Question

[Nghi Nguyễn]

Giúp mình với mọi người ạ?Mình cảm ơn!

Answer 1

[ILoveMath]

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\(\Rightarrow\dfrac{5}{2x}=1\)

\(\Rightarrow2x=5\\ \Rightarrow x=\dfrac{5}{2}\)

Answer 2

[Nguyễn Huy Tú]

a, đk \(2x^2\left(x-1\right)\ne0\Leftrightarrow x\ne0;x\ne1\)

b, \(A=\dfrac{5x\left(x-1\right)}{2x^2\left(x-1\right)}=\dfrac{5}{2x}=1\Rightarrow5=2x\Leftrightarrow x=\dfrac{5}{2}\left(tm\right)\)