Ok very interesting and I always wanted to try one of these and never did so now we will put you to the test since I have the answer along with the problem there can only be so many paths to figure out any math problem and my way is finding the cycle for all numbers come back in there unique cycles😮

Very lengthy procedure indeed Simply find out c/a You have ca Multiply U get c Then finf out a n b Add a+b+c This is done in 40vsec Any sum should b done in 1 minute only

Why this lengthy method

I want to meet with you sir

Camera is not good

b=100/a bc=200 C=200/b=200/100. X a=2a Ca=300 2a*a=300 a^2=150 a=150^1/2=12.25 b=100/12.25=8.16 C=300/12.25=24.49 a+b+c=44.90

1873

3^112^2 +4^4^2^2+7^72^2/3^11^2^1+4^4^2^17^72^1 1^11^1+2^22^2^1^1+1^11^1/3^1^1^1+2^22^21^1+1^11^1 1^11^1/3 1^11^2 3^2 (x ➖ 3x+2)

Cube the expression and then take the square root.

55√6/3

(3/2)^3/2 (3^1/2^1)^3^2 (1^1/1^1)^3^2 3^2;(x ➖ 3x+2)

10^10 2^52^5 1^12^1 2^1 (b ➖ 2a+1) 10^20 2^55^4 2^1^14 2^1^12^2 1^1^1^1^2 1^2 (c ➖ 2b+1) 10^30 2^55^6 2^1^13^2 1^1^13^2 3^2 (c ➖ 3a+2)

Thank you for explaining. [My method:] (ab)(bc)(ca)=6,000,000 ∴ (abc)^2=6,000,000 ∴ abc = ±1000√6 ∴ a = (abc)/(bc) = (±1000√6)/200 = ±5√6 b = (abc)/(ca) = (±1000√6)/300 = ±(10/3)√6 c = (abc)/(ab) = (±1000√6)/100 = ±10√6 ∴ a+b+c = ± (5 + 10/3 + 10) √6 = ±(55/3)√6 <<< If the problem condition requires "a, b, c: positive," the answer is only (55/3)√6. [Note: a>0, b>0, c<0] But I guess the problem condition requires "a, b, c: real number," Therefore, I think -(55/3)√6 [Note: a<0, b<0, c<0] is also one of the values. >>>

Great explanation

(a, b and c are of same sign. ab).(bc).(ca) = 6000000 =(abc)^2, so abc = 1000.sqrt(6) or abc = -1000.sqrt(6) *If abc = 1000.sqrt(6) then a = abc/bc = 5.sqrt(6), in the same way b = (10:3).sqrt(6) and c = 10.sqrt(6) Then a +b +c = (55/3).sqrt(6) *If abc = -1000.sqrt(6) then a, b, c are the opposite of the former values and a +b + c = (-55/3).sqrt(6)

10^4+9^4+19^4

Total 608. Let one side be a. The other is 343-a. So asquareplus 343-a the whole square. So you get a quadratic equation involving a. So it will have two roots. One of them is right.

👏👏

0

(x ➖ 5x+5)

(y ➖ 3x+2) (2) ➖ (5)= 3.(y ➖ 3x+3)

Thanks,the rules exponential is the leader

Sqrt[5Surd[5,31]Sqrt[5Surd[5,31]Sqrt[5Surd[5,31]Sqrt[5Surd[5,31]Sqrt[5Surd[5,31]]]]]]=5 x=5Surd[5,31]

130/2=60.10m^Log^m130/8= 1.50 {60.10+1.50}=61.60 61^1.5^6 1^1.5^3^2 5^1^3^2 1^1^3^2 3^2( m ➖ 3 m +2),

You should state that a, b, and c are natural numbers. Otherwise, the solutions set is infinite.

Great explanation

at 2:05 let f(x) = x^3 + x - 130, by RRT f(5) = 0 and using SDM, f(x) = (x - 5)(x^2 + 5x + 26)

Thanks

（２の11乗）＝2048，（2の5乗）＝32……よって，ａ＝11，ｂ＝5となる！ This question is very very easy ！

(√3 +1):(√3 - 1)=2+√3

3^25+5^5/3^25+5^5 3^5^5+ 1^1/3^5^5+1^1 3^1^1/3^1^1 1^1/3^1 3^1 (x ➖ 3x+1)

(2x2+x^2}= 2x^4 2^1x^2^2 1^1x^1^2 x^1^2 (x ➖ 2x+1){2y^2+x^2} 2yx^4 2^1yx^2^2 1^1yx^1^2 yx^1^2 (yx ➖ 2yx +1)

0

Nice Square Root Math Simplification: (√75 + √25)/(√75 - √25) = ? (√75 + √25)/(√75 - √25) = [(√25)(√3 + 1)]/[(√25)(√3 - 1)] = (√3 + 1)/(√3 - 1) = (√3 + 1)²/[(√3 - 1)(√3 + 1)] = (3 + 2√3 + 1)/(3 - 1) = (4 + 2√3)/2 = 2 + √3 Answer check: (√75 + √25)/(√75 - √25) = 2 + √3; Confirmed as shown Final answer: Simplified form; (√75 + √25)/(√75 - √25) = 2 + √3

Practice putting every base and exponent in standard notation.

I knew 16^18>18^16 but I just wanted to put them in Standard notation.

16^18>18^16

16^18=4722366482869645213696 18^16=121439531096594251776

_a² - b² = 9_ _ab = 3_ Let _x = a + b, y = a - b_ ∴ _xy = 9_ _x² - y² = 4ab = 12_ _x²_ is a root of _(t - x²)(t + y²) = 0_ ⇒ _t² - (x² - y²)t - (xy)² = 0_ ⇒ _t² - 12t - 81 = 0_ ⇒ _x² = t = 6 ± 3√13_ ⇒ _x² = 6 + 3√13 ( x ∊ _*_R_*_ )_ ⇒ *_x = ±√(6 + 3√13)_*

a = k = 41 → b = (k^2 - 1)/2 = 840 → c = (k^2 + 1)/2 = 841 → area ∆= 420(41) = 17220 = x; perimeter ∆ = a + b + c = 1772 (this way is well knowned since the Old Greek Plato 🙂)

Thanks Professor, wonderful and nice

(B ➖ 3a+3) (b ➖ 3a+1)

just eyeballing it we can see that 2016 can be written as 2048 - 32 => 2^11 - 2^5. Therefore we can say that a = 11 and b = 5

this feels like unnecessary, when you have big numbers you have to use calculator

(a,b) = 11,5

Honestly, though your calculation is good, 625*625*125-6 is simpler IMO

2016/2=108 2016/2= 108 {108 ➖ 108}=0

a = 11, b = 5 - a bit complicated, Sir.

This is awesome. Nice additon formulas, many thanks, Sir!

2016 ÷ 32 is 63. 32 is 2^5 63 = 64-1 = 2^6 - 1 2^a - 2^b assuming a > b 2^b (2^(a-b) -1) 2^5(2^6-1)=2016 Hence b = 5 and a=11