Dịch sang tiếng việt rồi giải
1. fraction\(\frac{6}{35}\)can be written as the product of two fractions of which the numerator and the denominator are one-digit postitve intergers
For instance: \(\frac{6}{35}=\frac{2}{3}.\frac{5}{7}\)
Look for different writtings.
2. Puzzle: there are a pairs of fractions that when we multiply or add them together, we recive the same results
For instance: the pair of fractions \(\frac{7}{3}\)and\(\frac{7}{4}\)has:
\(\frac{7}{3}.\frac{7}{4}=\frac{7.7}{3.4}=\frac{49}{12}\)
\(\frac{7}{3}+\frac{7}{4}=\frac{7.4+7.3}{12}=\frac{49}{12}\)
We ask you a riddle to find another pair of fractions having that property
So sánh:
A=\(\frac{4}{7}+5+\frac{3}{7^2}+\frac{5}{7^3}+\frac{6}{7^4}\) và B=\(\frac{5}{7^4}+5+\frac{6}{7^2}+\frac{4}{7}+\frac{5}{7^3}\)
\(\frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}-\left(\frac{-5}{6}\right)-\frac{6}{7}-\frac{-7}{8}+\frac{6}{7}-\frac{5}{6}+\frac{4}{5}-\frac{3}{4}+\frac{2}{3}-\frac{1}{2}\)
Cho A= \(\frac{4+\frac{4}{2012}-\frac{4}{2013}+\frac{4}{2014}-\frac{4}{2015}}{\frac{7}{2014}-\frac{7}{2015}+\frac{7}{2012}-\frac{7}{2013}+7}\)
Và B= \(\frac{1+2+2^2+...+2^{2013}}{2^{2015}-2}\)
Tính A - B
p/S: LM ƠN GIÚP TỚ VS :
Cho C= \(4+\frac{1}{7^6}+\frac{3}{7}+\frac{4}{7^2}+\frac{-441}{7^6}+\frac{27}{7^5}\)
D=\(\frac{147}{7^3}+4+\frac{35}{7^7}+\frac{4}{7^2}+\frac{27}{7^5}+\frac{-9}{7^9}\)
So sánh C với D
Không quy đồng mẫu số, hãy so sánh:
\(C=\frac{3}{7^2}+\frac{4}{7^3}+\frac{5}{7^4}\) và \(D=\frac{3}{7^3}+\frac{5}{7^2}+\frac{4}{7^4}\)
Tính giá trị của các biểu thức sau một cách hợp lý:
a)A=\(\frac{3^{10}.11+3^{10}.5}{3^9.2^4+2^{10}.13}+\frac{2^{10}.13+2^{10}.65}{2^8.104}\)
b)B=\(\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{3+\frac{3}{13}+\frac{3}{169}+\frac{3}{91}}{7+\frac{7}{13}+\frac{7}{169}+\frac{7}{91}}\)
Tính
\(M=\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{9}-\frac{2}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{9}-\frac{4}{11}}\)
Tính
A = \(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\)
B = \(\frac{4}{5}+\frac{4}{5^2}-\frac{4}{5^3}+...+\frac{4}{5^{200}}\)