The question: my work: I concluded that x^3((1/x)-1) equals to (-x^3/x), not x^2 - x^3
3 38175425 New member Joined Jun 20, 2020 Messages 35 Dec 7, 2022 #1 The question: my work: I concluded that x^3((1/x)-1) equals to (-x^3/x), not x^2 - x^3
skeeter Elite Member Joined Dec 15, 2005 Messages 3,216 Dec 7, 2022 #2 [imath]\dfrac{1}{x} - 1 \ne -\dfrac{1}{x}[/imath] use the distributive property of multiplication [imath]a(b-c) = ab - ac[/imath]
[imath]\dfrac{1}{x} - 1 \ne -\dfrac{1}{x}[/imath] use the distributive property of multiplication [imath]a(b-c) = ab - ac[/imath]
3 38175425 New member Joined Jun 20, 2020 Messages 35 Dec 7, 2022 #3 skeeter said: [imath]\dfrac{1}{x} - 1 \ne -\dfrac{1}{x}[/imath] use the distributive property of multiplication [imath]a(b-c) = ab - ac[/imath] Click to expand... thanks!
skeeter said: [imath]\dfrac{1}{x} - 1 \ne -\dfrac{1}{x}[/imath] use the distributive property of multiplication [imath]a(b-c) = ab - ac[/imath] Click to expand... thanks!
Steven G Elite Member Joined Dec 30, 2014 Messages 14,561 Dec 9, 2022 #4 \(\displaystyle Since\ \dfrac{1}{x}-\dfrac{2}{x}=\dfrac{-1}{x}\ it\ follows\ that\ \dfrac{1}{x}-1 \neq \dfrac{-1}{x}\)
\(\displaystyle Since\ \dfrac{1}{x}-\dfrac{2}{x}=\dfrac{-1}{x}\ it\ follows\ that\ \dfrac{1}{x}-1 \neq \dfrac{-1}{x}\)
