Ułamek piętrowy - Historia wersji

AndrzejSalwicki o 08:21, 5 cze 2024

2024-06-05T08:21:41Z

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2024-06-04T08:47:05Z

@@ Linia 8: / Linia 8: @@
 <math>  {\dfrac{\dfrac{\dfrac{2^{k_{r}}-1}{3}\cdot {2^{k_{r-1}}-1}}{3}\cdot 2^{k_{r-2}}-1 }{3} \cdot 2^{k_{r-3}} =n}?</math><br />
 lub <br />
-<math>  n= \dfrac{\Biggl(\dfrac{\Biggl[\begin{array}{c}A \\3  \end{array}}{\dots}\biggr] \Biggr)}{3}\cdot 2^{k_0} </math><br />
+<math>  n= \dfrac{\Biggl(\dfrac{\Biggl[\begin{array}{c}A \\3  \end{array}\Biggr]}{\dots} \Biggr)\cdot 2^{k_1}}{3}\cdot 2^{k_0} </math><br />
 <math> n=  \dfrac{\Biggl (\dfrac{\Biggl[\dfrac{ (\dfrac{2^{k_x}-1}{3})\cdot 2^{k_{x-1}}-1}{\genfrac{}{}{0pt}{0}{3}{\mathbf{\dots}}}\cdot 2^{k_{x-2}-1\Biggr ]}{3} \Biggr)\cdot 2^{k_1}-1  }
 {3}\cdot 2^{k_0}    </math>
 Wróć na stronę [[Obliczenia utemperowane]]

@@ Linia 8: / Linia 8: @@
 <math>  {\dfrac{\dfrac{\dfrac{2^{k_{r}}-1}{3}\cdot {2^{k_{r-1}}-1}}{3}\cdot 2^{k_{r-2}}-1 }{3} \cdot 2^{k_{r-3}} =n}?</math><br />
 lub <br />
-<math>  n= \dfrac{\Biggl(\dfrac{\begin{array}{c}A \\3  \end{array}}{3} \Biggr)}{3}\cdot 2^{k_0} </math><br />
+<math>  n= \dfrac{\Biggl(\dfrac{\Biggl[\begin{array}{c}A \\3  \end{array}}{\dots}\biggr] \Biggr)}{3}\cdot 2^{k_0} </math><br />
 <math> n=  \dfrac{\Biggl (\dfrac{\Biggl[\dfrac{ (\dfrac{2^{k_x}-1}{3})\cdot 2^{k_{x-1}}-1}{\genfrac{}{}{0pt}{0}{3}{\mathbf{\dots}}}\cdot 2^{k_{x-2}-1\Biggr ]}{3} \Biggr)\cdot 2^{k_1}-1  }
 {3}\cdot 2^{k_0}    </math>
 Wróć na stronę [[Obliczenia utemperowane]]

@@ Linia 8: / Linia 8: @@
 <math>  {\dfrac{\dfrac{\dfrac{2^{k_{r}}-1}{3}\cdot {2^{k_{r-1}}-1}}{3}\cdot 2^{k_{r-2}}-1 }{3} \cdot 2^{k_{r-3}} =n}?</math><br />
 lub <br />
-<math>  n= \dfrac{\Biggl(\dfrac{\\begin{array}{c}A \\3  \end{array}}{3} \Biggr)}{3}\cdot 2^{k_0} </math><br />
+<math>  n= \dfrac{\Biggl(\dfrac{\begin{array}{c}A \\3  \end{array}}{3} \Biggr)}{3}\cdot 2^{k_0} </math><br />
 <math> n=  \dfrac{\Biggl (\dfrac{\Biggl[\dfrac{ (\dfrac{2^{k_x}-1}{3})\cdot 2^{k_{x-1}}-1}{\genfrac{}{}{0pt}{0}{3}{\mathbf{\dots}}}\cdot 2^{k_{x-2}-1\Biggr ]}{3} \Biggr)\cdot 2^{k_1}-1  }
 {3}\cdot 2^{k_0}    </math>
 Wróć na stronę [[Obliczenia utemperowane]]

@@ Linia 8: / Linia 8: @@
 <math>  {\dfrac{\dfrac{\dfrac{2^{k_{r}}-1}{3}\cdot {2^{k_{r-1}}-1}}{3}\cdot 2^{k_{r-2}}-1 }{3} \cdot 2^{k_{r-3}} =n}?</math><br />
 lub <br />
 <math> n=  \dfrac{\Biggl (\dfrac{\Biggl[\dfrac{ (\dfrac{2^{k_x}-1}{3})\cdot 2^{k_{x-1}}-1}{\genfrac{}{}{0pt}{0}{3}{\mathbf{\dots}}}\cdot 2^{k_{x-2}-1\Biggr ]}{3} \Biggr)\cdot 2^{k_1}-1  }
 {3}\cdot 2^{k_0}    </math>
 Wróć na stronę [[Obliczenia utemperowane]]

@@ Linia 8: / Linia 8: @@
 <math>  {\dfrac{\dfrac{\dfrac{2^{k_{r}}-1}{3}\cdot {2^{k_{r-1}}-1}}{3}\cdot 2^{k_{r-2}}-1 }{3} \cdot 2^{k_{r-3}} =n}?</math><br />
 lub <br />
-<math> \color{red}n\color{black}=  \dfrac{\Biggl (\dfrac{\Biggl[\dfrac{ (\dfrac{2^{\color{blue}k_x}-1}{3})\cdot 2^{\color{blue}k_{x-1}\color{black}}-1}{\genfrac{}{}{0pt}{0}{3}{\mathbf{\dots}}}\cdot 2^{\color{blue}k_{x-2}\color{blue}}-1\Biggr ]}{3} \Biggr)\cdot 2^{\color{blue}k_1}\color{black}-1  }
+<math> n=  \dfrac{\Biggl (\dfrac{\Biggl[\dfrac{ (\dfrac{2^{k_x}-1}{3})\cdot 2^{k_{x-1}}-1}{\genfrac{}{}{0pt}{0}{3}{\mathbf{\dots}}}\cdot 2^{k_{x-2}-1\Biggr ]}{3} \Biggr)\cdot 2^{k_1}-1  }
-{3}\cdot 2^{\color{blue}k_0\color{black}}    </math>
+{3}\cdot 2^{k_0}    </math>
 Wróć na stronę [[Obliczenia utemperowane]]

@@ Linia 2: / Linia 2: @@
 <math> 2^{k_{r}}  =n?</math><br />
 lub<br />
-<math> \dfrac{2^{k_{r}}-1}{3}\cdot {2^{k_{r-1}}  =n?</math><br />
+<math> \dfrac{2^{k_{r}}-1}{3}\cdot 2^{k_{r-1}}  =n?</math><br />
 lub<br />
-<math>\dfrac{\dfrac{2^{k_{x}}-1}{3}  \cdot {2^{k_{x-1}}-1}}{3}  =n?</math><br />
+<math>\dfrac{\dfrac{2^{k_{r}}-1}{3}  \cdot {2^{k_{r-1}}-1}}{3} \cdot 2^{k_{r-2}} =n?</math><br />
 lub <br />
 <math>  {\dfrac{\dfrac{\dfrac{2^{k_{x}}-1}{3}\cdot {2^{k_{x-1}}-1}}{3}\cdot 2^{k_{x-2}}-1 }{3}  =n}?</math><br />

@@ Linia 1: / Linia 1: @@
 Czy każdą liczbę naturalną można przedstawić jako ułamek piętrowy następującej postaci?<br />
+<math> 2^{k_{x}}  =n?</math><br />
 <math> \dfrac{2^{k_{x}}-1}{3}  =n?</math><br />
 lub<br />