x = -1/2 or x = -2. Let p = √((2x+3)/(x+1)) and q = √((3x+2)/(x+1)) (*) with { (2x+3)/(x+1) ≥ 0 and (3x+2)/(x+1) ≥0 and x ≠ -1 } => x ≤ -3/2 and -2/3 ≤ x (**). Now from (*) we have p² + q² = (2x+3)/(x+1)+(3x+2)/(x+1) = 5(x+1)/(x+1) =5 => p² + q² = 5 (1). From the original equation and (*) we have p³ +q³ = 9 (2). We solve the system of (1) and (2). (1) (p+q)² - 2(pq) = 5 (1)' (2) (p+q)³ -3(pq)(p+q) = 9 (2)'. Put p+q = r and pq = t and the system of the (1)' and (2)' rewrite as r² - 2t = 5 (1)'' and r³-3rt = 9 (2)''. From (1)'' => 2t = r² - 5 . The (2)'' write as 2r³ -3r(r²-5) =18 or 2r³ -3r³+15r-18=0 or r³ - 15r +18 = 0 (r-3)(r²+3r-9)=0 => r=3 or r=(-3±3√5)/2.. From r=3 => t=2. So p+q =3 and pq =2 => p, q roots of the m² -3m+2=0 => m =1 or m=2 so (p,q)=(1,2) or (p,q)=(2,1). For p=1 => √((2x+3)/(x+1)) =1 => (2x+3)/(x+1) =1 => x=-2 accepted due to (**). For p=2 => √((3x+2)/(x+1)) = 2 => (2x+3)/(x+1) = 4 => x=-1/2 accepted due to (**). The solution x=-1/2 satisfies the given equation. We will work accordingly in the case that r = (-3±3√5)/2 .. 😅
> a^3/2+b^3/2 =9 =8+1 or 1+8; => {(2x+3)/(x+1)}^3/2=8 or1; =√(4^3) or√1^3=> (2x+3)(x+1)=4;=> 2x+3=4x+4;=> 2x=-1; x= -1/2; for (2x+3)/(x+1)=1;=> x = -2; verified for part b^3/2; ok; => x = -1/2 or -2 .
χ=/-1.θετω (2χ+3)/(χ+1)=α>0 και β=(3χ+2)/(χ+1)>0. Τοτε α+β=5. Η εξισωση που δινεται γραφεται (α^3)^(1/2)+(β^3)^(1/2)=9. Υψωνω στο τετραγωνο και εχω: α^3+β^3+2(α^3β^3)^(1/2)=81. (α+β)^3-3αβ(α+β)+2([(αβ)^3]^(1/2)=81. 125-15(αβ)+2[(αβ)^(1/2)]^3=81. Αν (αβ)^(1/2)=ψ>0 εχω: 2ψ^3-15ψ^2+44=0 (ψ-2)(2ψ^2-11ψ-22)=0 ψ=2 ή ψ=[11+(27×11)^(1/2)]/4 ψ>0. Ψ=2 (αβ)^(1/2)=2 αβ=4.αρα α+β=5 και αβ=4. α=1,β=4 ή α=4,β=1.χ=-2 ή χ=-1/2 Αν αβ=(11/8)[19+3(33)^(1/2)] και α+β=5 η εξισωση που προκυπτει δεν δινει πραγματικες λυσεις για τα α,β. Τελικα: χ=-2,-1/2.
x = -1/2 or x = -2.
Let p = √((2x+3)/(x+1)) and
q = √((3x+2)/(x+1)) (*) with
{ (2x+3)/(x+1) ≥ 0 and (3x+2)/(x+1) ≥0 and x ≠ -1 } =>
x ≤ -3/2 and -2/3 ≤ x (**).
Now from (*) we have
p² + q² = (2x+3)/(x+1)+(3x+2)/(x+1)
= 5(x+1)/(x+1) =5 => p² + q² = 5 (1).
From the original equation and (*) we have p³ +q³ = 9 (2).
We solve the system of (1) and (2).
(1) (p+q)² - 2(pq) = 5 (1)'
(2) (p+q)³ -3(pq)(p+q) = 9 (2)'.
Put p+q = r and pq = t and the system of the (1)' and (2)' rewrite as
r² - 2t = 5 (1)'' and r³-3rt = 9 (2)''.
From (1)'' => 2t = r² - 5 .
The (2)'' write as 2r³ -3r(r²-5) =18 or
2r³ -3r³+15r-18=0 or r³ - 15r +18 = 0 (r-3)(r²+3r-9)=0 =>
r=3 or r=(-3±3√5)/2..
From r=3 => t=2.
So p+q =3 and pq =2 => p, q roots of the m² -3m+2=0 => m =1 or m=2 so (p,q)=(1,2) or (p,q)=(2,1).
For p=1 => √((2x+3)/(x+1)) =1 =>
(2x+3)/(x+1) =1 => x=-2 accepted
due to (**).
For p=2 => √((3x+2)/(x+1)) = 2 =>
(2x+3)/(x+1) = 4 => x=-1/2 accepted due to (**).
The solution x=-1/2 satisfies the given equation.
We will work accordingly in the case that r = (-3±3√5)/2 .. 😅
A=Sqrt（（2X+3）/（X+1））， B=Sqrt（（3X+2）/（X+1））， A^3+B^3=9， A^2+B^2=（2X+3）/（X+1）+（3X+2）/（X+1）=5， A=2， B=1， Or A=1，B=2
> a^3/2+b^3/2 =9 =8+1 or 1+8; =>
{(2x+3)/(x+1)}^3/2=8 or1; =√(4^3) or√1^3=> (2x+3)(x+1)=4;=> 2x+3=4x+4;=> 2x=-1; x= -1/2; for
(2x+3)/(x+1)=1;=>
x = -2; verified for part b^3/2; ok; =>
x = -1/2 or -2 .
8:00 vʼ(mn)=(11+3vʼ33)/4≈7.06
mn≈50; Δ≈25-4•50≈-175
{6x^3+12}/{x^3+3}+{12x^3+6}/{x^3+3}={18x^3/3x^3+18x^3/3x^3}=36x^6/6x^6=6x^1 2^3x^1 2^1x^1 2x^1 (x ➖ 2x+1).
X=-1/2 or -2.
χ=/-1.θετω (2χ+3)/(χ+1)=α>0 και β=(3χ+2)/(χ+1)>0. Τοτε α+β=5. Η εξισωση που δινεται γραφεται (α^3)^(1/2)+(β^3)^(1/2)=9. Υψωνω στο τετραγωνο και εχω: α^3+β^3+2(α^3β^3)^(1/2)=81.
(α+β)^3-3αβ(α+β)+2([(αβ)^3]^(1/2)=81.
125-15(αβ)+2[(αβ)^(1/2)]^3=81.
Αν (αβ)^(1/2)=ψ>0 εχω:
2ψ^3-15ψ^2+44=0 (ψ-2)(2ψ^2-11ψ-22)=0
ψ=2 ή ψ=[11+(27×11)^(1/2)]/4 ψ>0.
Ψ=2 (αβ)^(1/2)=2 αβ=4.αρα
α+β=5 και αβ=4. α=1,β=4 ή α=4,β=1.χ=-2 ή χ=-1/2
Αν αβ=(11/8)[19+3(33)^(1/2)] και α+β=5
η εξισωση που προκυπτει δεν δινει πραγματικες λυσεις για τα α,β.
Τελικα: χ=-2,-1/2.