An Interesting Algebra Problem | Give It A Try!

[@ManojkantSamal]

^=read as to the power
*=read as square root
We know that
(a-b)^7=a^7-b^7-[7ab{(a^6+b^6)-3ab(a^3-b^3)+5a^2b^2(a-b)}].....eqn8
As per question
X^2+5x=5
X(x+5)=5
X+5=(5/x)...... Eqn1
Again
(X+5)^7 - {5^7/(x+5)^7}
=(x+5)^7 - {5^7/(5/x)^7}
=(x+5)^7 - {5^7×x^7/(5^7)}
=(x+5)^7- (x)^7
Let a=x+5, b=x
So,
a-b=x+5-x=5.....eqn2
ab=x(x+5)
=x^2+5x
=5..(As per question )
ab=5...eqn3
Now,
7ab=7×5=35...eqn4
a^6+b^6=(a^3)^2+(b^3)^2
=(a^3-b^3)^2+(2a^3b^3)
={(a-b)^3+3ab(a-b)}^2+(2a^3b^3)
=(125+75)^2 +(2×125)
=(200^2)+250
=40000+250
=40250
So,
a^6+b^6=40250......eqn5
3ab(a^3-b^3)=(3×5)×200
=15×200=3000....eqn6
5a^2b^2(a-b)=(5×25)×5
=125....eqn7
Put eqn4,eqn5, eqn6, eqn 7 in eqn8
(a-b)^7=a^7-b^7-[35{40250-3000+625}]
=a^7-b^7-(35×37875)
=a^7-b^7-1325625
So,
a^7-b^7=(a-b)^7+1325625
=(5^7)+1325625
=78125+1325625
=1403750
So,
a^7-b^7=1403750
(X+5)^7-(x)^7=1403750
(X+5)^7-{5^7/(x+5)^7}=1403750

[@gregevgeni1864]

F = 59375
x²+5x=5 x²+2•5x+5²-5x-5² -5=0 (x+5)²-5(x+5)-5=0
(x+5) -5 -5/(x+5) =0 =>
(x+5)-5/(x+5)=5 (*) because x≠-5.
From (*) =>
(x+5)³ -(5/(x+5))³ = 50 (1)
(x+5)⁴+(5/(x+5))⁴ = 1175 (2).
Multiplying the equalities (1), (2) by terms we get
(x+5)⁷-(5/((x+5))⁷-5³((x+5)+5⁴/(x+5) =
= 1175 • 50 =>
F -5³((x+5) -5/(x+5)) = 1175 •50 =>
F - 125•5=58750=> F = 59375

[@zawatsky]

x²=5x-5. x+5=5/x. x=5/(x+5), т. е. финальная задача - посчитать (5/х)⁷+х⁷. x²-5x+5=0. D=25-50=5, √D=√5. x=(1±√5)/2. [5÷(1±√5)/2]⁷=[10/(1±√5)]⁷. (1±√5)²=1±2√5+5=6±2√5.👈²=36±24√5+20=56±24√5. (1±√5)⁶=(1±√5)²(1±√5)⁴=(6±2√5)(56±24√5), причём только + на + или - на -. В общем, много выходит...

[@潘博宇-k4l]

E=41875

[@RealQinnMalloryu4]

{x^2+x^2 ➖ }{5x+5x ➖ }={x^4+10x^2}=10x^6 5^5x^6 2^3^2^3x^2^4 1^1^1^3x^2^2^2 3x^2 (x ➖ 3x+2) (x^7+35) ➖18125/(x^7+35)={35x^7 ➖ 18.125/35x^7}=18.090x^7/35x^7=6.018x^1 100^600^18 6^8x 6^2^3x 2^3^1^1^1x 2^3x (x ➖ 3x+2).

[@Quest3669]

X+5= 5/x so ? =(5/X))^7-X^7.
5/x--(x)= 5 so;
(625/x^4)+ x^4= 1175 and
((125/x^3)-(x^3)= 200 hence
?= 1175×200+625= 235000+ 625= 235625 soln.

[@SidneiMV]

x² + 5x = 5 => x + 5 = 5/x
E = (5/x)⁷ - 5⁷(x/5)⁷ = E
E = (5/x)⁷ - x⁷
5/x - x = 5
(5/x)² + x² = 35
(5/x)⁴ + x⁴ = 1175
(5/x - x)³ = 5³
(5/x)³ - x³ - 15(5/x - x) = 125
(5/x)³ - x³ = 200
[(5/x)³ - x³][(5/x)⁴ + x⁴] = (1175)(200)
(5/x)⁷ - x7 + 5³x - 5⁴/x = 235000
E - 5³(5/x - x) = 235000
E = 235000 + 625
*E = 235625*

[@ManojkantSamal]

Ans :: 143750.......(May be )
Explain later.....

[@Fjfurufjdfjd]

Θετω χ+5=α. Η σχεση που δινεται γραφεται:χ(χ+5)=5 (α-5)α=5. α^2-5α=5.ή
α^2-5=5α..ή (διαιρω με το α=/0 α-5/α=5.
Ψαχνω Ε=α^7-(5/α)^7
Παιρνω το γινομενο: [α^3-(5/α)^3][α^4+(5/α)^4]=[α^7-(5/α)^7]-α^4(5/α)^3+α^3(5/α)^4=
Ε-5^3(α)+(5^4/α)=Ε-125×5=Ε-625.
Αλλα α^3-(5/α)^3=....5^3-15×5=125-75=50 και
α^4+(5/α)^4=...[α^2+(5/α)^2]^2-50.
α^2+(5/α)^2=[α-(5/α)]^2+2α×5/α=25+10=35.
Εχω τελικα: 35^2-50=1225-50=1175.
50×1175=Ε-625..
58750=Ε-625 αρα Ε=58750+625=59375.

[@Fjfurufjdfjd]

hahaha το χριστουγεννιατικο δεντρο φταιει.

[@guyhoghton399]

*_x² + 5x = 5_*
⇒ _x + 5 = 5/x = y_ (say)
∴ *_y - 5/y_*_ = y - x = (x + 5) - x = _*_5_* ... ①
Let *_uₙ = yⁿ + (-5/y)ⁿ_*
Then *_uₘ₊ₙ = uₘuₙ - (-5)ⁿuₘ₋ₙ_*
_u₀ = 2, u₁ = 5_ (from ①)
*_u₇ = (x + 5)⁷ - 5⁷/(x + 5)⁷_* ... the value we seek
_u₂ = u₁₊₁ = u₁² - (-5)¹u₀ = 25 + 10 = 35_
_u₄ = u₂₊₂ = u₂² - (-5)²u₀ = 35² - 50 = 1175_
_u₃ = u₂₊₁ = u₂u₁ - (-5)¹u₁ = (35)(5) + (5)(5) = 200_
*_u₇ = u₄₊₃ = u₄u₃ - (-5)³u₁ = (1175)(200) + (125)(5) = 235,625_*