Matematyka.pl

"janusz47" pisze: \(\displaystyle{ a_{n+1}-a_{n} = (-1)^{n+1}+(-1)^{n+2}-(-1)^{n}-(-1)^{n+1}=}\)\(\displaystyle{ (-1)^{n+2}-(-1)^{n} = (-1)^{n} ( (-1)^{2} -1)= (-1)^{n}(1 -1)=(-1)^{n} \cdot 0 = 0}\)\(\displaystyle{ \Rightarrow (a_{n+1} -a_{n} = 0 ) \Rightarrow (a_{n+1} = a_{n}) dla \ n \in N_{+}.}\)

janusz47 pisze: \(\displaystyle{ t = 2 ^{x}> 0}\)
\(\displaystyle{ t ^{3} -8t ^{2} - 16t +128 = 0}\)

janusz47 pisze:
\(\displaystyle{ p(2b,1c) = \frac{ {7 \choose 2} {5 \choose 1} }{ {12 \choose 3} }}\)

janusz47 pisze: \(\displaystyle{ \lim_{x \to 0} \frac{tgx}{x} = \lim_{ x\to 0 } \frac{1}{cosx} \cdot \lim_{ x\to 0} \frac{sinx}{x} =1 \cdot 1 =1}\)

janusz47 pisze: \(\displaystyle{ y = x= t , t \in < -1, 1>}\)
\(\displaystyle{ \int_{-1}^{1} \left| t +t \right| \sqrt{1 + 1 }dt = 2 \sqrt{2} \int_{-1}^{0} (-t)dt +
2 \sqrt{2} \int_{0}^{1} t dt = \sqrt{2} + \sqrt{2} = 2 \sqrt{2}}\)

"janusz47" pisze: Podstaw \(\displaystyle{ z = x +i \cdot y}\) i skorzystaj z definicji wartości bezwględnej
liczby zespolonej

janusz47 pisze:twierdzenie Bezout

janusz47 pisze: \(\displaystyle{ \arc{\cos(x)} =u}\)
\(\displaystyle{ x = \cos(u)}\)
\(\displaystyle{ sin(u) = \sqrt{1 - x^{2}}}\)
\(\displaystyle{ \tan{\arc{\cos(x)}} = \tan(u) = \frac{\sin(u)}{\cos(u)}= \frac{\sqrt{1 - x^{2}}}{x}}\)