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204 without touching pen/paper or watch video. 204x208=(204-4)x(204+4)=204^2-4^2. So under sqrt is 204^2-16+16.

X-7, y-5, 7+5=12

If b-a<>3 ? If b-a=-3 ? Than (b/a)^3 =(2/3)^3 If b-a=9 (b/a)^3 = 2/3

x^4=2^x (x^4)^1/x=(2^x)^1/x (x^4)^1/x=2 (x^1/x)^4=2 x^1/x=2^1/4 x^1/x=(2^4)^1/16 x^1/x=16^1/16 x=16

А это единственное решение?

Call me crazy, but if you plug the numbers into a calculator, you don't get 1. (Unless my calculator skills are way off!)

i think you mean to say that the answers x &y are Rel numbers otherwise any pos. vlues of x, could work

(5555+4444)(5555-4444)= 9999• 1111= 11108889

what about the case where b-a <> 3?

x=27

(1/4)^x=x 1/4^x=x 4^x=1/x (4^x)^1/x=(1/x)^1/x 2^2=(1/x)^1/x 2=1/x x=1/2

8(a^a)(b^b) = 27(a^b)(b^a) 8(a/b)^a = 27(a/b)^b (a/b)^(a - b) = (27/8)¹ = (3/2)³ .. find (a, b) and calculate a² + b² ..

Solve for x integer. There are 2 other real solutions.

Очень классно!

இவ்வளவு மெதுவாக செய்தால் பிள்ளைகள் பார்க்க மாட்டார்கள்

х^5=1 х=√пятой степени из 1 х=1

x^1/x=16^1/16 x=16

1:48 só é verdade se a^a>=1

You can not say you found all the solutions just because you compared two sides and find the obvious solution.

3^(2√x) = 3^(3x/2) 3^(2√x) = 3^(3x/2) 2√x = 3x/2 4√x = 3x 16x = 9x² => 9x² - 16x = 0 x(9x - 16) = 0 *x = 0* 9x - 16 = 0 => *x = 16/9*

x=0; 16/9

Good work mr user❤

Just by looking at it i find x=2 and y=1

12 - ? 12=4*3= (2^2)*(3^1) (2^1)*(3^2)=2*9=18 18=18 x=2 y=1

√333=3√37 √a+√b=3√37. 則√a=2√37 b=√37 則a=148 b=37

(1/4)ˣ = x 1/4ˣ = x (x)4ˣ = 1 (x)eˣˡⁿ⁴ = 1 (xln4)(eˣˡⁿ⁴) = ln4 = 2ln2 (xln4)(eˣˡⁿ⁴) = (ln2)eˡⁿ² xln4 = ln2 2xln2 = ln2 => *x = 1/2*

amigo a sua conta esta errada, na segunda linha da solução você inverteu a fração e não mudou o sinal do exponte, quatro elevado a x ficaria quatro elevado a -x.

(2, 1).

nice

Sqr(333)=sqr(9*37)=sqr(9)*sqr (37)=3sqr(37) On peut écrire : 3 sqr(37)= sqr (37) + 2 sqr (37) 3 sqr (37) =sqr(37)+ sqr(4)*sqr (37)=sqr (37) +sqr(4*37)=sqr(37)+ sqr(148) sqr(37) + sqr(148)=sqr(333) a=37 b=148 a =148 b =37

x=6+y xy=36 (6+y)y=36 y^2+6y-36=0 y=[-6+-rq(36+144)]/2 y=[-6+-rq180]/2 y=[-6+-6•rq5]2 y=-3+-3•rq5 x=6+(-3+-3•rq5) x=3+-3rq5

Very easy to solve. No big formula required. X+Y = 12. XY=35. 35=35*1 or 17.5*2 or 7*5. First two are not suitable. So 7*5 =XY.. So X=7. Y=5. So 7*7=49. 5*5=25. 49-25=24. 7+5=12.

the answer should be exact. you should elaborate and check whether it's + or - in the end. imagine spaceX Starship is landing and gets a command: give ±5000 pounds of thrust woke math doesn't work

(x+3)⁴+(x+1)⁴=82 =81+1 =3⁴+1⁴ --> x=0 =(-1)⁴+(-3)⁴ --> x=-4 Therefore x={0,4} Let u=x+2 (x+3)⁴=[(x+2)+1]⁴ =(u+1)⁴ =u⁴+4u³+6u²+4u+1 (x+1)⁴=(u-1)⁴ =u⁴-4u³+6u²-4u+1 Their sum is 82=(x+3)⁴+(x+1)⁴ =2u⁴+12u²+2 41=u⁴+6u²+1 --> u⁴+6u²-40=0 a quadratic equation in x² which can be factored as (u²-4)(u²+10)=0 u²=4 or u²=-10 u=±2 or u=±isqrt(10) As x=u+2, x=-2±2 or x=-2±isqrt(10) x={0,-4,-2+isqrt(10),-2-isqrt(10)}

x=sqrt[3-sqrt{5+2sqrt(3)}] y=sqrt[3+sqrt{5+2sqrt(3)}] Note that x>0 --> x+y>0 xy=sqrt[9-{5+2sqrt(3)}] =sqrt[4-2sqrt(3)] =sqrt[{sqrt(3)-1}²] =sqrt(3)-1 x²=3-sqrt[5+2sqrt(3)] y²=3+sqrt[5+2sqrt(3)] x²+y²=6 (x+y)²=x²+y²+2xy =6+2[sqrt(3)-1] =4+2sqrt(3) =[sqrt(3)+1]² x+y=sqrt(3)+1

x=3±vʼ(log2/log5)≈3.656; 2.344

Too slow.

(x+3)^4+ (x+1)^4=82 ; i,e x^4+12x^3+54x^2+108x+81+x^4+ 4x^3+6x^2+4x+1=82 ; 2x^4+16x^3+60x^2+112x+82=82 ; i,e x^4+8x^3+30x^2+56x=0 ; i,e x(x^3+8x^2+30x+56)=0 ; i,e x(x+4)(x^2+4x+14)=0 ; i,e x= 0 or x= -4 or x= -2+i√10 or x=-2-i√10. . So , you are partially true. Because, x has four values . Two are real and other two are imaginary.

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There is a more direct route from 1:14 to 7:20 that makes better use of the substitution t = x + 2. Just expand the binomials: (t^4 + 4t^3 + 6t^2 + 4t + 1) + (t^4 - 4t^3 + 6t^2 - 4t + 1) = 82 2t^4 + 12t^2 + 2 = 82 t^4 + 6t^2 - 40 = 0

x = 0 and x = -4 are two solutions.

in complex plane the a, b, c would be the three cubic complex roots of any non-zero number.. but yea it doesn't have a solution over reals

5(sqrt3+sqrt2)

C.15

How would you solve this question for an exam? 😁

Позорьтесь решением не знать что х^2-6х+9=(х-3)^2

Если это отделение десятичной дроби, то он ставится внизу, а не посередине как знак умножения

0*64=0

👍

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