Question

ajutor îmi trebu urgent!!!​

Answer 1

Explicație pas cu pas:

12.

a) x ≠ -2

$\dfrac{1}{x + 1} + \dfrac{5}{x + 1} = \dfrac{1 + 5}{x + 1} = \dfrac{6}{x + 1}$

b) x ≠ -1; x ≠ 0

$\dfrac{x + 2}{ {x}^{2} + x} + \dfrac{x - 1}{ {x}^{2} + x} = \dfrac{x + 2 + x - 1}{x(x + 1)} = \dfrac{2x + 1}{x(x + 1)}$

13.

a) x ≠ 0

$\dfrac{x + 1}{ {x}^{2}} + \dfrac{^{x)} x - 1}{x} = \dfrac{x + 1 + x(x - 1)}{ {x}^{2}} = \dfrac{x + 1 + {x}^{2} - x}{ {x}^{2}} = \dfrac{{x}^{2} + 1}{ {x}^{2}} = 1 + \dfrac{1}{ {x}^{2}}$

c) x ≠ -1; x ≠ 0

$\dfrac{^{x)} x - 1}{x + 1} + \dfrac{^{x + 1)} 1}{x} = \dfrac{x(x - 1) + x + 1}{x(x + 1)} = \dfrac{ {x}^{2} - x + x + 1}{x(x + 1)} = \dfrac{ {x}^{2}+ 1}{x(x + 1)}$

14.

a) x ≠ 0; x ≠ ⅔

$\dfrac{^{3x - 2)} 1}{x} - \dfrac{^{x} 4}{3x - 2} = \dfrac{3x - 2 - 4x}{x(3x - 2)} = \dfrac{- x - 2}{x(3x - 2)} = - \dfrac{x + 2}{x(3x - 2)}$