Question

rezolvati in R ecuatia va roog

Answer 1

Trebuie să aducem fracțiile la același numitor și să rezolvăm ecuațiile obținute.
Rezolvarea unei ecuații de gradul 1
[tex]ax+b = 0\\~ x = \frac{-b}{a}[/tex]
Rezolvarea unei ecuații de gradul 2
[tex]ax^2+bx+c = 0\\~ \Delta = b^2-4ac\\~ x_{1} = \frac{-b+\sqrt{\Delta}}{2a}\\~ x_{2} = \frac{-b-\sqrt{\Delta}}{2a}[/tex](dacă delta e negativ, atunci ecuația nu are soluții reale)

11.
a)
[tex]\frac{3x^2}{x^2-2x+1}-\frac{5x}{x-1}+2 = 0\\~ \frac{3x^2}{(x-1)^2}-\frac{5x}{x-1}+2 = 0\\~ \frac{3x^2}{(x-1)^2}-\frac{5x(x-1)}{(x-1)^2}+\frac{2(x-1)^2}{(x-1)^2} = 0\\~ 3x^2-5x(x-1)+2(x-1)^2 = 0 \\~ 3x^2-5x^2+5x+2(x^2-2x+1) = 0\\~ -2x^2+5x+2x^2-4x+2 = 0 \\~ x+2 = 0 \\~ x = -2;[/tex]

b)
[tex]-\frac{5t^2}{(t+2)^2}+\frac{t}{t+2}+4 = 0\\~ -\frac{5t^2}{(t+2)^2}+\frac{t(t+2)}{(t+2)^2}+\frac{4(t+2)^2}{(t+2)^2} = 0\\~ -5t^2+t^2+2t+4(t^2+4t+4) = 0 \\~ -5t^2+t^2+2t+4t^2+16t+16 = 0 \\~ 18t+16 = 0 \\~ t = \frac{-16}{18} = \frac{-8}{9}\\[/tex]

c)
[tex]z^4+4z^2-5 = 0\\~ \text{Notez } z^2 = x \\~ \text{Ecuatia devine: }x^2+4x-5 = 0\\~ \Delta = 16+4*5 = 36\\~ x_{1} = \frac{-4+ \sqrt{36} }{2} = \frac{-4+6}{2} = 1\\~ x_{2} = \frac{-4-\sqrt{36}}{2} = \frac{-4-6}{2} = -5\\~ z^2= x_{1} \implies z_1 = 1, z_2 = -1\\~ z^2=x_{2} \implies z_3=\sqrt{5}, z_4 = -\sqrt{5}[/tex]

d)
[tex]4x^2+5x+1=0 \\~ \Delta = 25-16 = 9\\~ x_{1} = \frac{-5+\sqrt{9}}{2} = \frac{-5+3}{2} = -1\\~ x_{2} = \frac{-5-\sqrt{9}}{2} = \frac{-8}{2} = -4[/tex]

e)
[tex]\frac{2x}{x^2-16}-\frac{3}{x+4}= \frac{5-x}{x-4}\\~ \frac{2x}{(x+4)(x-4)}-\frac{3}{x+4}= \frac{5-x}{x-4}\\~ \frac{2x}{(x+4)(x-4)}-\frac{3(x-4)}{(x+4)(x-4)}= \frac{(5-x)(x+4)}{(x-4)(x+4)}\\~ 2x-3(x-4)=(5-x)(x+4)\\~ 2x-3x+12 = 5x+20-x^2-4x\\~ x^2+2x-3x+4x-5x+12-20 = 0\\~ x^2-2x-8=0\\~ \Delta = 4+32 = 36\\~ x_1 = \frac{-2+\sqrt{36}}{2} = \frac{-2+6}{2} = 2\\~ x_2 = \frac{-2-\sqrt{36}}{2} = \frac{-2-6}{2} = -4[/tex]

f)
[tex]\frac{3}{2x-1}-\frac{4x}{x+2}=5\\~ \frac{3(x+2)}{(2x-1)(x+2)}-\frac{4x(2x-1)}{(2x-1)(x+2)}=\frac{5(2x-1)(x+2)}{(2x-1)(x+2)}\\~ 3(x+2)-4x(2x-1)=5(2x-1)(x+2)\\~ 3x+6-8x^2+4x=5(2x^2+4x-x-2)\\~ -8x^2+7x+6=10x^2+15x-10\\~ -18x^2-8x+16 = 0\\~ 9x^2+4x-8=0\\~ \Delta = 16+4*8*9= 304\\~ x_{1} = \frac{-4+\sqrt{304}}{18} = \frac{-4+4\sqrt{19}}{18}=\frac{4(\sqrt{19}-1)}{18}=\frac{-2(1-\sqrt{19})}{9}\\~ x_2 = \frac{-4-4\sqrt{19}}{18} = \frac{-2(1+\sqrt{19})}{9}[/tex]