Professor, diga-me com sinceridade: essa questão foi preparada por você? (Professor, tell me honestly: was this question prepared by you?).

2^1lx= (3+-√5)/2 or X=1/[ log_2 (3+-√5)/2]

Θετω 2^(1/χ)=α>0 και εχω α^4-α^2+1=2α^3+2α; α^4-2α^3-α^2-2α+1=0 ; (α^2-3α+1)(α^2+α+1)=0 στο δευτερο τριωνυμο ειναι Δ=-3<0. Αρα α=(3+-(5)^(1/2))/2>0.(1/χ)×log_2(2)= log_2[3+-(5)^1/2]/2=......χ=1/[(log_2(3+-5^1/2))-1]

X= 12- 2√ 11

Rearranging the equation (√x³/4) - √x = (5/(√x +4)) => (√x)(√x² - 4)(√x + 4) = 20 => (√x)(√x + 2)(√x - 2)(√x + 4) = 20 With √x + 1 = t //* alternatively (x + 2√x)(x + 2√x - 8) = 20 and substitution *// (t - 3)(t - 1)(t + 1)(t + 3) = 20 => (t² - 1)(t² - 9) = 20 => (t² - 5 + 4)(t² - 5 - 4) = 20 => (t² - 5)² - 16 = 20 => (t² - 5)² = 36 => t² - 5 = ±6 => t² = 11, -1 => t = ±√11 => √x + 1 = ±√11, ±i => √x = - 1 ± √11, -1 ± i => x = 12 ± 2√11 (for Real x)

Η εξισωση γινεται y^4+4y^3-4y^2-16y-20=0με y=χ^(1/2)οπου χ>0 και αρα y>0 ; ριζα χ+4>0. Η εξισωση γραφεται: (y^2+2y)^2-8(y^2+2y)-20=0 αν y^2+2y=ω>0 εχω ω^2-8ω-20=0 ω=10 ή ω=-2 απορριπτεται. Αρα y^2+2y=10 y>0 αρα y=-1+(11)^(1/2) και τελικα χ=y^2=(-1+ριζα 11)^2 ; χ=12-2(11)^(1/2)

{y=-x/2±√(1-3/4x²), 2-0,75x²-+1,5x√(1-3/4x²)=±(1,5x-+√(1-3/4x²))√(5-3/4x²),z=-x/2±√(5-3/4 x²);{(√((11+6√3)/39);-√((11+6√3)/39)/2+√((41-6√3)/52);-√((11+6√3)/39)/2+√((249-6√3)/52))(;;)(;;)(;;))} x+y+z=±(√((41-+6√3)/52)+√((249-+ 6√3)/52))=±√(5±2√3)

{x+x ➖}{x+x ➖}/{4x+4x ➖ } ➖ {5x+5x ➖ }/{x+x ➖ }+{4+4 ➖ }={x^2+x^2}/8x^2+ ➖ 10x^2/(x^2+8}={2x^2/8x^2 ➖ 10x^2/8x^2}=8x/{0+0 ➖} x^{0+x^0 ➖ }=8x/1x^1=8x^1 2^3x^1 1^1^2x^1 2x^1 (x ➖ 2x+1).

Let t = √x. Then, t^4+4t^3-4t^2-16t-20 =0. Let us write this as (t^2+at-10)(t^2+bt+2)=0. Comparing, we see that a=b=2. So t^2+2t-10=0, whence t = -1+/-√11. t^2+2t+2=0 does not give real solutions. So, √x= -1 +/- √11 > x = 12 +/- 2 √11.

x^5;(x^5 ➖ 320) ➖ 2=( x^5 )^2➖ 315 ➖ 2={x^25 ➖ 315} ➖ 2=290 ➖(2)^2={290 ➖ 4}=286 10^20^2^43 10^2^102^43^1 2^5^2^2^5^2^1^1 1^1^1^1^1^2 1^2 (x ➖ 2x+1).(x)^2 ➖ (22)^2={x^2 ➖484}=482 10^40^2^42 10^40^2^6^7 10^4^10^2^6^7 2^5^4^2^5^2^6^7^1 1^1^4^1^1^1^3^21^1 2^2^3^2 1^13^2 3^2 (x ➖ 3x+2).

864/{312+24=}= 864/336=2.192 2.10^10^2^46 1.2^5^2^5^12^23 1^1^1^1^2^23^1 2^1^1 2^1 (x ➖ 2x+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 ab^1^2 (ab ➖ 2ab+1).{a+a ➖ }+{3b+3b ➖ }={a^2+6b^2}= 6ab^4 3^2ab^2^2 3^1ab^1^2 3ab^2 (ab ➖ 3ab+2).

(√72)/(√26 + √2) = [(√72).(√26 - √2)] / [(√26 + √2).(√26 - √2)] (√72)/(√26 + √2) = [(6√2).(√26 - √2)] / [(√26)² - (√2)²] (√72)/(√26 + √2) = [(6√2).(√26 - √2)]/[26 - 2] (√72)/(√26 + √2) = [(6√2).(√26 - √2)]/24 (√72)/(√26 + √2) = [(√2).(√26 - √2)]/4 (√72)/(√26 + √2) = [(√52) - (√4)]/4 (√72)/(√26 + √2) = [2√13 - 2]/4 (√72)/(√26 + √2) = (√13 - 1)/2 [(√72)/(√26 + √2)]² = (√13 - 1)/2]² [(√72)/(√26 + √2)]² = (13 - 2√13 + 1)/4 [(√72)/(√26 + √2)]² = (14 - 2√13)/4 [(√72)/(√26 + √2)]² = (7 - √13)/2 [(√72)/(√26 + √2)]⁴ = [(7 - √13)/2]² [(√72)/(√26 + √2)]⁴ = (7 - √13)²/2² [(√72)/(√26 + √2)]⁴ = (49 - 14√13 + 13)/4 [(√72)/(√26 + √2)]⁴ = (62 - 14√13)/4 [(√72)/(√26 + √2)]⁴ = (31 - 7√13)/2 [(√72)/(√26 + √2)]⁸ = [(31 - 7√13)/2]² [(√72)/(√26 + √2)]⁸ = (961 - 434√13 + 637)/4 [(√72)/(√26 + √2)]⁸ = (1598 - 434√13)/4 [(√72)/(√26 + √2)]⁸ = (799 - 217√13)/2 [(√72)/(√26 + √2)]¹² = [(√72)/(√26 + √2)]⁸.[(√72)/(√26 + √2)]⁴ [(√72)/(√26 + √2)]¹² = [(799 - 217√13)/2].[(31 - 7√13)/2] [(√72)/(√26 + √2)]¹² = (799 - 217√13).(31 - 7√13)/4 [(√72)/(√26 + √2)]¹² = (24769 - 5593√13 - 6727√13 + 19747)/4 [(√72)/(√26 + √2)]¹² = (44516 - 12320√13)/4 [(√72)/(√26 + √2)]¹² = 11129 - 3080√13 → compare to: a + b√13 → a = 11129 → b = - 3080

解答ぐらい簡素にしろよ。

х=(2у±6√(1-у²))/5 у=±1 -+1,68у√(1-у²)-5,76у²+72/25 1,68(-1+2y²)(1-y²)-⁰,⁵-5,76*2y=0 1-2500/49y²+2500/49y⁴=0, y/(y²-1/2)≥0 y²=(25±24)/50=49/50; 1/50 y=±7/5/√2;±1/5/√2 y=7/5/√2; -1/5/√2+*-*+>y max=3

Substitute x-21=y

{x²/³+y²/³=a,x²/³y²/³=b;{b=0,5a²-17/2,a³-51a-130=0; a=(65+√(65²-17³))¹/³+(..-..)..=2√17 cos(1/3arccos(65/√17³)), n=0 b=34cos²(1/3arccos(65/√17³))-8,5 {y=±(√17cos(1/3arccos(65/√17³)) -+√(-8,5cos(2/3arccos(65/√17 ³))))³/², x=±(√17cos(1/3arccos(65/√17³))± √(-8,5cos(2/3arccos(65/√17³))))³/².

⁵√[(x - 20)⁵ - 2] = x - 22 {⁵√[(x - 20)⁵ - 2]}⁵ = (x - 22)⁵ (x - 20)⁵ - 2 = (x - 20 - 2)⁵ (x - 20)⁵ - 2 = [(x - 20) - 2]⁵ → let: p = x - 20 p⁵ - 2 = [p - 2]⁵ p⁵ - 2 = (p - 2)².(p - 2)².(p - 2) p⁵ - 2 = (p² - 4p + 4).(p² - 4p + 4).(p - 2) p⁵ - 2 = (p⁴ - 4p³ + 4p² - 4p³ + 16p² - 16p + 4p² - 16p + 16).(p - 2) p⁵ - 2 = (p⁴ - 8p³ + 24p² - 32p + 16).(p - 2) p⁵ - 2 = p⁵ - 8p⁴ + 24p³ - 32p² + 16p - 2p⁴ + 16p³ - 48p² + 64p - 32 - 2 = - 10p⁴ + 40p³ - 80p² + 80p - 32 - 10p⁴ + 40p³ - 80p² + 80p - 30 = 0 p⁴ - 4p³ + 8p² - 8p + 3 = 0 → the aim, if we are to continue effectively, is to eliminate terms to the 3rd power p⁴ - 4p³ + 8p² - 8p + 3 = 0 → let: p = z - (b/4a) → where: b is the coefficient for p³, in our case: - 4 a is the coefficient for p⁴, in our case: 1 p⁴ - 4p³ + 8p² - 8p + 3 = 0 → let: p = z - (- 4/4) → p = z + 1 (z + 1)⁴ - 4.(z + 1)³ + 8.(z + 1)² - 8.(z + 1) + 3 = 0 (z + 1)².(z + 1)² - 4.(z + 1)².(z + 1) + 8.(z² + 2z + 1) - 8z - 8 + 3 = 0 (z² + 2z + 1).(z² + 2z + 1) - 4.(z² + 2z + 1).(z + 1) + 8z² + 16z + 8 - 8z - 8 + 3 = 0 (z⁴ + 2z³ + z² + 2z³ + 4z² + 2z + z² + 2z + 1) - 4.(z³ + z² + 2z² + 2z + z + 1) + 8z² + 16z + 8 - 8z - 8 + 3 = 0 (z⁴ + 4z³ + 6z² + 4z + 1) - 4.(z³ + 3z² + 3z + 1) + 8z² + 8z + 3 = 0 z⁴ + 4z³ + 6z² + 4z + 1 - 4z³ - 12z² - 12z - 4 + 8z² + 8z + 3 = 0 z⁴ + 2z² = 0 z².(z² + 2) = 0 First case: z² = 0 z = 0 → recall: p = z + 1 p = 1 → recall: p = x - 20 x = p + 20 → x = 21 Second case: (z² + 2) = 0 z² + 2 = 0 z² = - 2 z² = 2i² z = ± i√2 → recall: p = z + 1 p = 1 ± i√2 → recall: p = x - 20 x = p + 20 x = 1 ± i√2 + 20 → x = 21 ± i√2

let t = x - 22, (x - 20)^5 - 2 = (x - 22)^5, => (t + 2)^5 - t^5 - 2 = 10(t^4 + 4t^3 + 8t^2 + 8t + 3) = 10(t + 1)^2*(t^2 + 2t + 3) = 0 => t = { -1, -1±√2i } => x = { 21, 21±√2i }

Acho uma grande falta quando o professor não indica o conjunto em questão.(I think it's a big mistake when the teacher doesn't indicate the set in question.)

Let x-20=a and [(x-20)^5 -2]^1/5=b. Thus, a-b=2 and a^5-b^5=2 > (a^5-b^5)/(a-b) = a^4+b^4+ab(a^2+b^2)+a^2hb^2=1. Let ab=t. Given a-b=2, a^2+b^2=2(t+2) and a^4+b^4=2(t^2+8t+8). So, 2(t^2+8t+8) + 2t(t+2) + t^2=1 > 5t^2+20t+15=0 > (t+1)(t+3)=0 > t=-1, -3. t=ab=-3 > a^2-2a+3=0, with no real solutions. t=-1 > (a-1)^2=0 > a=x-20 = 1 > x=21.

X1=21, X2=21+(2)^(1/2)i, X3=21-(2)^(1/2)i.

X= 21

*=read as square root ^=read as to the power *72/(*26+*2) =(6.*2)/{*2(*13+1)} =6/(*13+1)={6(*13-1)}/12 (multiply both numerator &denominator by *13-1) =(*13-1)/2.......eqn1 Squaring eqn1 {(13+1)-2.*13}/4 =2(7-*13)/4=(7-*13)/2...eqn2 Squaring eqn2 {(49+13)-14.*13}/4 =2(31-7.*13)/4 =(31-7.*13)/2.....eqm3 Take the cube of eqn3 Numerator (31)^3-(7.*13)^3+3.317.*13(7.*13-31) =29791-4459.*13+59241-20181.*13 =89032-24640.*13 =8(11129-3080.*13) Denominator 2^3=8 So, 8(11129-3080.*13)/8 =11129-3080.*13 As per question a+(b.*13)=11129-3080.*13 So, a=11129 b.*13=-3080.*13 b=-3080 So a+3b=11129+{3×(-3080)} =11129-9240 =789

x = a+20 , a⁵-2 = y⁵=> a⁵-y⁵ = 2, y = x-22 = a-2=> a-y = 2 => a²+y² = 4 +2ay ,(a-y)³ =a³-y³-3ay(a-y) a³-y³ = 8+6ay => (a³-y³)(a²+y²) = (8+6ay)(4 +2ay)=> a⁵-y⁵-a²y²(y-a) = 12a²y²+40ay+32 10a²y²+40ay+30 = 0 = 10(a²y²+4ay+3) = 10 (ay+3)(ay+1), y = a-2 => a(a-2)+3 = 0 =(a-1)²+2 >0 or a(a-2)+1 = 0 => a = 1 => x = a+20 = 21

The first one, root 2, may be

Let x = √72/[√2(√13+1)] = 6/(√13+1) = 1/2(√13-1). So, x^3=2√13-5, x^6=77-20√13 and x^12 = 11,129-3080√13 = a +b√13. Thus, a =11,129 and b = -3080 > a+3b=1889.

69412

Shorts... √x+1/√x=198 from l. H. S for f(14) . Cubing r. h. s √x+1/√x+3z= z^3 where z is r. Hs z^3-3z-198= 0 gives z= 6

This is a piecewise defined function the derivative is not just the one you stated. You have to break it into two cases for x>0 and x<0.

2^2<3^3 (n ➖ 2n+2).(n ➖ 3 n+3). 4^4<5^5( n➖ 2n+1)().( n ➖ 5n+1).6^6<7^7( n ➖ 2n+1).( n ➖ 3n+1).

(x)^2 ➖ (1)^2/{x+x+x+ xx ➖ x ➖ x ➖ x}={x^2 ➖ 1}/x^4=xx^1/x^4=x^4 (x ➖ 4x+4). {x+x+xx ➖ x ➖ x}+{1+1+11 ➖ 1 ➖ 1}/{x+x+x+x+x+xx ➖ x ➖ x ➖ x ➖ x ➖ x}={x^3+3}/x^6=3x^3/x^6=3x^2 (x ➖3x+2).(14)^2=196 14^14 2^7^2^7 1^1^2^1 2^1 (x ➖ 2x+1).

We are given f(x^1/4 - x^(-1/4))= x^1/6 + x^(-1/6). Let t = x^1/12. So, f(t^3-1/t^3)= t^2+1/t^2. Let E=t^2+1/t^2. If t^3-1/t^3=14, (t-1/t)(E+1)=14. But, (t-1/t)^2 = E-2. So (E-2)^1/2(E+1) = 14 > (E-2)(E^2+2E+1) = 196 > E^3-3E-2 = 196 > E^3-3E= 198 > E=6. Thus, f(14) = 6.

The function x^(1/x) has a maximum at x=e with is roughly 2.72. 2^1/2=4^1/4 is less than e^1/e. So 3^1/3 which is roughly e^1/e, is greater than 2^1/2 and 4^1/4. Beyond x=4, the function x^1/x is a monotonically decreasing one. So, the largest number is 3^1/3 among those given.

(9-5)^1/2 ÷(3-5)=1; 7-5=2; <=> (9-6,6)^1/2 /(3- 6,6)≠7-6,6 <=> (2,4)^1/2 ÷ -3,6 =1,54÷-3,6= -0,4 7- 6,6=0,4 {-0,4≠04} <=>x=(2,569)^2

Golly that was easier than it looks. Or maybe I have been watching @infyGyan videos for quite some time!!!

2-x/(x-6)•(x-12)-x^2/81=0 <=> 2- |-0,9|/(|-0,9| -6)•(|-0,9|-12) = (1,1)^1/2 ÷ (89,01)^1/2 = 1,04/9,43 =1/9 <=> x=|-0,9|

f(14)=2^2+2=6 η σχεση γινεται: f[ α^3-1/(α)^3]=f[α^2+1/(α^2)] αλλα α^2+(1/α^2)=(α-1/α)^2+2 και α^3 -(1/α^3)=(α-1/α)^3+3(α-1/α) και (α-1/α)^3+3(α-1/α)=14 α-1/α=2 f(14)=2^2+2=6

x = 1/(√(13)+√(23))+ 1/(√3 + √(13)) = (√(23)- √(13))/( 23 - 13) + (√(13)- √3 )/( 13 - 3) = ( √(23) -√3)/10 x √ 5 = ( √(23) -√3)/ √ (20) 1 /( x √ 5) = √ (20) /( ( √(23) -√3) = ( √(23) +√3)/ √ (20) Hereby (x √ 5 )+ 1/( x √ 5 ) = √ (23/5) (x √ 5 )^2 + 1/ ( x √ 5) ^2 .= 23/5 - 2 = 13/5 (x √ 5 )^4 + 1/ (x √ 5)^4 = (13/ 5) ^ 2 - 2 = 119/25 (x √ 5 )^8 + 1/ (x √ 5)^8 = (119/25)^2 - 2 = 20.65

F(14)=8

F(4)=6.