Question

Rezolvaţi ecuația 1 supra x² + 6 x + 9 + 1 supra x ² + 7 x + 12 + 1 supra x ² + 9 x + 20 = x + 3 supra x + 5 , x aparține R \ { - 5, - 4, - 3} ​

Answer 1

Răspuns:

Explicație pas cu pas:

$\frac{1}{x^2+6x+9} + \frac{1}{x^2+7x+12} + \frac{1}{x^2+9x+20} = \frac{x+3}{x+5}$

$\frac{1}{(x+3)^2} + \frac{1}{(x+3)(x+4)} + \frac{1}{(x+4)(x+5)} = \frac{x+3}{x+5}$

$\frac{(x+4)(x+5)}{(x+3)^2(x+4)(x+5)} + \frac{(x+3)(x+5)}{(x+3)^2(x+4)(x+5)} + \frac{(x+3)^2}{(x+3)^2(x+4)(x+5)} = \frac{x+3}{x+5}$

$\frac{(x+4)(x+5)+(x+3)(x+5)+(x+3)^2}{(x+3)^2(x+4)(x+5)} = \frac{x+3}{x+5}$

$\frac{x^2 + 9x + 20+ x^2 + 8x + 15 + x^2 + 6x + 9}{(x+3)^2(x+4)(x+5)} = \frac{x+3}{x+5}$

$\frac{3x^2 + 23x +44}{(x+3)^2(x+4)(x+5)} = \frac{x+3}{x+5}$

$\frac{(3x+11)(x+4)}{(x+3)^2(x+4)(x+5)} = \frac{x+3}{x+5}$

$\frac{(3x+11)}{(x+3)^2} = x+3$

$(x+3)^3=3x+11$

$x^3 + 9x^2 + 27x + 27 = 3x + 11$

$x^3 + 9x^2 + 24x + 16 = 0$

$x^3 + x^2 + 8x^2+ 8x + 16x + 16 = 0$

$x^2(x + 1) + 8x(x+1) + 16(x + 1) = 0$

$(x+1)(x^2 + 8x + 16) = 0$

$(x+1)(x+4)^2 = 0$

x + 1 = 0 ⇒ x = -1

(x + 4)² = 0 ⇔ x+4 = 0 ⇒ x = - 4  nu este solutie, deoarece x∈ R \ {-5 , -4 , -3}

Solutia: x = -1

Answer 2

$\it \dfrac{1}{x^2+6x+9}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}=\dfrac{x+3}{x+5} \Rightarrow\\ \\ \\ \Rightarrow \dfrac{1}{(x+3)^2}+\dfrac{1}{(x+3)(x+4)}+\dfrac{1}{(x+4)(x+5)}=\dfrac{x+3}{x+5} \Rightarrow\\ \\ \\ \Rightarrow\dfrac{1}{(x+3)^2}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}=\dfrac{x+3}{x+5} \Rightarrow\\ \\ \\ \Rightarrow \dfrac{1}{(x+3)^2}+\dfrac{1}{x+3}=\dfrac{x+3}{x+5}+\dfrac{1}{x+5} \Rightarrow\dfrac{x+4}{(x+3)^2}=\dfrac{x+4}{x+5} \Rightarrow$

$\it \Rightarrow \dfrac{1}{(x+3)^2}=\dfrac{1}{x+5} \Rightarrow (x+3)^2=x+5\\ \\ \\Not\breve am\ x+4=t,\ \ t\ne0\ \ \ \ (*)\\ \\ Ecua\c{\it t}ia\ devine:\\ \\ (t-1)^2=t+1 \Rightarrow t^2-2t+1-t-1=0 \Rightarrow t^2-3t=0 \Rightarrow\\ \\ \Rightarrow t(t-3)=0\ \stackrel{(*)}{\Longrightarrow} t-3=0 \Rightarrow t=3 \Rightarrow x+4=3 \Rightarrow x=-1 \ (solu\c{\it t}ia)$