrac{5x\left(x-1\right)}{2x^2\left(x-1\right)}=\dfrac{5}{2x}=1\Rightarrow5=2x\Leftrightarrow x=\dfrac{5}{2}\left(tm\right)\)
a) Để A xác định thì
\(2x^3-2x^2\ne0\Leftrightarrow2x^2\left(x-1\right)\ne0\Leftrightarrow x\ne0;1\)
b)\(\dfrac{5x^2-5x}{2x^3-2x^2}=\dfrac{5x\left(x-1\right)}{2x^2\left(x-1\right)}=\dfrac{5}{2x}\)
Tại x=1 , giá trị của A là :\(\dfrac{5}{2\cdot1}=\dfrac{5}{2}\)
