Question

Fie expresia $E= \frac{1}{a+b} : \frac{3a-5}{ a^{2}- b^{2} } + \frac{a+b}{3a-5}$
a) Aduceti la o forma mai simpla expresia:
b)Determinati DVA a raportului obtinut:
c)Determinati E(2)
Ajutatima va rog

Answer 1

[tex]E=\dfrac{1}{a-b}:\dfrac{3a-5}{a^2-b^2}+\dfrac{a+b}{3a-5}\\ E=\dfrac{1}{a-b}\cdot \dfrac{a^2-b^2}{3a-5}+\dfrac{a+b}{3a-5}\\ E=\dfrac{1}{a-b}\cdot \dfrac{(a-b)(a+b)}{3a-5}+\dfrac{a+b}{3a-5}\\ E=\dfrac{a+b}{3a-5}+\dfrac{a+b}{3a-5}\\ E=\dfrac{2(a+b)}{3a-5}\\ D.V.A,a\in \mathhbb{R}-\left\{\dfrac{5}{3}\right\}\\ \text{Nu stiu la ce tereferi cand zici E(2),deoarece e o expresie cu doua}\\ \text{variabile.}[/tex]