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{ (4+1/16)÷(4-1/16) } × 16 { 64+1 }÷{ 64 - 1 } 65 ÷ 63 => 65/63

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01) 242 = 2 * 121 02) 242^2 = (2 * 121)^2 = 4 * 121^2 03) 242^2 - 121^2 = 4 * 121^2 - 121^2 = 121^2 * (4 - 1) = 121^2 * 3 04) (3 * 121^2) / (242 * 121) = 3 * 121 / 242 05) (3 * 121) / (2 * 121) = 3 / 2 06) ANSWER : 3 / 2

It is 65 over 63 because the 16s cancel each other out.

242^2 - 121^2 = (242+121).(242-121) 363.121 363.221. ------------ = 363/242 = 3/2

No need to substitute here. (121²*2²-121²)/(121*2*121)=121²*(4-1)/(121²*2)=3/2. At first sight the solution would be smaller than 1, but no

(3x+2)² = 9x²+10x-6 9x²+12x+4 = 9x²+10x-6 2x = -10 x = -5

x²/2 + 4 = 3x x²/2 -3x + 4 = 0　　　 x² - 6x + 8 = 0 　　　　　(x - 2)(x - 4) = 0 x = 2, 4

25!=a 26!=26*25!=26a 27!=27*26*25!=702a (702a+26a)/a=728a/a=728

(99^201+99^200)/(99^200)=100

27^x = a a+a+a=27 3a=27 a=9 27^x = 9 (3^3)^x=3^2 3^3x=3^2 3x=2 x=2/3

³√3 = 3^1/3 √27 = √(3^3) = 3^3/2 3^1/3 / 3^3/2 = 3^1/3-3/2 = 3^2/6-9/6 = 3^-7/6 = 3^x x = -7/6

3^√x/2 = 3^2 √x/2 = 2 √x = 4 x = 16

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Best music and presentation of any math channel!

Зачем куча сложных манипуляций От первого уравнения отнять 2ху (х-у) ^2=13-2ху 25-13=-2ху В итоге элементарная система ху=-6 х-у=5 Решение труда не составит

thank you very much)

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Ну здесь проще простого. Степенную дробь 25¹²⁵/125²⁵ представляем в виде степени с одинаковыми основаниями равными 5; т.е. 25¹²⁵=5²⁵⁰; 125²⁵=5⁷⁵. Теперь выполним действия с одинаковыми основаниями: 5²⁵⁰/5⁷⁵=5¹⁷⁵. Значит 25¹²⁵/125²⁵=5¹⁷⁵.

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Math Olympiad System of Equations: x² + y² = 13, x - y = 5; x, y = ? y = x - 5, x² + (x - 5)² = 13, 2x² - 10x + 25 = 13, 2(x² - 5x + 6) = 0 x² - 5x + 6 = (x - 2)(x - 3) = 0, x - 2 = 0 or x - 3 = 0 x = 2, y = x - 5 = 2 - 5 = - 3 or x = 3, y = x - 5 = 3 - 5 = - 2 Answer check: x = 2, y = - 3: x² + y² = 2² + (- 3)² = 13, x - y = 2 - (- 3) = 5; Confirmed x = 3, y = - 2: x² + y² = 3² + (- 2)² = 13, x - y = 3 - (- 2) = 5; Confirmed Final answer: x = 2, y = - 3 or x = 3, y = - 2

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A Nice Olympiad Factorial Problem: (√9)!/√(5!) = ? (√9)!/√(5!) = 3!/√[(5)(4)(3!)] = (√3!)/√[(5)(4)] = (√6)/(2√5) = (√30)/10

At first sight you would never say the solution is such a large number. But it is indeed

112²-88² = (112+88)(112-88) 112²-88² = (200)(24) 112²-88² = 16*100*3 √(112²-88²) = √(16*100*3) √(112²-88²) = √(16*100)*√3 √(112²-88²) = 40√3

√x = 4, x= 16

Way too many steps, 3!/sqrt(5!) × (sqrt(5!)/sqrt(5!)) =(3!×sqrt(5!))/5! =sqrt(5!)/5×4 =sqrt (120)/20 =sqrt(30)/10

40√3.

The answer is 5^175

5^2×125/5^3×25=5^250/5^75=5^175

4!/(24)^1/2=24/2×6^1/2=12×6^_1/2

6/(120)^1/2 => 6/2(30)^1/2 = 3/(30)^1/2 ~= 3/5,447

(( n + 1) ! + n! ) / ( n - 1)! = n ( n + 1 + 1) = (n + 2) n

25!(26×27+26)/25!==26(27+1)=26×28=

factorization! that's it. 40√3

Me da (1/5)^25

I come for the penmanship. I stay for the relaxing jazz. The math itself is tertiary.

5¹⁷⁵: 25¹²⁵/125²⁵ =(5²)¹²⁵/(5³)²⁵ =[5^(2×125)]/[5^(3×25)] =5²⁵⁰/5⁷⁵ =5^(250-75) =5¹⁷⁵ QED

112 + 88 = 200 112 - 88 = 24 200 = 2.100 24 = 2^3.3 √2^4.3.10^2 40√3

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X³ = 3³ x=3

Very nice ,good teacher

40x(sq rt 3).

excelente apoyo mutuo

What xd

25!(26*27+26)/25!= 26(27+1) (27-1)(27+1) 27^2-1 728