Strona 1 z 1
Oblicz
: 15 maja 2015, 14:54
autor: tobi505
\(a)\frac{3 \frac{1}{3} \cdot [5-(2 \frac{1}{2}- \frac{5}{2})] }{3 \frac{2}{3}+ \frac{3}{8}:(3 \frac{1}{2}- \frac{1}{8}) } \cdot 6 \frac{4}{5}\)
\(b)3+2,(9)\)
\(c)0,1(2) + 0,(5)\)
Re: Oblicz
: 15 maja 2015, 14:58
autor: eresh
tobi505 pisze:\(a)\frac{3 \frac{1}{3} \cdot [5-(2 \frac{1}{2}- \frac{5}{2})] }{3 \frac{2}{3}+ \frac{3}{8}:(3 \frac{1}{2}- \frac{1}{8}) } \cdot 6 \frac{4}{5}\)
\(\frac{3 \frac{1}{3} \cdot [5-(2 \frac{1}{2}- \frac{5}{2})] }{3 \frac{2}{3}+ \frac{3}{8}:(3 \frac{1}{2}- \frac{1}{8}) } \cdot 6 \frac{4}{5}=\\
=\frac{\frac{10}{3}[5-0]}{\frac{11}{3}+\frac{3}{8}:(3\frac{4}{8}-\frac{1}{8})}\cdot \frac{34}{5}=\\
=\frac{\frac{50}{3}}{\frac{11}{3}+\frac{3}{8}\cdot\frac{8}{27}}\cdot\frac{34}{5}=\\
=\frac{\frac{50}{3}}{\frac{33}{9}+\frac{1}{9}}\cdot\frac{34}{30}=\frac{50}{3}\cdot\frac{9}{34}\cdot\frac{34}{5}=30\)
Re: Oblicz
: 15 maja 2015, 15:01
autor: eresh
tobi505 pisze:
\(b)3+2,(9)\)
\(2,(9)=x\\
2,9999999...=x\;\; \bez \cdot 10\\
29,999999...=10x\\
29,99999999...-2,9999999....=10x-x\\
27=9x\\
x=\frac{27}{9}=3\\
3+2,(9)=3+3=6\)
Re: Oblicz
: 15 maja 2015, 15:06
autor: eresh
tobi505 pisze:
\(c)0,1(2) + 0,(5)\)
\(0,1222222222....=x\;\; \bez \cdot 10\\
1,222222222....=10x\\
1,22222222....-0,1222222222=10x-x\\
1,1=9x\\
\frac{11}{10}=9x\\
x=\frac{11}{90}\\
0,1(2)=\frac{11}{90}\)
\(0,55555555555....=x\\
5,5555555555....=10x\\
5=9x\\
x=\frac{5}{9}\\
0,(5)=\frac{5}{9}\)
\(0,1(2)+0,(5)=\frac{11}{90}+\frac{5}{9}=\frac{11}{90}+\frac{50}{90}=\frac{61}{90}\)