Question Number 123608 by malwan last updated on 26/Nov/20
$${sir}\:{TINKUTARA} \\ $$$${what}\:{is}\:{the}\:{problem}\:{with}\:{brackets} \\ $$$${in}\:{red}\:{color}\:? \\ $$$$\left.\mathrm{1}\right)\:\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{5}{x}} \:\: \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{5}{x}} \\ $$
Commented by JDamian last updated on 26/Nov/20
I think the problem is the way the app matches a pair of brackets: the first ) and the following ( are a pair having nothing in between. Then the ( and the last ) are not paired. If you insert a fraction between them, you'll see that the size of them will be increased according to the size of the fraction.
Commented by benjo_mathlover last updated on 26/Nov/20
$$\left(\frac{\mathrm{1}}{\mathrm{5}{x}}\right)^{\frac{\mathrm{1}}{\mathrm{5}{x}}} \\ $$
Commented by malwan last updated on 27/Nov/20
$${thank}\:{you} \\ $$$$\mathrm{1}−\:\:\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{5}{x}} \\ $$
Commented by mr W last updated on 28/Nov/20
$$\left.\right)\:\left(\:{will}\:{be}\:{treated}\:{as}\:\left(\:\right)\right. \\ $$$$\left.\mathrm{1}\right)\:\:\frac{{A}}{{B}}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{5}{x}} \:\: \\ $$$$\left.{so}\:{don}'{t}\:{use}\:\mathrm{1}\right)\:{but}\:\left(\mathrm{1}\right)\:{instead} \\ $$
Answered by Lordose last updated on 26/Nov/20
$$\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{5x}} \\ $$
Answered by $@y@m last updated on 27/Nov/20
$$\left(\mathrm{1}\right)\:\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{5}{x}} \:\: \\ $$$$\left.{Just}\:{give}\:{opening}\:{parenthesis}\:{before}\:\mathrm{1}\right) \\ $$
Commented by malwan last updated on 27/Nov/20
$${thank}\:{you} \\ $$