Question

Iar nu mi-a iesit :(​

Answer 1

Răspuns:

Explicație pas cu pas:

[x/(2x-1) +5/(2x+1) +(11x+3)/(1-4x²)]  /  [(x+2)/(2x+1)] =

[x/(2x-1) +5/(2x+1) -(11x+3)/(4x²-1)]  /  [(x+2)/(2x+1)] =

din + am facut  -(-)

acuma numitorul comun 2 (4x²-1)                     =

[ x(2x+1) +5(2x-1) -(11x+3) ] /(4x²-1)  / [(x+2)(2x-1)/(2x+1)]/(2x-1) =

[2x²+x+10x-5-11x-3] /  [(x+2)(2x-1)] =

(2x²-8) / (2x²+3x-2)=

[2(x-2)(x+2)] / [(x+2)(2x-1)] =

2(x-2) / (2x-1) =(2x-4)/(2x-1)= 1 - 3/(2x-1)

Answer 2

b) $\it (\frac{x}{2x - 1} + \frac{5}{2x + 1} + \frac{11x + 3}{1 - 4x^{2} }) : \frac{x + 2}{2x + 1}  = ( \frac{x}{2x - 1} + \frac{5}{2x + 1} - \frac{11x + 3}{4x^{2} - 1 } ) \cdot \frac{2x + 1}{x + 2} = ( ~^{2x + 1)} \frac{x}{2x - 1} +  ~^{2x - 1)} \frac{5}{2x + 1} - \frac{11x + 3}{(2x - 1)(2x + 1)} ) \cdot \frac{2x + 1}{x + 2} =  \frac{x(2x + 1) + 5(2x - 1) - 11x - 3}{(2x - 1)(2x + 1)} \cdot \frac{2x + 1}{x  + 2} = \frac{2x^{2} + x + 10x - 5 - 11x - 3}{(2x - 1)(2x + 1)} \cdot \frac{2x + 1}{x + 2} = \frac{2x^{2} - 8}{(2x - 1)(2x + 1)} \cdot \frac{2x + 1}{x + 2} = \frac{2(x^{2} - 4) }{(2x - 1)(x + 2)} = \frac{2(x - 2)(x + 2)}{(2x - 1)(x + 2)} = \frac{2(x - 2)}{2x - 1} = \frac{2x - 4}{2x - 1}$