Let $123456789$ be $x$
Therefore, the equation will be
$x^{2}-(x-1)(x+1)$
$=x^{2}-(x^{2}-1)$
$=x^{2}-x^{2}+1$
$=1$

why can't like this!

= 123456789^2 - (123456788 x 123456790)

= 123456789^2 - (123456789 - 1) (123456789 + 1)

= 123456789^2 - (123456789^2 - 1)

= 123456789^2 -123456789^2 + 1

= 1

here you go, it's one:)

Let 123456789 be a basic number between other two, doesnt matter wich number you choose, it's a math property

So we have:

3 X 3 - ( 2 X 4) = 9 - 8 = 1

4 X 4 - (3 X 5) = 16 - 15 = 1

5 X 5 - (4 X 6) = 25 - 24 = 1

And so on...

Ignoring most of the numbers above, we can try this expression: ( 89 * 89 ) - ( 88 * 90 ) = ? [ 88 * 90 =7920 , 89 * 89 =7921 ] therefore the answer is 1 in this case. so no matter how big the number is, if the other digits are same then they can be ignored and the smaller numbers that are there can be used to get the answer easily

let 123456789 = x , then 123456788 = x-1 & 123456790 = x+1 x^2 - ( x - 1 )( x + 1 ) = x^2 - ( x^2 - x + x - 1 ) = 1

123456789 × 123456789 = 15241578750190500 123456788 × 123456790 = 15241578750190500 15241578750190500 - 15241578750190500=0

It's pretty simple if you write it as a difference of squares formula.
(Difference of squares => A^2-B^2=(A+B)(A-B)
123456789 = a
a^2 - (a-1)*(a+1)
=a^2 - (a^2 - 1^2)
=a^2-a^2+1
=1

p2 - (p - 1)(p + 1) = p2 - p2 + 1 = 1, Therefore 123456789^2 - (123456789+1)(123456789 +1) = 123456789^2 - 123456789^2 + 1 = 1

Put the 123456787 in evidence 123456789 = 123456787 + 2 Rest 2 123456788 = 123456787 + 1 Rest 1 1234567890 = 123456787 + 3 Rest 3

2² - ( 1 * 3 ) 4 - ( 3 ) 4 - 3 = 1

OBS:. If you put 123456788 in evidence, the 123456789 will be "1²" and when use ( - ) = "NEGATIVE ANSWER"

(A+D)(A-D) = A^{2} - D^{2}
Adding D^{2} and subtracting (A+D)(A-D) to both sides, we get:
A^{2} - (A+d)(A-d)=D^{2}
123456789^{2} - (123456789+1)(123456789-1) = 1^{2}
We get:
123456789^{2} - (123456790)(123456788) = 1

$123456789^{2}-(123456788\times 1234567890)$

Each number in parentheses differs by 1 from 123456789

$=123456789^{2}-((123456789-1)\times(123456789+1)$

The product can be written as the difference of two squares

$=123456789^{2}-(123456789^{2}-1^{2})$

$=123456789^{2}-123456789^{2}+1^{2}$

$=1^{2}$

$=1$

Let us consider three consecutive numbers 9,10,11. then 10^2-(9*11) =100-(99) =1.

$a^{2}-1= (a-1) \times (a+1)$

$123456789^{2}-1= 123456788 \times 123456790$

$0= -123456789^{2}+1-(123456788 \times 123456790)$

$0= 123456789^{2}-1-(123456788 \times 123456790)$

$1= 123456789^{2}-(123456788 \times 123456790)$

$=\boxed{1}$

Another way

(123456788+1)^2-123456790(123456788) Expanding the brackets give

123456788^2+2(123456788)(1)+1^2 - 123456790(123456788) =123456788^2 - (123456788)(123456788) +1 =(123456788)^2-123456788^2+1 =1

/ ( let's, 123456789=123456700+89; 123456788=123456700+88; 123456790=123456700+90 now, imagine 123456700 =a So, (a+89)^2 - {(a+88)(a+90)} =a^2 + 178a + 7921 -a^2 - 178a - 7920 =1 / )

Let x = 123456789

substitute x to the given, 123456789 - (123456788 * 123456790) x^2 - (x-1)(x+1) x^2 - (x^2 - 1^2) x^2 - x^2 +1 1

x=123456789; 123456789²-(123456788 x 123456790)=x²-[(x-1)(x+1)]= x²-[x²-1]=x²-x²+1=0+1=1

(1234567)^2 x 89 - (1234567 x 1234567) x (88+90) = 1

Let 123456789123456789 be xx Therefore, the equation will be x^{2}-(x-1)(x+1)x 2 −(x−1)(x+1) =x^{2}-(x^{2}-1)=x 2 −(x 2 −1) =x^{2}-x^{2}+1=x 2 −x 2 +1 =1=1

set $123456789=a$ , the expression can be changed to $a^2-((a-1)(a+1))=1$

Let the no. 123456789 be x.
Therefore, it implies that 123456788 = 123456789-1 = x-1
and 123456790 = 123456789+1 = x+1
So, it forms the equation, x^2 - (x-1)(x+1) = x^2 - x^2 + 1 =1

$123456789^2 - (123456788 \times 123456790)$

$= 123456789^2 - (123456789 - 1)(123456789 + 1)$

$= 123456789^2 - (123456789^2 - 1^2)$

$= 123456789^2 - 123456789^2 + 1^2$

$= 1^2$

$= 1$

$x = 123456789^2 - (123456788\times 123456790)$

$x = 123456789^2 - ((123456788 + 1 - 1)\times (123456790 + 1 - 1))$

$x = 123456789^2 - ((123456789 - 1)\times (123456789 + 1))$

$x = 123456789^2 - (123456789^2 + 123456789 - 123456789 - 1)$

$x = 123456789^2 - (123456789^2 - 1)$

$x = 123456789^2 - 123456789^2 + 1$

$x = 1$

$123456789^2-(123456790 \times 123456788)=123456789^2-(123456789^2-1)=123456789^2-123456789^2+1=\boxed{\large{1}}$

$x = 123456789$

$x - 1 = 123456788$

$x + 1 = 123456790$

$x^2 - (x - 1) (x + 1)$

$x^2 - (x^2 - 1^2)$

$x^2 - x^2 + 1^2$

$1^2$

$1$

$123456789^2 - (123456789 - 1) (123456789 + 1)$

$123456789^2 - (123456789^2 - 1^2)$

$123456789^2 - 123456789^2 + 1^2$

$1^2$

$1$

We know that $x^{2} = (x-1)(x+1)+1$ , since $(x-1)(x+1) = x^{2}-1$ .

Thus, $123456788 \times 123456790 = 123456789^{2}$

So, the expression $123456789^{2} - 123456788 \times 123456790 = \boxed{1 }$ .

Because the number 123456700 is common for the three numbers involved in the problem [(123456700+89), (123456700+88), (123456700+90)],

(123446700)^2-(123456700×123456700)=0

Therefore, the only numbers that we need to worry about is 89, 88, and 90 So..

89^2-(88×90)=1

Why not this?

think about it more and you will get the answer. also online calculators do not work.

We're talking abou 3 numbers right in a row. So put it in more manageable terms. 1, 2, 3 for example. 2^2 - (3x1)= 4-3=1. Test it to confirm. 9, 10, 11= 100 - 99= 1. With the equation b^2 - (a x c) where a, b, c are three integers in a row, 1 will always be the answer.