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Andreutzakriss
a fost răspuns

Rezolvati ecuatiile:
a)6(x²+√3)=√108
b)(x-√6)²=24
c)x²-2x=8
e)x²-8x=-7
f)[tex] \dfrac{x+ \sqrt{3} }{3 \sqrt{3} }= \dfrac{- \sqrt{3} }{x- \sqrt{3} } [/tex]
g)[tex] \dfrac{ x^{2} +4}{4}= \dfrac{ x^{2} +5x}{5} [/tex]
h)(3x-2)²-2(2x+5)=2(9-26x)
i)12(x-3)²=(6+2√3)²

Răspuns :

a) 6(x²+√3) = 6√3
x²+√3=√3 ⇒ x²=0 ⇒ x = 0

b) (x-√6)² = 24
x-√6=+-√24= +- 2√6
x₁-√6=+2√6 ⇒ x₁=3√6
x₂-√6= -2√6 ⇒ x₂= -√6

c)x²-2x=8
x²-2x-8=0
Δ=4+32
Δ=36 ⇒ √Δ=6
x₁ = (2+6)/2 =4
x₂ = (2 -6)/2  =-2

e) x²-8x= -7
x²-8x +7 = 0
x²-7x -x +7 =0 
x(x-1) - 7(x-1) = 0
(x-1)(x-7) = 0
x₁ =1
x₂ =7

f) x+√3 / (3√3) = -√3 / (x-√3) 
 x² - 3 = - 9
 x² = -6 ∉ |R   {in C, x are valorile x₁ =√6i si x₂ =-√6i }

g) (x²+4)/4 = (x² + 5x)/5 ⇒
5*(x²+4)= 4*(x² + 5x) ⇒
5x² + 20 =4x² +20x
x² - 20x +20 = 0
Δ=400 - 80
Δ=320=8√5
x₁ = (20-8√5)/2= 2(5-2√5)
x₂ = (20+8√5)/2 = 2(5+2√5)

h)(3x-2)²-2(2x+5)=2(9-26x)
9x²-12x+4-4x-10=18-52x
9x²-16x-6=18-52x
9x²+36x-24=0
Δ=1296 + 864
Δ=12√15
x₁ = (- 36 + 12√15)/2*9= 2(√15-3)/3
x₂ = (- 36 - 12√15)/2*9 = -2(3+√15)/3

i)12(x-3)²=(6+2√3)² | :12
(x-3)²=2(2+√3)    { (6+2√3)²=36+ 24√3 + 12 }
x₁-3=√[2(2+√3)] ⇒ x₁=√[2(2+√3)] +3
x₂-3= - √[2(2+√3)] ⇒ x₂= - √[2(2+√3)] + 3

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