T: Zastosowanie przekształcen algebraicznych, wszystko podane w załączniku pozdro ;).... Question from @Free10124

October 2018 1 10 Report

T: Zastosowanie przekształcen algebraicznych, wszystko podane w załączniku pozdro ;)

1a)

$\frac{1}{\sqrt{2}-(\sqrt{3}+1)}\cdot \frac{\sqrt{2}+(\sqrt{3}+1)}{\sqrt{2}+(\sqrt{3}+1)}=\frac{\sqrt{2}+\sqrt{3}+1}{2-(\sqrt{3}+1)^2}=\frac{\sqrt{2}+\sqrt{3}+1}{2-3-2\sqrt{2}-1}=-\frac{\sqrt{2}+\sqrt{3}+1}{2+2\sqrt{2}}=\\\\=-\frac{\sqrt{2}+\sqrt{3}+1}{2+2\sqrt{2}}\cdot \frac{2-2\sqrt{2}}{2-2\sqrt{2}}=-\frac{2\sqrt{2}+2\sqrt{3}+2-4-2\sqrt{6}-2\sqrt{2}}{4-8}=\frac{2\sqrt{3}-2\sqrt{6}-2}{4}=\\\\=\frac{\sqrt{2}-\sqrt{3}-1}{2}$

1b)

$\frac{2}{(1-\sqrt{2})-\sqrt{3}}\cdot \frac{(1-\sqrt{2})+\sqrt{3}}{(1-\sqrt{2})+\sqrt{3}}=\frac{2(1-\sqrt{2}+\sqrt{3})}{(1-\sqrt{2})^2-3}=\frac{2(1-\sqrt{2}+\sqrt{3})}{-2\sqrt{2}}=\\\\=\frac{\sqrt{2}-\sqrt{3}-1}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{2-\sqrt{6}-\sqrt{2}}{2}$

1c)

$\frac{1}{(\sqrt{2}+\sqrt{3})+\sqrt{5}}\cdot \frac{(\sqrt{2}+\sqrt{3})-\sqrt{5}}{(\sqrt{2}+\sqrt{3})-\sqrt{5}}=\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{(\sqrt{2}+\sqrt{3})^2-5}=\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2+\sqrt{6}+3-5}=\\\\=\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{\sqrt{6}}\cdot \frac{\sqrt{6}}{\sqrt{6}}=\frac{\sqrt{12}+\sqrt{18}-\sqrt{30}}{6}=\frac{2\sqrt{3}+3\sqrt{2}-\sqrt{30}}{6}$

$\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{98}+\sqrt{99}}+\frac{1}{\sqrt{99}+\sqrt{100}}=\\\\=\frac{\sqrt{1}-\sqrt{2}}{(\sqrt{1}+\sqrt{2})(\sqrt{1}-\sqrt{2})}+\frac{\sqrt{2}-\sqrt{3}}{(\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3})}+\frac{\sqrt{3}-\sqrt{4}}{(\sqrt{3}+\sqrt{4})(\sqrt{3}-\sqrt{4})}+...+\\\\+\frac{\sqrt{98}-\sqrt{99}}{(\sqrt{98}+\sqrt{99})(\sqrt{98}-\sqrt{99})}+\frac{\sqrt{99}-\sqrt{100}}{(\sqrt{99}+\sqrt{100})(\sqrt{99}-\sqrt{100})}=$

$=\frac{\sqrt{1}-\sqrt{2}}{1-2}+\frac{\sqrt{2}-\sqrt{3}}{2-3}+\frac{\sqrt{3}-\sqrt{4}}{3-4}+...+\frac{\sqrt{98}-\sqrt{99}}{98-99}+\frac{\sqrt{99}-\sqrt{100}}{99-100}=\\\\=-\sqrt{1}+\sqrt{2}-\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{4}-...-\sqrt{98}+\sqrt{99}-\\\\-\sqrt{99}+\sqrt{100}=-\sqrt{1}+\sqrt{100}=\begin{cases} -1+10=9\\-1-10=-11\\ 1+10=11 \\ 1-10=-9 \end{cases}$

$(\sqrt{1+x^2}+x)^-^1=\sqrt{1+x^2}-x\\\\ L=(\sqrt{1+x^2}+x)^-^1=\frac{1}{\sqrt{1+x^2}+x}\cdot \frac{\sqrt{1+x^2}-x}{\sqrt{1+x^2}-x}=\frac{\sqrt{1+x^2}-x}{1+x^2-x^2}=\\\\=\sqrt{1+x^2}-x=P$

$\frac{x}{y}=\frac{1+\sqrt{5}}{2} \ \ \ \Rightarrow \ \ \ \frac{x}{y}=\frac{x+y}{x}\\\\ \frac{x}{y}=\frac{x+y}{x}\\\\ \frac{x}{y}=1+\frac{y}{x}\\\\ \frac{x}{y}-\frac{y}{x}=1$

Korzystamy z założenia:

$\frac{1+\sqrt{5}}{2}-\frac{2}{1+\sqrt{5}}\stackrel{?}{=}1\\\\ L=\frac{1+\sqrt{5}}{2}-\frac{2}{1+\sqrt{5}}=\frac{(1+\sqrt{5})^2-4}{2+2\sqrt{2}}=\frac{1+2\sqrt{5}+5-4}{2+2\sqrt{2}}=\frac{2+2\sqrt{2}}{2+2\sqrt{2}}=1=P$