Відео

{x^4+x^4 ➖} {97+97 ➖}{ 56+56 ➖} +{3+3 ➖ }x^16+184+112+6}=302x16 10^30^2x^16 10^3^10^2x^4^4 2^5^3^2^5^2x2^2^2^2 1^1^3^1^1^1x^1^1^1^2 2 3x^2 (x ➖ 3x+2) .

{144x^2+140x^2}/4+({45184 ➖ 4900}/4 284x^4 /4+{4x+4x ➖ }1884/= ={284/4+8x^2}+1884x/4={292x^4/4+1884x^4/4}=1176x^4/8/147 7^7^7 3^4^3^4^3^4 1^2^2^1^2^23^2^2 1^1^1^13^1^2 3^2 (x ➖ 3x+2). {28x^4+20x^4}/8x^4=58x^4/8x^4=7.2x^1 3^4.2x^1 3^2^2.2x^2 3^1^1.2x^1 3.2x (x ➖ 3x+2).

Above example solved by another method! √97+56√3=(7+√48)^2, 7+√48=(2+√3)^2 x^4 =(2+√3)^4 x=2+√3 x3=(2+√3)^3=8+3√3+12√3+18 x^3=26+15√3

पूरा सोच कर पढाईये

at 4:00 if we add first and third equation then x^2 + x = 110 (x+11)(x-10) = 0 x = - 11 , 10 now y = -1-x hence y= 10 , -11

Pura pagal ho gaya hai .angrejo ke jamane ka ek badshah pagal ho gaya thha Rahul vohi lag rahe hain .

(1)/(11x)^2 ➖ (111)^2 ➖ (1)^2/(111x)^2 ➖ (11)^2=1/{121x^2 ➖ 1.121}➖ 1/{1.121x^2 ➖ 121}={1/1000x^2 ➖ 1/1000x^2}={0+0 ➖}/{0+0 ➖ }{x^0+x^0 ➖}=1/1x^1=1x1 (x ➖ 1x+1).

{y^30+y^30 ➖}+{y^20+y^20 ➖}={y^60+y^40}=y^100 y^10^10 y^2^5^2^5 y^1^1^2^1 y^2^1 (y ➖ 2y+1).

A = 5/4 , b =6/3

Bhadiya hai.❤❤❤❤❤

X=1

For solving {1/x^3 - 1/(x+1)^3}=7/8, I resolved it to (1/x - 1/(x+1))(1/x^2 + 1/x(x + 1) + 1/(x +1)^2)= 7/8 as I used the formula for a^3 - b^3. {1/x(x +1)}{1/x^2 + 1/(x+1)^2 + 1/x(x +1)} = 7/8.{1/x(x +1)}{(1/x - 1/(x+1))^2 + 2/x(x + 1) + 1/x(x + 1)} = 7/8 using a^2 + b^2 = (a - b)^2 + 2ab. {1/x(x + 1)}{(1/x(x + 1))^2 + 3/x(x + 1)} = 7/8 Putting 1/x(x + 1) = a. a(a^2 + 3a) = 7/8. a^3 + 3a^2 = 7/8. 8a^3 + 24a^2 = 7. bringing everything to lhs 8a^3 + 24a^2 - 7 = 0. Here if we put a = 1/2, 8(1/2)^3 + 24(1/2)^2 - 7 = 8(1/8) + 24(1/4) - 7 = 1 + 6 - 7 = 0. So we divide by a - 1/2 using synthetic division. This gave (a - 1/2)(4a^2 + 4a + 17) = 0 So a = 1/2. Other roots are (-7 + sqrt(21))/4,(-7 - sqrt(21))/4. Putting back 1/x(x + 1) = 1/2. x(x + 1) = 2 x^2 + x = 2 x^2 + x - 2 = 0 x^2 + 2x - x - 2 = 0 x(x + 2) -1(x +2) = 0 (x - 1)(x + 2) = 0 x = 1,-2

Sir I have a suggestion for you...I think you must stop uploading every day and start solving some harder problems which are unique too...Your videos seems so repeative and boring...That is why I also stopped watching your videos...Start taking inspirations from "Mind your decisions" , "Prime newtons" etc.

A little error at the end . Answer should be 39603. Thanks.

(x^10)^2 ➖ (3x^5)^2+1= {x^100 ➖ 9x^25}+1=9x^75+{1+1 ➖ }={9x^75+2}=9x^77 9x^7^11 9x^7^11^1 3^2x^7^1^1^1 3^1x^1^1 3x^1 (x ➖ 3x+1). (x^110 )^2➖ (x^55)^2+1= {x^1.100 ➖ x^3025}+1x^1875+{1+1 ➖ }={x^1875+2}=x^1877 x^3^6^1^1x^3^3^2^1 x^1^3^2^1 x^3^2 (x ➖ 3x+2).

Very excellent problem really watched more than once to help in increasing the reach!! 💥💥💥💫💥

10^30 10^3^10 2^5^3^2^5 1^1^3^2^1 3^2(ab ➖ 3ab+2). 10^40 10^4^10 2^5^4^2^5 1^1^4^1^1 2^2^1^1 1^2^1^1 21(ac ➖ 2ac+1) . 10^50 10^5^10 2^5^5^2^5 1^1^12^1 2^1 (bc ➖ 2bc+1).

{5/a+5/a ➖}+{6/b6/b ➖}={10/a+12/b}=22/ab 2^11/ab 2^11^1/ab 2^1^1/ab2^1/ab (ab ➖ 2ab+1).

Just some observation trial and error Got the answer to be a=1 b=3

Good method

Nice video

(1)^2/(x^3)^2 ➖ (1)^2/((x^3+3)={1/x^9 ➖ 1/3x^3={0+0 ➖}/3x^6=1/3x^6 =3x^6 3x^3^2 1x^1^2 1x^2(x ➖ 2x+1).

3 is wrong, it will be 3/2. Please amend it.

(x+1)^3 - x^3 = 7{ (x^2+x)^3 }/8 8 {3(x^2+x) +1} -7 (x^2+x)^3 = 0 7t^3 -24t- 8 = 0 where t = x^2+x 7t^3 -28t +4t-8 = 0 7t (t+2)((t-2) +4 (t-2) = 0 (t-2) (7t^2+14t+4) =0 now use x^2+x= t

x=0,-1,1,3

It’s in my head.

1/(x^3)-1/(x+1)^3=7/8 x=-2 x=1 x=(-1/2)±(1/14)iSqrt[147±28Sqrt[21] ]=-0.5±0.1(0.714285 recurring)iSqrt[147±28Sqrt[21]]

[(2x² - 13x + 15)/(x - 3)]² + [(2x² - 11x + 9)/(x - 3)]² = 4 → where: x ≠ 3 [(2x² - 13x + 15)² + (2x² - 11x + 9)²]/(x - 3)² = 4 (2x² - 13x + 15)² + (2x² - 11x + 9)² = 4.(x - 3)² 4x⁴ - 26x³ + 30x² - 26x³ + 169x² - 195x + 30x² - 195x + 225 + 4x⁴ - 22x³ + 18x² - 22x³ + 121x² - 99x + 18x² - 99x + 81 = 4x² - 24x + 36 8x⁴ - 96x³ + 382x² - 564x + 270 = 0 4x⁴ - 48x³ + 191x² - 282x + 135 = 0 → the aim, if we are to continue effectively, is to eliminate terms to the 3rd power. Let: x = z - (b/4a) → where: b is the coefficient for x³, in our case: - 48 a is the coefficient for x⁴, in our case: 4 4x⁴ - 48x³ + 191x² - 282x + 135 = 0 → let: x = z - (- 48/16) → x = z + 3 4.(z + 3)⁴ - 48.(z + 3)³ + 191.(z + 3)² - 282.(z + 3) + 135 = 0 4.(z + 3)².(z + 3) - 48.(z + 3)².(z + 3) + 191.(z² + 6z + 9) - 282z - 846 + 135 = 0 4.(z² + 6z + 9).(z² + 6z + 9) - 48.(z² + 6z + 9).(z + 3) + 191z² + 1146z + 1719 - 282z - 711 = 0 4.(z⁴ + 6z³ + 9z² + 6z³ + 36z² + 54z + 9z² + 54z + 81) - 48.(z³ + 3z² + 6z² + 18z + 9z + 27) + 191z² + 864z + 1008 = 0 4.(z⁴ + 12z³ + 54z² + 108z + 81) - 48.(z³ + 9z² + 27z + 27) + 191z² + 864z + 1008 = 0 4z⁴ + 48z³ + 216z² + 432z + 324 - 48z³ - 432z² - 1296z - 1296 + 191z² + 864z + 1008 = 0 4z⁴ - 25z² + 36 = 0 → by completing the square: 4z⁴ - 25z² = [2z² - (25/4)]² - (25/4)² [2z² - (25/4)]² - (25/4)² + 36 = 0 [2z² - (25/4)]² - (625/16) + (576/16) = 0 [2z² - (25/4)]² - (49/16) = 0 [2z² - (25/4)]² = 49/16 [2z² - (25/4)]² = (7/4)² 2z² - (25/4) = ± 7/4 2z² = (25/4) ± (7/4) 2z² = (25 ± 7)/4 z² = (25 ± 7)/8 First case: z² = (25 + 7)/8 z² = 32/8 z² = 4 z = ± 2 → recall: x = z + 3 x = ± 2 + 3 → x = 1 → x = 5 Second case: z² = (25 - 7)/8 z² = 18/8 z² = 9/4 z = ± 3/2 → recall: x = z + 3 x = 3 ± (3/2) → x = 9/2 → x = 3/2

Thank you for explaining. As for the final answer, -3/2 + 3 = 3/2. (In this video, -3/2 + 3 = 3. I guess it is a mistake.)

Nice problem Let y = sum of expressions in brackets divided by 2 Or y = (2x^2 -12 x +12)/ (x-3) then 2x^2 = y( x-3)+12 x-12 Numerator of first bracket is y(x-3)+3-x =( x-3)(y-1) Numerator of second bracket is y(x-3)+x-3 = (x-3)(y+1) Equation reduces to (y+1)^2+(y -1)^2 = 4 hence y = 1 , - 1 when y = 1 2 x^2 - 13 x+15 = 0 x = 5 , 3/2 when y = -1 2x^2-11 x+9 = 0 x = 1 , 9/2

(4x^4 ➖169+ 30)/(x^2 ➖9)(x^2 ➖ 9)+(4x^4 ➖ 121x^2+18)/(x^2 ➖ 9) =(165x^4+30)/x^0+x^0 ➖)+(117x^2+18)/{x^0+x^0 ➖ }={195x^4/x^1/+135x^2/x^1}=230x^6/x^2=230x^3 10^20^30x^3 10^20^3^10x^3 10^2^10^3^10x^3 2^5^2^2^5^3^2^5x^3 1^1^1^1^1^1^2^1x^3 2x^3 (x ➖ 3x+2).

121/4

X=1 , hit and trial😅

Nice

[(√24)/(√30 + √6)] = [(2√6)/({√6 * √5} + √6)] [(√24)/(√30 + √6)] = [(2)/({√5} + 1)] [(√24)/(√30 + √6)] = 2/(√5 + 1) [(√24)/(√30 + √6)] = 2.(√5 - 1)/[√5 + 1).(√5 - 1)] [(√24)/(√30 + √6)] = 2.(√5 - 1)/[5 - 1] [(√24)/(√30 + √6)] = 2.(√5 - 1)/4 [(√24)/(√30 + √6)] = (√5 - 1)/2 [(√24)/(√30 + √6)]² = [(√5 - 1)/2]² [(√24)/(√30 + √6)]² = (√5 - 1)²/2² [(√24)/(√30 + √6)]² = (5 - 2√5 + 1)/4 [(√24)/(√30 + √6)]² = (6 - 2√5)/4 [(√24)/(√30 + √6)]² = (3 - √5)/2 [(√24)/(√30 + √6)]⁴ = [(3 - √5)/2]² [(√24)/(√30 + √6)]⁴ = (3 - √5)²/2² [(√24)/(√30 + √6)]⁴ = (9 - 6√5 + 5)/4 [(√24)/(√30 + √6)]⁴ = (14 - 6√5)/4 [(√24)/(√30 + √6)]⁴ = (7 - 3√5)/2 [(√24)/(√30 + √6)]⁸ = [(7 - 3√5)/2]² [(√24)/(√30 + √6)]⁸ = (7 - 3√5)²/2² [(√24)/(√30 + √6)]⁸ = (49 - 42√5 + 45)/4 [(√24)/(√30 + √6)]⁸ = (94 - 42√5)/4 [(√24)/(√30 + √6)]⁸ = (47 - 21√5)/2 [(√24)/(√30 + √6)]¹⁶ = [(47 - 21√5)/2]² [(√24)/(√30 + √6)]¹⁶ = (47 - 21√5)²/2² [(√24)/(√30 + √6)]¹⁶ = (2209 - 1974√5 + 2205)/4 [(√24)/(√30 + √6)]¹⁶ = (4414 - 1974√5)/4 [(√24)/(√30 + √6)]¹⁶ = (2207 - 987√5)/2 [(√24)/(√30 + √6)]²⁴ = [(√24)/(√30 + √6)]¹⁶. [(√24)/(√30 + √6)]⁸ [(√24)/(√30 + √6)]²⁴ = [(2207 - 987√5)/2].[(47 - 21√5)/2] [(√24)/(√30 + √6)]²⁴ = (2207 - 987√5).(47 - 21√5)/4 [(√24)/(√30 + √6)]²⁴ = (103729 - 46347√5 - 46389√5 + 103635)/4 [(√24)/(√30 + √6)]²⁴ = (207364 - 92736√5)/4 [(√24)/(√30 + √6)]²⁴ = 51841 - 23184√5 → compared to: a - b√5 → you can deduce that: a = 51841 b = 23184 a - 2b = 51841 - (2 * 23184) a - 2b = 51841 - 46368 a - 2b = 5473

(Sqrt[x]-5)/(Sqrt[x]-4)-(Sqrt[x]-6)/(Sqrt[x]-5)=(Sqrt[x]-7)/(Sqrt[x]-6)-(Sqrt[x]-8)/(Sqrt[x]-7) x=121/4=30 1/4=30.25

(x ➖ 5)^2/,(x ➖ 4)^2= (x^2 ➖ 25)/(x^2 ➖ 16)/={x^0+x^0 ➖}/{x^0+x^0 ➖}=x^1/x^1 (x ➖ 6)^2/(x ➖ 5)^2= (x^2 ➖ 36)/(x^2 ➖ 25)={x^0+x^0 ➖}/{x^0+x^0 ➖}={x^1/x^1 ➖ x^1/x^1}={x^0+x^0 ➖ }/(x^0+x^0 ➖}=x^1/x^1 (x ➖ 7)^2/(x ➖ 6)^2=(x^2 ➖ 49)/(x^2 ➖ 36)={x^0+x0 ➖}/{x^0+x^0 ➖}={x^1/x^1 ➖ x^1/x^1}={x^0+x^0 ➖}/{x^0+x^0 ➖ }=x^1/x^1 (x ➖ 8)^2/(x ➖ 7)^2= (x^2 ➖ 64)/(x^2 ➖ 49)={x^0+x^0 ➖ }/{x^0+x^0 ➖}={x^1/x^1 ➖ x^1/x^1}={x^0+x^0 ➖}/{x^0+x^0 ➖}=x^1/x^1 (x ➖ 1x+1).

576/{72+144}=576/216 =2.64 2.4^16 2.4^4^4 2.2^2^2^2^2^2 1.1^1^1^1^1^2 1^2(x ➖ 2x+1) .v(a)^2 ➖(b)^2={a^2 ➖ b^2}={ab^0+ab^0 ➖ }=a^1(5)^2=25 {ab1+25}=25a^1 5^5ab1 1^5ab^1 5ab^1 (ab ➖ 5ab+1).

{x^2+9}/18+(x^2 ➖ 1)/18=9x^2/18+{x^0+x^0➖ }/18={9x^2/18+x^1/18}=9x^3/36=4x^3 2^2x^3 1^2x^1 2x^1 (x ➖ 2x+1) (x^9+9x^3)/12=9x12/4= 2.1x^12 2.1x^3^4 2.1x^3^2^2 1.1x^3^1^2 x^3^2 (x ➖ 3x+2).

√(72)/(√(26) + √2) = ( √(13 ) - 1)/2 = x, say x ^2 + x = 3 x^6 + x^3 + 3 x ^3 * ( x^2 + x) = 27 x^6 + x^3 + 9 x ^3 = 27 x^6 + 10 x^3 = 27 x ^ 12 = 27 - 10 x ( 3 - x) = 27 - 30 x + 10 ( 3 - x) = 57 - 40 x = 57 - 20 ( √ (13) - 1) = 77 - 20 √ (13)

Another Golden Ratio problem. √24/(√30+√6)=2/(1+√5)=1/Φ. Fibonacci showed that Φ^24=46368Φ+28657 φ=1/Φ=(√5-1)/2. Fibonacci also showed that φ^24=28657-46368φ Thus, a=28657+46368/2=51841, b=46368/2=23184, and a-2b=51841-2*23184=5473

Respected Sir, Good morning

*= read as square root Y=(3+*5)/8, or (3-*5)/8may be As per question Y^(1/4) +y^(1/2)=1 Let t=y^(1/4)) so Y^(1/2)=t^2 According to the question t^2+t-1=0 Here, a=1, b=1, c=-1 D=*{(b)^2-4ac} =*(1+4)=*5 So, t=*{-b±D}/2a={(-1)±*5}/2 {Y^(1/4)}^2={-1±*5}^2/2^2 Y^(1/2)=(-1±*5)^2/4 =(1+5-2.*5)/4 =(6-2.*5)/4=(3-*5)/2 Y={(3)^2+(*5)^2-2.3.*5}/4 =(9+5-6*.*5)/16)=(14-6.*5)/4 =(7-3.*5)/2 Y also (7+3.*5)/2 Now (-1+*5)^2=1+5-2.*5=6-2.*5 =2(3-*5) (-1-*5)^2=1+5+2.*5=6+2.*5 =2(3+*5) Hence, Y=2(3-*5)/16 or 2(3+*5)/16 =(3-*5)/8 or (3+*5)/8

{2021x+2021x ➖ }/{2022x+2022x ➖}= 4042x^2/4044x^2:+{2022x+2022x ➖ }/{2023x+2023x ➖}={4042x^3/4044x^2+4044x^2/4046x^2}=8088x^4/8090x^4{ 2023x+2023x ➖ }/{2024x+2024x ➖}={8088x^4/8090x^4+4046x^2/4048x^2}=16314x^6/16138x^6+2024x+{2024x+2024x ➖ }/2025x+2025x ➖}={161314x^6/16138x^6+4048x^2/4050x^2}=165 384x^8/20.188+{2025x+2025x ➖ }/{2021x+2021x ➖ }={165.384'x^8/20.188x^8+4050x^2/4042x^2}=169.434x^10/24.230x^10 >5 4^4^3^2.2^2^3^2^2x^2^5/1^2^2.1^3^10x^2^5 2^2^2^2^1^1.1^1^1^1^1x^1^1/1^1.3^2^5x^1^1 1^1^1^1/3^2^5^1x^1^1 3^2^1^1x^1^1 3^2x (x ➖ 3x+2).

(x ➖ 6)^2/{2020x+2020x ➖} ×(x ➖ 5)^2/{2021x+2021x ➖ }+(x ➖ 4)^2/{2022x+2022x ➖ }=(x^6 ➖ 36)/4040x^2+(x^5 ➖ 5)/4042x^2+(x^4 ➖ 16)/4044x^2={x^0+x^0 ➖/4040x^2+(x^0+x^0 ➖}/4042x^2+{x^0+x^0 ➖}/4044x^2={x^1/4040x^2+x^1/4042x^2+x^1/4044x^2}=x^3/8126x^6 x^3/100^800^10^10^2^13x^6 x^3/10^10^10^10^10^80^10^10^2^13x^6 x^3/10^10^10^10^10^8^10^10^10^2^13x^6 x^3/2^5^2^5^2^5^2^5^2^5^8^2^5^2^5^2^5^2^13^1x^6x^3/1^1^1^1^1^1^1^1^1^1^2^3^1^1^1^1^1^1^1^1^1x^3^2 x^1/1^1^1x,^3^2 x^3^2 (x ➖ 3x+2).

{49x+49x ➖}/{16x+16x ➖ }+98x^2/32x2:+{49x+49x ➖ }/{32x+32x ➖ }+{49x+49x ➖ }/{64x+4x ➖ }=+{98x/32x^2+98x^2/64x^+98x^2/128x^2}=294x^6/224x^6=1.70x^1 2^35x 2^5^7x2^1^1x 2^1x (x ➖ 2x+1) {1x+1x ➖ }/{2x+2x ➖ }+{1x+1x ➖/{32x+32x ➖ }+{1x+1x ➖ }/{64x+64x ➖ }/=6x^6/196x^6 3^2x^3^2/2^98x^6 1^1x^1^1/2^48x^6 /2^3^16x^6 /2^34^4x^3^2 /1^1^2^2^2^2x^3^2 /1^1^1^1x^3^2 x^3^2(x ➖ 3x+2 ).

864/{312+24}=864/336 2.92 2. 2^46 2.2^23 2.2^23^1 2.2^1^1 1.2^1 1.2 (b ➖ 2a+1).{a+a ➖}+{b+ b ➖}+{13+13 ➖ }={a^2+b^2+26}= 26ab^4 2^13ab^4 2^13^1ab^4 2^1^1ab^2^2 1ab^1^2 1ab^2 (b ➖ 2a+1). {a+a ➖ }+{3b+3b ➖ }= {a^2+6b^2}=6ab^4 3^2ab^2^2 3^1ab^1^2 3ab^2 (b ➖ 3a+2).

Y=(7-(3×(5^(1/2)))/2

a(b+c)=300...eqn1 b(a+c)=400.....eqn2 c(a+b)=500.......eqn3 a(b+c)=300 ab+ac=300...eqn4 b(a+c)=400 ab+bc=400...eqn5 So, Eqn5=eqn4+100 ab+bc=ab+ac+100 bc-ac=100 c(b-a)=100 c(a-b)=-100.....eqn6 Eqn3 ÷ eqn6 C(a+b)/c(a-b)=500/-100 (a+b)/(a-b)=-5 a+b=-5(a-b) a+b=-5a+5b a+5a=5b-b 6a=4b a=(4/6)b=(2/3)b...... Eqn7 b=(3/2)a Again C(a+b)=500 ac+bc=500...eqn8 Eqn8=eqn5+100 ac+bc=ab+bc+100 ac-ab=100 a(c-b)=100...eqn9 Eqn1 ÷ eqn9 a(b+c)/a(c-b)=300/100 (b+c)/(c-b)=3 b+c=3(c-b) b+c=3c-3b b+3b=3c-c 4b=2c C=2b So, C=2b=2{(3/2)a C=2b=3a Now eqn1 a(b+c)=300 a{(3/2)a +3a}=300 a{(9/2)a}=300 a(9a)=300×2 9a^2=600 a^2=600/9 a=(10.*6)/3.(*=read as square root ) So, b=3a/2={3(10.*6)/3}/2 =(10.*6)/2 C=3a=3{(10.*6)/3} =10.*6 Hence, a+b+c={(10.*6)/3}+{(10.*6)/2}+(10.*6) ={2(10.*6)+3(10.*6)+6(10.*6)}/6 ={11(10.*6)}/6 =(55.*6)/3