Question

помогите по Алгебре ​

Answer 1

Объяснение:

1)

$1.\ \frac{1}{2}+0,125-\frac{1}{6}=\frac{1}{2}+\frac{1}{8} -\frac{1}{6}=\frac{1*12+1*3-1*4}{24}=\frac{12+3-4}{24}=\frac{11}{24} .\\2.\ 6,4:\frac{80}{3}=6\frac{2}{5} *\frac{3}{80} =\frac{32}{5}*\frac{3}{80}=\frac{32*3}{5*80}=\frac{2*3}{5*5}=\frac{6}{25} .\\3.\ \frac{11}{24}*\frac{6}{25} =\frac{11*6}{24*25} =\frac{11}{4*25}=\frac{11}{100}=0,11.\\4.\ 0,11+\frac{1}{8} =0,11+0,125=0,235.$

2)

$1.\ 3\frac{1}{3}+2,5=\frac{10}{3}+2\frac{1}{2}= \frac{10}{3}+\frac{5}{2}=\frac{10*2+5*3}{6}=\frac{20+15}{6} =\frac{35}{6} .\\2.\ 2,5-1\frac{1}{3} =2\frac{1}{2} -\frac{4}{3}=\frac{5}{2}-\frac{4}{3}=\frac{5*3-4*2}{6} =\frac{15-8}{6}=\frac{7}{6} .\\3.\ \frac{35}{6}:\frac{7}{6}=\frac{35}{6}*\frac{6}{7}=\frac{35*6}{6*7}=5.\\4.\ 4,6-2\frac{1}{3}=4\frac{3}{5} -\frac{7}{3} =\frac{23}{5} -\frac{7}{3}=\frac{23*3-7*5} {15}=\frac{69-35}{15}=\frac{34}{15}.$

$5.\ 4,6+2\frac{1}{3}=4\frac{3}{5}+\frac{7}{3}=\frac{23}{5}+\frac{7}{3} =\frac{69+35}{15}=\frac{104}{15} .\\6.\ \frac{34}{15}:\frac{104}{15}=\frac{34}{15}*\frac{15}{104}=\frac{34*15}{15*104} =\frac{17}{52} .\\7.\ 5*\frac{17}{52} =\frac{85}{52}.\\8.\ \frac{85}{52}*5,2= \frac{85}{52}*\frac{52}{10} =\frac{85*52}{52*10}=\frac{17}{2} =8,5.$

$9.\ \frac{1}{7} -0,125 =\frac{1}{7} -\frac{1}{8}=\frac{8-7}{56}=\frac{1}{56} .\\10.\ 0,05:\frac{1}{56}=\frac{1}{20}*\frac{56}{1} =\frac{56}{20} =\frac{14}{5}=2,8.\\11.\ 2,8+5,7=8,5.\\12.\ 8,5:8,5=1.$