Question

Dau coroana...măcar unul dintre ele. ​

Answer 1

Răspuns:

Explicație pas cu pas:

a) (x-5)/(x-2) *(x+5)/x-7)  =(x+5)(x-5)/(x+2)(x-7)=( x^2-25)/(x+2)(x-7)

sau

( x^2-25)/( x^2-5*x-14)

Answer 2

$\frac{1}{2x} - \frac{x {}^{3} - x }{x {}^{2} - 2x + 1 } : \frac{x {}^{4} + 4x {}^{3} }{2x - 2} \\ \\ = \frac{1}{2x} - \frac{x(x {}^{2} - 1)}{x {}^{2} - 2\cdot x\cdot1 + 1 {}^{2} } : \frac{x {}^{3}(x + 4) }{2x - 2} \\ \\ = \frac{1}{2x} - \frac{\not x(x - 1)(x + 1)}{(x - 1) {}^{2} } \cdot \frac{2x - 2}{x {}^{\not3} (x + 4)} \\ \\ = \frac{1}{2x} - \frac{x + 1}{x - 1} \cdot \frac{2(x - 1)}{x {}^{2}(x + 4) } \\ \\ = \frac{1}{2x} - (x + 1)\cdot \frac{2}{x {}^{2} (x + 4)} \\ \\ = \frac{1}{2x} - \frac{2x + 2}{x {}^{2} (x + 4)} \\ \\ = \frac{x(x + 4) - 2(2x + 1)}{2x {}^{2}(x + 4) } \\ \\ = \frac{x {}^{2} + 4x - 4x - 4}{2x {}^{3} + 8x {}^{2} } \\ \\ = \frac{x {}^{2} - 4}{2x {}^{3} + 8x {}^{2} }$