\frac{1}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1} warto zapamiętać
bo
\frac{1}{x}-\frac{1}{x+1}=\frac{x+1-x}{x(x+1)}=\frac{1}{x(x+1)}
Rozwiązanie
dziedzina D: \ x \in \mathbb R - \{5,6,7,8,9\}
\frac{1}{(x+5)(x+6)}+\frac{1}{(x+6)(x+7)}+\frac{1}{(x+7)(x+8)}+\frac{1}{(x+8)(x+9)}=1
{\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}+\frac{1}{x+7}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+9}=1}
\frac{1}{x+5}-\frac{1}{x+9}=1
\frac{1}{x+5}=1+\frac{1}{x+9}
\frac{1}{x+5}=\frac{1+x+9}{x+9}
\frac{1}{x+5}=\frac{10+x}{x+9} proporcja - mnożę “na krzyż”
(x+5)(x+10)=x+9
x^2+10x+5x+50-x-9=0
x^2+14x+41=0
a = 1, b = 14, c = 41
\Delta = 14^2-4\cdot 1\cdot 41=32
\sqrt\Delta=\sqrt{32}=\sqrt{16\cdot 2}=4\sqrt2
x_1=\frac{-14-4\sqrt2}{2}=\frac{2(-7-2\sqrt2}{2}=-7-2\sqrt2
x_2=\frac{-14+4\sqrt2}{2}=\frac{2(-7+2\sqrt2}{2}=2\sqrt2-7
x_1=-7-2\sqrt2
x_2=2\sqrt2-7