Your Calculator Might Explode

$\large \dfrac{{9999}^{999}-{9999}^{998}}{{9999}^{998}}$

$\large=\dfrac{{9999}^{999}}{{9999}^{998}}-\dfrac{{9999}^{998}}{{9999}^{998}}$

$\large=9999-1=\color{#20A900}{ \boxed {9998}}$

$\frac{9999^{999}-9999^{998}}{9999^{998}}$

$=\frac{9999^{998} \times(9999-1)}{9999^{998}}$

$=\frac{9999^{998}\times9998}{9999^{998}}=\frac{9998}{1}=\boxed{\color{#D61F06}{9998}}$

Your head won't blast. But your calculator totally will.

$\frac{9999^{999}-9999^{998}} { 9999^{998}}$

$= \frac{9999^{998} \times (9999-1)} {9999^{998}}$

The ' $9999^{998}$ ' cancels out to give $(9999-1) = 9998$

It's easy all of them have the same base so you can subtract the exponents in a division, so the first term of the subtraction after distribuying the division, is 9999^1 and the second term is just 1 so 9999 - 1 = 9998

The same as X raised to the power of n. You first take out the LCD of the two terms on the numerator And then you cancel the factors. Then you are left with 9999-1. The answer is 9998.

$\frac{9999^{999}-9999^{998}}{9999^{998}} = \frac{9999^{998} \times (9999 - 1)}{9999^{998}} = 9999-1 = 9998$

Using the DoTS property gives us 9999^998(9999 - 1) - cancelling the 9999^998. It remains like 9999 - 1, giving us 9998.

In my mind, I did a similar method to what many people below have described

Let A = 9999^998

(9999)A - A

= --------------- A

A (9999 - 1)

= --------------- A

= 9999 - 1

= 9998

• $\frac{9999^{999}-9999^{998}}{9999^{998}}$

• $\frac{9999^{998}*(9999-1)}{9999^{998}}$

• 9999-1

• 9998

$a^{n}-a^{n-1} = (a-1)a^{n-1}$

The above can be proved through simple distribution on the right side. What we have here is:

$\frac{a^{n}-a^{n-1}}{a^{n-1}} = \frac{(a-1)a^{n-1}}{a^{n-1}} = (a-1)$

In this case a is 9999 and n is 999. Regardless, the answer is a - 1, which is 9999 - 1, which is 9998.

Let A=9999^9998

(9999A - A) / A

9998A / A

9998

I split it into two fractions, 9999^999/9999^998 and -9999^998/9999^998. Using the index quotient rule, the first fraction can be evaluated to 9999^1, which is 9999. The second fraction’s denominator and numerator are equal (keep in mind it is negative) meaning it is 1 and so it is 9999 + -1 = 9998

imagine...

$\frac{(2^3 - 2^2)}{2^2}$ = a

$2^3 - 2^2 = 2^2 * a$

$2^3 = (2^2 * a) + 2^2$

$2^3 = (a+1) * 2^2$

$2^3 / 2^2 = a+1$

$2^{3-2} = a+1$

back to question:

$9999^{999-998} = a+1$

$9999^1 = a+1$

$9999-1 = a$

$9998$

{(9999^999)-(9999^998)}/9999^998}={(9999^998)(9999-1)}/9999^998=9999-1=9998

$\frac {9999^{999} - 9999^{998}}{9999^{998}} = \frac {9999^{998}\times(9999 - 1)}{9999^{998}}$ $= 9999 - 1 = 9998$

( (9999^999) - (9999^998) ) / (9999^998) = ( (9999^998)(9999-1) ) / (9999^998) = 9999-1 = 9998

So easy ( 9999^999 - 9999^998 ) / ( 9999^998) = ( 9999^998( 9999 - 1 ) ) / (9999^998) = 9999 -1 = 9998

9999^{999}-9999^{998}/9999^{998} Taking 9999^{998} common in the numerator, we get, 9999^{998}(9999-1)/9999^{998) Cancelling out the like terms, we get, 9999-1=9998

= (9999^998 (9999-1)) / 9999^998 = 9999-1 = 9998

(9999^999 - 9999^998 )/9999^998

=[9999^998×(9999-1)]/9999^998

=9999-1 =9998

9999^9998(9999-1)/9998 = 9999-1 = 9998

(9999^999 - 9999^998)/9999^998 = 9999^998(9999 - 1) /999^998 = 9999 - 1 = 9998

(9999^999-9999^998)/9999^998 = (9999^999/9999^998)-1 = 9999-1 = 9998

x = 9999

-> (x^999 - x^998)/(x^998)

-> (x^999/x^998) - (x^998/x^998)

-> x - 1

= 9999 - 1

= 9998

{9999~998(9999-1)}/9999~998 = 9999-1 =9998

well i just slove it in my head~ no sol hahahaha XD

(9999^999 - 9999^998 )/(9999^998) = ( 9999^999)/ (9999^998) - (9999^998)/(9999^998) = 9999^(999-998) - 9999^(998-998) = 9999^1 - 9999^0 = 9999 - 1= 9998

The same as X raised to the power of n. You first take out the 9999 raise to power 8 of the two terms on the numerator And then you cancel the factors. Then you are left with 9999-1. The answer is 9998.

9999^999 = 9999 * 9999^998 Therefore 9999^999 - 9999^998 = 9998 * 9999^9998. So if you divide by 9999^998 the you're left with only 9998