Question Number 200970 by Mastermind last updated on 27/Nov/23
Answered by JDamian last updated on 27/Nov/23
$${Really}? \\ $$
Answered by mr W last updated on 27/Nov/23
$$\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{2}{x}−\mathrm{1}}=\frac{\mathrm{1}+\left(\mathrm{2}{x}\right)^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{x}−\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}{x}−\mathrm{1}}+\left(\mathrm{2}{x}+\mathrm{1}\right) \\ $$$$\Rightarrow{A}=\mathrm{2}{x}+\mathrm{1} \\ $$
Answered by Rasheed.Sindhi last updated on 28/Nov/23
$$\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{2}{x}−\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}{x}−\mathrm{1}}+{A}\Rightarrow{A}=\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{2}{x}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{x}−\mathrm{1}} \\ $$$$\Rightarrow{A}=\frac{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{x}−\mathrm{1}}=\frac{\left(\mathrm{2}{x}+\mathrm{1}\right)\cancel{\left(\mathrm{2}{x}−\mathrm{1}\right)}}{\cancel{\left(\mathrm{2}{x}−\mathrm{1}\right)}}=\mathrm{2}{x}+\mathrm{1} \\ $$