(1) 2^p + 2^q + 2^r = 42 We can consider the sum (1) as a binary number in the notation of which there are 3 ones. power 2: 7654 3210 42 = 0010 1010 42 has exactly three ones in its binary notation Therefore p = 5, q = 3, r= 1

42 = 32 + 8 + 2 =2^5 + 2^3 +2^1 {p,q, r} ={ 1,3,5}

A=5 B=1

Thank you 😋❤️

X={3.3027; -0.3027}

Solving equations and their systems in integers generally leads to limiting the number of combinations to check. In this problem, we are unlucky that 30 has so many divisors, but we have a different favorable circumstance. The equation represents a hyperbola passing through the point (0, 0). Given the form of the problem, we reject the solution (0,0). It also has asymptotes: a horizontal asymptote at y = 5/7 and a vertical asymptote at x = 6/7. From this (it's enough to sketch the hyperbola), it follows that the solutions are pairs where x and y are positive, and, as can be easily checked, one of them must be equal to 1. It is sufficient to check by substituting x = 1, which gives y = 5, and x = 2. For x = 2, the value of y is less than 2, and the function is positive and decreasing, so the only option for an integer solution is y = 1. This gives x = 5.

2log4 (x) +2log3 (x) =1/2.... log4 (x) +log3 (x) =1/4... log3 (x) =log3 (4) /(4* (1+log3 (4))... log(x)/log(3)=0.139. X=1.164 and -1,164.

(1) 5/a + 6/b = 7 (a,b from Z+) Obviously 5/a + 6/b < 7 when a > 5 (a,b from Z+) Therefore a can be (2) a = 1,2,3, 4 or 5 From (1) 6/b = 7 - 5/a = (7a - 5) / a => b = 6[a/(7a - 5)] Test all a from (2) a 6[a/(7a - 5)] b Result 1 6[1/(7*1 - 5] = 6/2 = 3 3 (1,3) solution 2 6[2/(7*2 - 5] = 12/9 - no solution 3 6[3/(7*3 - 5] = 18/16 - no solution 4 6[4/(7*4 - 5] = 24/23 - no solution 5 6[5/(7*5 - 5] = 30/30 = 1 1 (5,1) solution

X={-0.6826}

{p,q,r}={1;3;5}

Sorry,instant answer is wrong.Will try seriously.

Instant answer is x=1

42=(binary)101010=2^5+2^3+2^1 hence r=5, q=3, p=1

Thank you very much for explaining how to solve difficult equations. I can find the solution to an equation more easily than I can write down the proof with steps, like some gifted people who just guess the answers but can’t explain it with proof

I think not much time is provided for solving such simple problems. Hence, hit and trial will be the best technique to adopt, as suggested by others here.

let x-1/x=u , u^3+u=30 , u^3+u-30=0 , by faktor. , add -3u^2 , 3u^2 , u^3-3u^2 +3u^2-9u +10u-30=0 , (u-3)(u^2+3u+10)=0 , u=3 , x-1/x=3 , x^2-3x-1=0 , x= (3+V13)/2 , (3-V13)/2 , /// for complex , u^2+3u+10=0 , u=(-3 +/- V(9-40))/2 , u=(-3 +/- i*V31)/2 , x-1/x= (-3 + i*V31)/2 , (-3 - i*V31)/2 , ///

Too long solution, we can see a the blink of eye that a=5,b=1 and/or a=1,b=3

X = - 0.2 only

4:40 wrong value for x correct value is *W(-5ln5)/(-5ln5)*

x = ± 5⁵ˣ [ I ] x = 5⁵ˣ x5⁻⁵ˣ = 1 -5x5⁻⁵ˣ = -5 (-5xln5)e⁻⁵ˣˡⁿ⁵= -5ln5 -5xln5 = W(-5ln5) *x = W(-5ln5)/(-5ln5)* [ II ] x = -5⁵ˣ -x = 5⁵ˣ -x5⁻⁵ˣ = 1 -5x5⁻⁵ˣ = 5 (-5xln5)e⁻⁵ˣˡⁿ⁵= 5ln5 -5xln5 = W(5ln5) = ln5 *x = -1/5*

125.

72-21|7/4 is not 21(7)-82/4 is. 😅😂 all not, but 203(7)-533/16 is!! You are always truth!!

This is not a simplification but a complication

A bullshit Test delivers bullshit results.

I don't see how x^7= [203(7)^.5]/16 - 533/16 represents a simplification of the original statement.

we get , x^4-40x^2-x+380=0 , add 5x^3 , -5x^3 , by fakt. , (x-5)(x^3+5x^2-15x-76)=0 , solu. , x=5 , test , sqrt(20+sqrt(20+5))=sqrt(20+5) , --> x>0 , --> V25=5 , x=5 , same , OK , /// is not part of the basic equation x^3+5x^2-15x-76=0 , (x+4)(x^2+x-19)=0 , x= -4 , x^2+x-19=0 , x=(-1 +/- V77)/2 , x= -4 , (-1+V77)/2 , (-1-V77)/2 , not a solution to the basic task /// ,

You're working too hard, in my opinion. Let x = (√7 - 1)/2. Compute x², which can be written as x² = (3/2) - x. Now just recursively use that formula: x³ = x•x² = x•[(3/2) - x] = (3/2)x - x² = (3/2)x - ((3/2) - x) = (5/2)x - 3/2. Now just compute x⁷ = x(x³)² = x•[(5/2)x - 3/2]² then substitute back in for x after doing the straightforward simplification.

Once again: x=u+v: y=u-v x+y=2u=8 => u=4 (u+v)*(u-v)=u^2-v^2=16-v^2=64 64-16+v^2=0 +>v^2= - 48 v=+/-4i*sqrt(3) x=+4i*sqrt(3) y= 4 - 4i*sqrt(3) x=-4i*sqrt(3) y= 4 + 4i*sqrt(3)

x+y=1; i,e x=1-y ; 3^(1-y)-3^y=3 ; i,e [3/(3^y)]-3^y-3=0 ; Let a=3^y ; i,e 3-a^2-3a=0 ; a^2+3a-3=0 ; i,e a= (-3+√21)/2 or a= (-3-√21)/2 ; i,e 3^y= (-3+√21)/2 ; i,e y= [log (√21-3)-log2]/log3 or y= [log(-√21-3)-log2]/log3 & x=[log3-log(√21-3)+log2]/log3 or x= [log3-log(-√21-3)+log2]/log3. x and y each have an imaginary value.

Looking at the thumbnail for about half a minute, one concludes that x= 5 is the solution. The uniqueness is more difficult. You see, x cannot be much larger than sqrt(20) >4. let's try 5, es stimmt.

Ответ 24.

I once made a graphical solution using Excel and came up with the following solutions: X(1) = -0.435 and x(2) = 0.768

5^(10x)=x^2 x=-1/5=-0.2 x = -(W_n(-5 log(5)))/(5 log(5)), n element Z

We can use complex number formula

Square both sides and arrange you will get Sqrt(20+x) = x^2-20 Let a = sqrt (20+x) So a^2 = 20+x Let b= x^2-20 a-b=0-----eq1 a^2+b=x^2+x ----eq2 eq1+eq2: a^2+a=x^2+x By comparison: a^2=x^2 and a=x So sqrt(20+x)=x 20+x =x^2 x^2-x-20=0 By solving this quadratic equation you will get x= 5 or x=-4 but after verifying , x=5 is the valid solution.

(-1-Sqrt[77])/2=-0.5-0.5Sqrt[77]

-0.5+0.5Sqrt[77]~3.88748219369606103020319415370815478043793841377725179546384781489138232310965314083784657853435287796882549362049529012083432768762587387860733544256135657551151461205014820344532368711712776176803036213530176358415020714866728425630060509305236186000663384649623637666349970666203236511622609556042607924070675399240656605822767316195771544351937244268796910297368157570989041682711680249796176054418217927284352878786865754465672359878928261038577484408497395285153072963173752168281595651295751884866310088361036894620916490334718260310895044274490891209534991607458382400939970216010505771732233074230938952570052430960731094897251686303418127455463928184732956423786504478699242496607346336970427365103995794204369559740351121349649278297724223841868718961144727187194024110059176099150563221377883101165322110984828007470129630272204111508562228477993157311096046318647945291472121503884267851107342247486442355939020256590363971992269971027477759570120022441866224141724714023504766759748534993915617669135331279735810658754652765...

x=(-0.5+0.5Sqrt[77])=(-1+Sqrt[77])/2

Sqrt[20+Sqrt[20+x]]=x x=5 It’s in my head.

let f(n) = x^n + y^n f(0) = 2 f(1) = 2 since x + y = 2 f(n+2) = f(1)*f(n+1) - xy*f(n) let xy = a f(n+2) = 2*f(n+1) - a*f(n-2) f(0) = 2 f1) = 2 f(2) = 4 - 2a f(3) = 8 - 6a f4) = 16 - 16a + 2a^2 f(5) = 32 - 40a + 10a^2 = 152 10a^2 - 40a + 120 = 0 a^2 - 4a + 12 = 0 (a - 6)*(a + 2) = 0 a = 6 or -2 ❶xy = 6 x + y = 2 x + 6/x = 2 x^2 - 2x + 6 = 0 x = 1 ± sqrt(5) ❷xy = -2 x + y = 2 x - 2/x = 2 x^2 - 2x - 2 = 0 x = 1 ± sqrt(3)

Harvard University Admission Question: x⁵ + y⁵ = 152, x + y = 2; x, y =? (x + y)² = x² + y² + 2xy = 2² = 4, x² + y² = 2(2 - xy) (x + y)(x² + y²) = x³ + y³ + xy(x + y) = 4(2 - xy), x³ + y³ = 4(2 - xy) - 2xy = 2(4 - 3xy) (x² + y²)(x³ + y³) = x⁵ + y⁵ + (x²y²)(x + y) = 4(2 - xy)(4 - 3xy) 4(2 - xy)(4 - 3xy) - 2x²y² = x⁵ + y⁵ = 152, 2(8 - 10xy + 3x²y²) - x²y² = 76 5x²y² - 20xy + 16 - 76 = 5(x²y² - 4xy - 12) = 0, x²y² - 4xy - 12 = (xy + 2)(xy - 6) = 0 xy + 2 = 0; xy = - 2 or xy - 6 = 0; xy = 6 xy = - 2: (x - y)² = (x + y)² - 4xy = 2² - 4(- 2) = 12 = (2√3)², x - y = ± 2√3; x + y = 2 2x = 2 ± 2√3, x = 1 ± √3; y = 2 - x = 2 - (1 ± √3) = 1 -/+ √3 xy = 6: (x - y)² = (x + y)² - 4xy = 2² - 4(6) = - 20 = (2i√5)², x - y = ± 2i√5; x + y = 2 2x = 2 ± 2i√5, x = 1 ± i√5; y = 2 - x = 2 - (1 ± i√5) = 1 -/+ i√5 Answer check: x = 1 ± √3, y = 1 -/+ √3: (1 ± √3)⁵ = 1 ± 5√3 + 10(√3)² ± 10(√3)³ + 5(√3)⁴ ± (√3)⁵ x⁵ + y⁵ = (1 ± √3)⁵ + (1 -/+ √3)⁵ = 2(1 + 30 + 45) = 152; Confirmed x + y = (1 ± √3) + (1 -/+ √3) = 2; Confirmed x = 1 ± i√5, y = 1 -/+ i√5: (1 ± i√3)⁵ = 1 ± i√5 + 10(i√5)² ± 10(i√5)³ + 5(i√5)⁴ ± (i√5)⁵ x⁵ + y⁵ = (1 ± i√5)⁵ + (1 -/+ i√5)⁵ = 2(1 - 50 + 125) = 152; Confirmed x + y = (1 ± i√5) + (1 -/+ i√5) = 2; Confirmed Final answer: x = 1 + √3, y = 1 - √3; x = 1 - √3, y = 1 + √3; Two complex value roots, x = 1 + i√5, y = 1 - i√5 or x = 1 - i√5, y = 1 + i√5

(√5 + 2)^(x) + (√5 - 2)^(x) = 5 Let's calculate: 1/(√5 - 2) = 1/(√5 - 2) = (√5 + 2)/[(√5 - 2).(√5 + 2)] = (√5 + 2)/[5 + 2√5 - 2√5 - 4] = (√5 + 2)/[5 - 4] = √5 + 2 → so we can see that: 1/(√5 - 2) = √5 + 2 (√5 + 2)^(x) + (√5 - 2)^(x) = 5 → recall the previous result [1/(√5 - 2)]^(x) + (√5 - 2)^(x) = 5 → let: a = √5 - 2 [1/a]^(x) + a^(x) = 5 [1/a^(x)] + a^(x) = 5 → let: a^(x) = z → where: z > 0 (1/z) + z = 5 (1/z) + (z²/z) = 5 (1 + z²)/z = 5 1 + z² = 5z z² - 5z + 1 = 0 Δ = (- 5)² - (4 * 1) = 25 - 4 = 21 z = (5 ± √21)/2 First case: z = (5 + √21)/2 Recall: a^(x) = z (√5 - 2)^(x) = (5 + √21)/2 Ln(√5 - 2)^(x) = Ln[(5 + √21)/2] x.Ln(√5 - 2) = Ln[(5 + √21)/2] x = Ln[(5 + √21)/2] / Ln(√5 - 2) → x = [Ln(5 + √21) - Ln(2)] / Ln(√5 - 2) Second case: z = (5 - √21)/2 Recall: a^(x) = z (√5 - 2)^(x) = (5 - √21)/2 Ln(√5 - 2)^(x) = Ln[(5 - √21)/2] x.Ln(√5 - 2) = Ln[(5 - √21)/2] x = Ln[(5 - √21)/2] / Ln(√5 - 2) → x = [Ln(5 - √21) - Ln(2)] / Ln(√5 - 2)

3^(x + y) = 3 3^(x + y) = 3^(1) x + y = 1 y = 1 - x 3^(x) - 3^(y) = 3 ← as 3 is a positive number, you can say that: x > y 3^(x) - 3^(y) = 3 → recall: y = 1 - x 3^(x) - 3^(1 - x) = 3 3^(x) - [3^(1) * 3^(- x)] = 3 3^(x) - [3 / 3^(x)] = 3 → let: 3^(x) = a → where: a > 0 a - (3/a) = 3 a² - 3 = 3a a² - 3a - 3 = 0 Δ = (- 3)² - (4 * - 3) = 9 + 12 = 21 a = (3 ± √21)/2 → recall the condition: a > 0 a = (3 + √21)/2 → recall: 3^(x) = a 3^(x) = (3 + √21)/2 Ln[3^(x)] = Ln[(3 + √21)/2] x.Ln(3) = [Ln(3 + √21) - Ln(2)] → x = [Ln(3 + √21) - Ln(2)] / Ln(3) Recall: y = 1 - x y = 1 - { [Ln(3 + √21) - Ln(2)] / Ln(3) } y = [Ln(3) - Ln(3 + √21) + Ln(2)] / Ln(3) } → y = [Ln(6) - Ln(3 + √21)] / Ln(3)

(√5 + 2)^(x) + (√5 - 2)^(x) = 5 Let's calculate: 1/(√5 - 2) = 1/(√5 - 2) = (√5 + 2)/[(√5 - 2).(√5 + 2)] = (√5 + 2)/[5 + 2√5 - 2√5 - 4] = (√5 + 2)/[5 - 4] = √5 + 2 → so we can see that: 1/(√5 - 2) = √5 + 2 (√5 + 2)^(x) + (√5 - 2)^(x) = 5 → recall the previous result [1/(√5 - 2)]^(x) + (√5 - 2)^(x) = 5 → let: a = √5 - 2 [1/a]^(x) + a^(x) = 5 [1/a^(x)] + a^(x) = 5 → let: a^(x) = z → where: z > 0 (1/z) + z = 5 (1/z) + (z²/z) = 5 (1 + z²)/z = 5 1 + z² = 5z z² - 5z + 1 = 0 Δ = (- 5)² - (4 * 1) = 25 - 4 = 21 z = (5 ± √21)/2 First case: z = (5 + √21)/2 Recall: a^(x) = z (√5 - 2)^(x) = (5 + √21)/2 Ln(√5 - 2)^(x) = Ln[(5 + √21)/2] x.Ln(√5 - 2) = Ln[(5 + √21)/2] x = Ln[(5 + √21)/2] / Ln(√5 - 2) → x = [Ln(5 + √21) - Ln(2)] / Ln(√5 - 2) Second case: z = (5 - √21)/2 Recall: a^(x) = z (√5 - 2)^(x) = (5 - √21)/2 Ln(√5 - 2)^(x) = Ln[(5 - √21)/2] x.Ln(√5 - 2) = Ln[(5 - √21)/2] x = Ln[(5 - √21)/2] / Ln(√5 - 2) → x = [Ln(5 - √21) - Ln(2)] / Ln(√5 - 2)

Bullshit as usual

(x + y)² = x² + y² + 2xy → given: x + y = 2 4 = x² + y² + 2xy (x + y)³ = (x + y)².(x + y) (x + y)³ = (x² + 2xy + y²).(x + y) (x + y)³ = x³ + x²y + 2x²y + 2xy² + xy² + y³ (x + y)³ = x³ + y³ + 3x²y + 3xy² x³ + y³ = (x + y)³ - 3x²y - 3xy² x³ + y³ = (x + y)³ - 3xy.(x + y) → given: x + y = 2 x³ + y³ = 8 - 6xy (x + y)⁵ = (x + y)².(x + y)².(x + y) (x + y)⁵ = (x² + 2xy + y²).(x² + 2xy + y²).(x + y) (x + y)⁵ = (x⁴ + 2x³y + x²y² + 2x³y + 4x²y² + 2xy³ + x²y² + 2xy³ + y⁴).(x + y) (x + y)⁵ = (x⁴ + 6x²y² + 4x³y + 4xy³ + y⁴).(x + y) (x + y)⁵ = x⁵ + x⁴y + 6x³y² + 6x²y³ + 4x⁴y + 4x³y² + 4x²y³ + 4xy⁴ + xy⁴ + y⁵ (x + y)⁵ = x⁵ + y⁵ + 5x⁴y + 5xy⁴ + 10x³y² + 10x²y³ x⁵ + y⁵ = (x + y)⁵ - 5x⁴y - 5xy⁴ - 10x³y² - 10x²y³ x⁵ + y⁵ = (x + y)⁵ - 5xy.(x³ + y³) - 10x²y².(x + y) → given: x + y = 2 x⁵ + y⁵ = 32 - 5xy.(x³ + y³) - 20x²y² → recall: x³ + y³ = 8 - 6xy x⁵ + y⁵ = 32 - 5xy.(8 - 6xy) - 20x²y² x⁵ + y⁵ = 32 - 40xy + 30x²y² - 20x²y² x⁵ + y⁵ = 32 - 40xy + 10x²y² → given: x⁵ + y⁵ = 152 152 = 32 - 40xy + 10x²y² 10x²y² - 40xy - 120 = 0 x²y² - 4xy - 12 = 0 → let: z = xy z² - 4z - 12 = 0 Δ = (- 4)² - (4 * - 12) = 16 + 48 = 64 = 8² z = (4 ± 8)/2 z = 2 ± 4 First case: z = 6 → recall: z = xy → xy = 6 Second case: z = - 2 → recall: z = xy → xy = - 2 Resume: x + y = 4 ← this is the sum S xy = 6 or xy = - 2 ← this is the product P So x & y are the solution of the following equation: a² - Sa + P = 0 First case: xy = 6 and (x + y) = 2 a² - 2a + 6 = 0 Δ = (- 2)² - (4 * 6) = 4 - 24 = - 20 = 20i² a = (2 ± i√20)/2 a = (2 ± 2i√5)/2 a = 1 ± i√5 → x = 1 + i√5 Recall: y = 2 - x y = 2 - 1 - i√5 → y = 1 - i√5 → x = 1 - i√5 Recall: y = 2 - x y = 2 - 1 + i√5 → y = 1 + i√5 Second case: xy = - 2 and (x + y) = 2 a² - 2a - 2 = 0 Δ = (- 2)² - (4 * - 2) = 4 + 8 = 12 a = (2 ± √12)/2 a = (2 ± 2√3)/2 a = 1 ± √3 → x = 1 + √3 Recall: y = 2 - x y = 2 - 1 - √3 → y = 1 - √3 → x = 1 - √3 Recall: y = 2 - x y = 2 - 1 + √3 → y = 1 + √3

{x^3+x^3 ➖ }{20+20 ➖ } /{x^3+x^3 ➖ +2+2 ➖ }={ x^6+40}/{x^6+4}= 40x^6/4x^6 =10x^1 2^5x^1^1 2^5^1x 2^1^1x 2^1x (x ➖ 2x+1).

At 11:00 you say 1/x =(- b +- V(b^2 - 4ac))/2a , the next line you say: 1/x =2c/( - b +- V(b^2 - 4ac)) , you explain why.

X+y-1=0 from the second

3^x=3/3^y from 3^ (x+y) =3. Then (3/3^y) -3^y=3. Solution: x=1.213 and y=-0.213.