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x2 equals minus 2.55; but since x must be greater or less than 2.23, then this means that x2 is NOT an answer.

two type of X

Raise both sides of the equation to the sixth power (1 + √(1+x))³ = x² Let у = √(1+x) => x = y² - 1 Then (1 + y)³ = (y² - 1)² = (y - 1)² * (y + 1)² or y + 1 = 0 => x = 0 (1 + y) = (y - 1)² => y² - 3y = 0 => x = -1, x = 8 remains after verifying x = 8

97!(98×99+98)/97!=98×(99+1)=98×100 =9800

22×21×20!/20!(21-1)=22×21/20=262/20=231÷10

6×5×4×3×2/6×4×2=5×3=15

8×6×4×2×8×5×2/8×7×6×5×4×3×2=2×8/7×3=16/21

Sorry no differency between x and n! It is no x!

Amene and interesting!

Actually, the solution is not working. There is no clear definition in which order you have to calculate x^x^x^x^x.... - if you do x^x, then again ^x, again ^x, the "solution" is wrong. Check it out!

This proves that if there is an answer it is the square root of 2. It doesn't prove it is an answer.

I am not the 1 in 100, but I appreciate the problem and its clever solution.

Sir, Is the answer not (sqr(7) + sqr(3)) / 2? In your answer you subtract sqr(3) instead. I see the error happening at about 8:10

I don't suppose you need the 12 square on the left to solve the question

2:51 excellent

My way of solution ▶ 9⁶ + 9⁵ + 9⁴+9³+9²+9 = 9⁵(9+1) + 9³(9+1) + 9(9+1) = 9⁵*10 + 81*9*10+9*10 = 9²+9²*9*10+729*10+90 = 81²*9*10+ 7290+90 = (80+1)²*90+7380 = (6400+160+1)*90+7380 = 6561*90+7380 = 65610*9+7380 = 590.490 +7380 = 597.870

The initial approach is very clever. I'm struggling to understand how I could develop an intuition how to apply it.

Hay