Matematika - březen 2023

Řeěení:

\( 25^{\sin x} =5^{\frac{1}{\cos x}} \)

\(  (25^{\sin x} )^{\cos x}=5 \)

\(  5^{2\sin x \cos x}=5 \)

\(  5^{\sin 2x }=5 \)

\( \sin 2x =1 \)

\( 2x=\frac{\pi}{2}+2k\pi \)

\( x=\frac{\pi}{4}+k\pi \) 🤗

Řeěení:

\( 1+\frac{1}{k^2}+\frac{1}{(k+1)^2}=\frac{k^2\cdot(k+1)^2+(k+1)^2+k^2}{k^2\cdot(k+1)^2} = \frac{k^2(k+1)^2+ k^2+2k+1+k^2}{k^2\cdot(k+1)^2}= \)

\( =\frac{k^2(k+1)^2+2k(k+1)+1}{k^2\cdot(k+1)^2}=\frac{(k(k+1)+1)^2}{k^2\cdot(k+1)^2} =(\frac{k(k+1)+1}{k(k+1)})^2 = (1+\frac{1}{k(k+1)})^2 \)

 Pak

\(  \sum_{k=1}^{k=99} (\sqrt{ 1+\frac{1}{k^2} +\frac{1}{(k+1)^2}} = \sum_{k=1}^{k=99} (\sqrt{ (1+\frac{1}{k(k+1)})^2} = \sum_{k=1}^{k=99} (1+\frac{1}{k(k+1)}) =\sum_{k=1}^{k=99} 1 + \sum_{k=1}^{k=99} \frac{1}{k(k+1)} = 99+1-\frac{1}{100}=100-\frac{1}{100} \)

Tedy

\( n=100 \) 🤣

Řeěení:

\( x+\frac{1}{x}=-1 \) tedy \( x^2+x+1=0 \)

To je v R ireducibilní, zkusme C.

\( x=\frac{-1+i\sqrt{3}}{2}=\frac{-1}{2}+i\frac{\sqrt{3}}{2}=\cos \frac{2\pi}{3}+i\sin\frac{2\pi}{3} \)

Dále

\( \frac{1}{x} = \cos \frac{4\pi}{3}+i\sin\frac{4\pi}{3} = \cos \frac{2\pi}{3}-i\sin\frac{2\pi}{3} \)

 No a umocníme a sečteme

\( x^{91}+\frac{1}{x^{91}}=\cos \frac{2\cdot 91\pi}{3}+i\sin\frac{2\cdot 91\pi}{3}+\cos \frac{2\cdot 91\pi}{3}-i\sin\frac{2\cdot 91\pi}{3} =\cos \frac{2\pi}{3} +\cos \frac{2\pi}{3}+0=-1 \)

A pokud neznáme komplexní čísla:

  \( x^2+x+1=0 \)

\( x^3-1=(x-1)(x^2+x+1) =0 \)

\(x^3=1\)

\( x^{91} = x\cdot x^{90}= x\cdot (x^3)^{30}=x \)

\( x^{91}+\frac{1}{x^{91}} = x+\frac{1}{x}=-1 \)

Řeěení:

Pokud \( \arccos{\frac{3}{17}} =\phi \) pak \( \cos\phi=\frac{3}{17} \) 

Pak \( \sin\frac{\phi}{2}=\pm\sqrt{\frac{1-\cos\phi}{2}} \) a spojme to dohromady

\( \sin\frac{\phi}{2}=\pm\sqrt{\frac{1-\frac{3}{17}}{2}}=\pm\sqrt{\frac{\frac{14}{17}}{2}} =\pm\sqrt{\frac{7}{17}} \) 

\( \sin{ \left( \frac{\arccos{\frac{3}{17}}}{2} \right) } = \pm\sqrt{\frac{7}{17}} \)

 😐

Řeěení:

 🤨

Řešení:

\( 2^{xy}=3^y \) pak  

 \( 12^y=3^y\cdot2^{2y}=2^{xy}\cdot2^{2y}=2^{xy+2y}=8 \)

\( xy+2y=3 \)

 😀