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a)\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}......\frac{99}{100}\)
\(=\frac{1.2.3.4.....99}{2.3.4.5.6.....100}\)
\(=\frac{1}{100}\)
b) Tương tự như câu a
\(\left(a\right)\frac{34-x}{30}=\frac{5}{6}\)
\(\frac{34-x}{30}=\frac{25}{30}\)
34 - x = 25
x = 34 - 25 = 9
\(\left(b\right)\frac{x+13}{34}=\frac{12}{17}\)
\(\frac{x+13}{34}=\frac{24}{34}\)
x + 13 = 24
x = 24 - 13 = 11
\(\left(c\right)\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{9}\right)+\left(x+\frac{1}{27}\right)+\left(x+\frac{1}{81}\right)=\frac{56}{81}\)
\(4x+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}=\frac{56}{81}\)
Đặt \(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)
Ta có : \(3A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\)
\(3A-A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}-\frac{1}{3}-\frac{1}{9}-\frac{1}{27}-\frac{1}{81}\)
\(2A=1-\frac{1}{81}=\frac{80}{81}\)
\(A=\frac{80}{81}\div2=\frac{40}{81}\)
\(\Rightarrow4x+\frac{40}{81}=\frac{56}{81}\)
\(4x=\frac{56}{81}-\frac{40}{81}\)
\(4x=\frac{16}{81}\)
\(x=\frac{16}{81}\div4=\frac{4}{81}\)
\(1\frac{1}{2}.1\frac{1}{3}.1\frac{1}{4}...1\frac{1}{9}\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{10}{9}\)
\(=\frac{10}{2}=5\)
DẤU . LÀ DẤU NHÂN NHA BẠN
\(1\frac{1}{2}\times1\frac{1}{3}\times1\frac{1}{4}\times1\frac{1}{5}\times1\frac{1}{6}\times1\frac{1}{7}\times1\frac{1}{8}\times1\frac{1}{9}\)
\(=\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times\frac{6}{5}\times\frac{7}{6}\times\frac{8}{7}\times\frac{9}{8}\times\frac{10}{9}\)
\(=\frac{3\times4\times5\times6\times7\times8\times9\times10}{2\times3\times4\times5\times6\times7\times8\times9}=\frac{10}{2}=5\)
~ Hok tốt ~
\(\frac{464}{466}x\frac{43556}{58343}=\frac{464x43556}{466x58343}=\frac{20209984}{27187838}=0,743346491\)