Question

помогите пожалуйста сделать все задания
тому 50 баллов все задания НА ЛИСТОЧКЕ
я понимаю что я в 8 классе но это мне надо​

Answer 1

$\frac{22}{24}=\frac{11}{12};\\\\\frac{6}{24}=\frac{1}{4};\\\\\frac{15}{24}=\frac{5}{8};\\\\\frac{60}{90}=\frac{2*3*10}{3*3*10}=\frac{2}{3}.$

$\frac{3}{8}+\frac{1}{6}=\frac{3}{2*4}+\frac{1}{2*3}=\frac{3*3}{2*4*3}+\frac{1*4}{2*3*4}=\\\\=\frac{9}{24}+\frac{4}{24}=\frac{9+4}{24}=\frac{13}{24}.$

$0,5-\frac{4}{25}+\frac{3}{20}=\frac{5}{10}-\frac{4}{25}+\frac{3}{20}=\\\\=\frac{5}{2*5}-\frac{4}{5*5}+\frac{3}{2*2*5}=\frac{5*2*5}{2*5*2*5}-\frac{4*2*2}{5*5*2*2}+\frac{3*5}{2*2*5*5}=\\\\=\frac{50}{100}-\frac{16}{100}+\frac{15}{100}=\frac{50-16+15}{100}=\frac{49}{100}.$

$6-x=2\frac{3}{4}\\\\x=6-2\frac{3}{4}\\\\x=4-\frac{3}{4}\\\\x=3+\frac{4}{4}-\frac{3}{4}\\\\x=3\frac{1}{4}.$

$4,5-(x+2\frac{3}{8})=1\frac{1}{6}\\\\4\frac{1}{2}-x-2\frac{3}{8}=1\frac{1}{6}\\\\4\frac{1}{2}-1\frac{1}{6}-2\frac{3}{8}=x\\\\4+\frac{1}{2}-1-\frac{1}{6}-2-\frac{3}{8}=x\\\\1+\frac{1}{2}-\frac{1}{6}-\frac{3}{8}=x\\\\\frac{2}{2}+\frac{1}{2}-\frac{1}{6}-\frac{3}{8}=x\\\\\frac{3}{2}-\frac{1}{6}-\frac{3}{8}=x\\\\\frac{3}{2}-\frac{1}{2*3}-\frac{3}{2*2*2}=x\\\\\frac{3*2*2*3}{2*2*2*3}-\frac{1*2*2}{2*3*2*2}-\frac{3*3}{2*2*2*3}=x\\\\\frac{36}{24}-\frac{4}{24}-\frac{9}{24}=x$

$\frac{36-4-9}{24}=x\\\\\frac{23}{24}=x\\\\x=\frac{23}{24}.$

$\frac{1}{6}<x+1\frac{1}{3}<5\frac{1}{3}\\\\\frac{1}{6}-1\frac{1}{3}<x+1\frac{1}{3}-1\frac{1}{3}<5\frac{1}{3}-1\frac{1}{3}\\\\\frac{1}{6}-1\frac{1}{3}<x<5\frac{1}{3}-1\frac{1}{3}\\\\\frac{1}{6}-1-\frac{1}{3}<x<4\\\\\frac{1}{6}-1-\frac{1*2}{3*2}<x<4\\\\\frac{1}{6}-1-\frac{2}{6}<x<4\\\\\frac{1}{6}-\frac{6}{6}-\frac{2}{6}<x<4\\\\\frac{1-6-2}{6}<x<4\\\\\frac{-7}{6}<x<4\\\\\frac{-6-1}{6}<x<4\\\\\frac{-6}{6}-\frac{1}{6}<x<4\\\\-1\frac{1}{6}<x<4.$