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cái đầu =\(\frac{127}{128}\)vì:
1/2+1/4=3/4 mà 3/4 =1-1/4
1/2+1/4+1/8=7/8 mà 7/8=1-1/8
ta suy ra cách làm: Tổng dãy phân số trên bằng 1 trừ cho phân số cuối
=> Tổng dãy trên =1-1/128= 127/128
Bài giải
\(1\frac{1}{3}\text{ x }2\frac{1}{4}\text{ x }\text{ }3\frac{1}{5}\text{ x }4\frac{1}{6}\text{ x }5\frac{1}{7}\text{ x }6\frac{1}{8}\text{ x }7\frac{1}{9}\text{ x }8\frac{1}{10}\)
\(=\frac{4}{3}\text{ x }\frac{9}{4}\text{ x }\frac{16}{5}\text{ x }\frac{25}{6}\text{ x }\frac{36}{7}\text{ x }\frac{49}{8}\text{ x }\frac{64}{9}\text{ x }\frac{81}{10}\)
\(=\frac{4\text{ x }9\text{ x }16\text{ x }25\text{ x }36\text{ x }49\text{ x }64\text{ x }81}{3\text{ x }4\text{ x }5\text{ x }6\text{ x }7\text{ x }8\text{ x }9}\)
\(=\frac{4\text{ x }3^2\text{ x }4^2\text{ x }5^2\text{ x }6^2\text{ x }7^2\text{ x }8^2\text{ x }9^2}{3\text{ x }4\text{ x }5\text{ x }6\text{ x }7\text{ x }8\text{ x }9}\) ( Nếu bạn chưa học lũy thừa thì bạn không cần viết bước này )
\(=4\text{ x }3\text{ x }4\text{ x }5\text{ x }6\text{ x }7\text{ x }8\text{ x }9\)
\(=725760\)
Bài giải
\(1\frac{1}{3}\text{ x }2\frac{1}{4}\text{ x }\text{ }3\frac{1}{5}\text{ x }4\frac{1}{6}\text{ x }5\frac{1}{7}\text{ x }6\frac{1}{8}\text{ x }7\frac{1}{9}\text{ x }8\frac{1}{10}\)
\(=\frac{4}{3}\text{ x }\frac{9}{4}\text{ x }\frac{16}{5}\text{ x }\frac{25}{6}\text{ x }\frac{36}{7}\text{ x }\frac{49}{8}\text{ x }\frac{64}{9}\text{ x }\frac{81}{10}\)
\(=\frac{4\text{ x }9\text{ x }16\text{ x }25\text{ x }36\text{ x }49\text{ x }64\text{ x }81}{3\text{ x }4\text{ x }5\text{ x }6\text{ x }7\text{ x }8\text{ x }9}\)
\(=\frac{4\text{ x }3^2\text{ x }4^2\text{ x }5^2\text{ x }6^2\text{ x }7^2\text{ x }8^2\text{ x }9^2}{3\text{ x }4\text{ x }5\text{ x }6\text{ x }7\text{ x }8\text{ x }9}\) ( Nếu bạn chưa học lũy thừa thì bạn không cần viết bước này )
\(=4\text{ x }3\text{ x }4\text{ x }5\text{ x }6\text{ x }7\text{ x }8\text{ x }9\)
\(=725760\)
a= (\(\frac{2}{5}\)+\(\frac{2}{9}\)+\(\frac{2}{11}\)\(\times\)\(\frac{5}{7}\)\(+\frac{7}{9}\)\(+\frac{7}{11}\)\()\)
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 :\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(\Rightarrow4A=4.\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\right)\)
\(\Rightarrow4A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}=\frac{32}{64}+\frac{16}{64}+\frac{8}{64}+\frac{4}{64}+\frac{2}{64}+\frac{1}{64}\)
\(\Rightarrow4A=\frac{32+16+8+4+2+1}{64}=\frac{63}{64}\)
\(\Rightarrow A=\frac{63}{64}:4=\frac{63}{256}\)
\(\Rightarrow\frac{63}{256}.x=1\)
\(\Leftrightarrow x=1:\frac{63}{256}=\frac{256}{63}\)
1. 3/7
2. 1 - 1/2 - 1/4 - 1/8 - 1/16
=16/16 - 8/16 - 4/16 - 2/16 - 1/16
= 1/16
sorry you,offline today i k 5 times. tomorrow his appointment k offline ! thank kiu (>o<)
quy đồng lên
\(\frac{40}{70}< \frac{x}{70}< \frac{50}{70}\)
x thuộc {41;42;43;44;45;46;47;48;49}
\(\frac{41}{70};\frac{42}{70}\frac{43}{70}\frac{44}{70}\frac{45}{70}\frac{46}{70}\frac{47}{70}\frac{48}{70}\frac{49}{70}\)
A,\(\frac{7}{13}+\frac{1}{7}+\frac{18}{30}=\frac{49}{91}+\frac{13}{91}+\frac{18}{30}=\frac{62}{91}+\frac{18}{30}=\frac{1860}{2730}+\frac{1638}{2730}=\frac{583}{455}\)
B,\(\left(\frac{20}{21}+\frac{3}{14}\right)-\frac{2}{3}=\left(\frac{280}{294}+\frac{63}{294}\right)-\frac{2}{3}=\frac{7}{6}-\frac{2}{3}=\frac{21}{18}-\frac{12}{18}=\frac{9}{18}=\frac{1}{2}\)
C,\(=\frac{17}{5}+\frac{31}{7}=\frac{119}{35}+\frac{155}{35}=\frac{234}{35}\)
=7(1/2 +1/4 +1/8+1/16+1/32+1/64+1/128)
Xét 1/2+1/4+1/8+1/16+1/32+1/64+1/128
= 1+1/2+1/4+1/8+1/16+1/32+1/64 - 1/2-1/4-1/8-1/16-1/32-1/64-1/128
=1-1/128
=127/128
Vậy tổng ban đầu là:7 * 127/128=889/128