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Bài giải
a, \(\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}\right)\text{ : }\left(\frac{1}{6}+\frac{1}{10}-\frac{1}{15}\right)=\left(\frac{5}{30}+\frac{3}{30}+\frac{2}{30}\right)\text{ : }\left(\frac{5}{30}+\frac{3}{30}-\frac{2}{30}\right)=\frac{1}{3}-\frac{1}{5}=\frac{2}{15}\)
b, \(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)\text{ : }\left(\frac{1}{4}-\frac{1}{5}\right)=\left(\frac{60}{120}-\frac{40}{120}+\frac{30}{120}-\frac{24}{120}\right)\text{ : }\left(\frac{5}{20}-\frac{4}{20}\right)=\frac{13}{60}\text{ : }\frac{1}{20}=\frac{13}{3}\)
Ta có :
a, \(\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}\right)\text{ : }\left(\frac{1}{6}+\frac{1}{10}-\frac{1}{15}\right)=\left(\frac{5}{30}+\frac{3}{30}+\frac{2}{30}\right)\text{ : }\left(\frac{5}{30}+\frac{3}{30}-\frac{2}{30}\right)=\frac{1}{3}-\frac{1}{5}=\frac{2}{15}\)
b,
\(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)\text{ : }\left(\frac{1}{4}-\frac{1}{5}\right)=\left(\frac{60}{120}-\frac{40}{120}+\frac{30}{120}-\frac{24}{120}\right)\text{ : }\left(\frac{5}{20}-\frac{4}{20}\right)=\frac{13}{60}\text{ : }\frac{1}{20}=\frac{13}{3}\)
\(\dfrac{1}{3}\cdot\dfrac{1}{4}+\dfrac{7}{4}:3-\dfrac{2}{3}\)
\(=\dfrac{1}{12}+\dfrac{7}{12}-\dfrac{8}{12}\)
\(=\dfrac{0}{12}=0\)
\(=1+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)
\(=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}\)
\(=\frac{2187}{2187}+\frac{729}{2187}+\frac{81}{2187}+\frac{27}{2187}+\frac{9}{2187}+\frac{3}{2187}+\frac{1}{2187}\)
\(=\frac{2187+729+81+27+9+3+1}{2187}\)
\(=\frac{3037}{2187}\)
Đúng 100%
Bài 1:
a: x+1/2=5/6
nên x=5/6-1/2=1/3
b: x+1/4=3/4
nên x=3/4-1/4=2/4=1/2
c: x+3/10=1/2
nên x=1/2-3/10=5/10-3/10=1/5
d: x+1/4=3/8
nên x=3/8-1/4=3/8-2/8=1/8
\(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{729.3}\)
\(A=1+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)
=> \(3A=3+1+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^6}\)
=> \(3A-A=3+1+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^6}-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)\)
<=> \(2A=3-\frac{1}{3^7}=\frac{3^8-1}{3^7}\)
=> \(A=\frac{3^8-1}{2.3^7}\)
\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)......\left(1+\frac{1}{9}\right)\left(1+\frac{1}{10}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{10}{9}.\frac{11}{10}=\frac{3.4.5......10.11}{2.3.4.....9.10}\)
\(=\frac{11}{2}\)
Bằng 11/2 nhé