Question

[Trần Hà Tiên]

\(\frac{7}{4x.}\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)=22\)

\(\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\right).x=1\)

Answer 1

[Nguyễn Thị Quỳnh]

a, \(\frac{7}{4x}\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)=22\)

\(\frac{7}{4x}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)=22\)

\(\frac{7}{4x}\left[33.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\right]=22\)

\(\frac{7}{4x}\left[33.\left(\frac{35}{420}+\frac{21}{420}+\frac{14}{420}+\frac{10}{420}\right)\right]=22\)

\(\frac{7}{4x}\left[33.\frac{4}{21}\right]=22\)

\(\frac{7}{4x}.\frac{44}{7}\)=22

\(\frac{11}{x}=22\)

x=11:22

x=\(\frac{1}{2}\)

b,\(\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\right).x=1\)

Đặt A\(=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)

Ta có :\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)

\(\Rightarrow4A=4.\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\right)\)

\(\Rightarrow4A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}=\frac{32}{64}+\frac{16}{64}+\frac{8}{64}+\frac{4}{64}+\frac{2}{64}+\frac{1}{64}\)

\(\Rightarrow4A=\frac{32+16+8+4+2+1}{64}=\frac{63}{64}\)

\(\Rightarrow A=\frac{63}{64}:4=\frac{63}{256}\)

\(\Rightarrow\frac{63}{256}.x=1\)

\(\Leftrightarrow x=1:\frac{63}{256}=\frac{256}{63}\)