Question

Se considera functia f:(0,infinit) -> R,f(x) = 1-1/x^2. Calculati produsul p = f(2) ori (f3) ori ... f(9)​

Answer 1

$\displaystyle f:(0,\infty)\to\mathbb{R},\quad f(x) = 1-\dfrac{1}{x^2} = \dfrac{x^2-1}{x^2} = \dfrac{(x-1)(x+1)}{x\cdot x} \\ \\\\ P = f(2)\cdot f(3)\cdot ...\cdot f(9) = \prod\limits_{k=2}^{9}\dfrac{(k-1)(k+1)}{k\cdot k} =$

$\\\displaystyle=\Bigg(\prod\limits_{k=2}^{9}\dfrac{k-1}{k}\Bigg)\cdot \Bigg(\prod\limits_{k=2}^{9}\dfrac{k+1}{k}\Bigg) =$

$\\=\Bigg(\dfrac{1}{\not 2}\cdot \dfrac{\not 2}{\not 3}\cdot \dfrac{\not 3}{\not 4}\cdot ...\cdot \dfrac{\not 7}{\not 8}\cdot \dfrac{\not 8}{9}\Bigg)\cdot \Bigg(\dfrac{\not 3}{2}\cdot \dfrac{\not 4}{\not 3}\cdot \dfrac{\not 5}{\not 4}\cdot ...\cdot \dfrac{\not 9}{\not 8}\cdot \dfrac{10}{\not 9}\Bigg)=$

$\\=\dfrac{1}{9}\cdot\dfrac{1}{2} \cdot \dfrac{10}{1} = \dfrac{10}{18} = \boxed{\dfrac{5}{9}}$

Answer 2

Răspuns:

5/9

Explicație pas cu pas:

$f(x)=1-\frac{1}{x^{2}}=\frac{x^{2} -1}{x^{2}} \\p=f(2)*f(3)*...*f(8)*f(9)=\frac{3}{4}*\frac{8}{9}*\frac{15}{16}*\frac{24}{25}*\frac{35}{36} *\frac{48}{49} *\frac{63}{64}*\frac{80}{81} = \frac{5}{9}$