tính: \(3,5.\frac{4}{49}-\left[2,\left(4\right).2,\left(45\right)\right]:\left(-8,4\right)\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(10,\left(3\right)+0,\left(4\right)-8,\left(6\right)\)
\(=\frac{31}{3}+\frac{4}{9}-\frac{26}{3}\)
\(=\left(\frac{31}{3}-\frac{26}{3}\right)+\frac{4}{9}=\frac{5}{3}+\frac{4}{9}=\frac{15}{9}+\frac{4}{9}=\frac{19}{9}\)
b) \(\left[12,\left(1\right)-2,3\left(6\right)\right]:4,\left(21\right)\)
\(=\left[\frac{109}{9}-\frac{71}{30}\right]:\frac{139}{33}\)
\(=-\frac{52}{45}:\frac{139}{33}=-\frac{52}{45}\cdot\frac{33}{139}=-\frac{572}{2085}\)(số xấu quá)
c) \(3\frac{1}{2}\cdot\frac{4}{49}-\left[2,\left(4\right)\cdot2\frac{5}{11}\right]:\frac{-42}{53}\)
\(=\frac{7}{2}\cdot\frac{4}{49}-\left[\frac{22}{9}\cdot\frac{27}{11}\right]\cdot\frac{-53}{42}\)
\(=\frac{2}{7}-6\cdot\left(-\frac{53}{42}\right)=\frac{2}{7}-\left(-\frac{53}{7}\right)=\frac{2}{7}+\frac{53}{7}=\frac{55}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Xét số hạng tổng quát:
\(k^4+\frac{1}{4}=\left(k^4+2\cdot\frac{1}{2}\cdot k^2+\frac{1}{4}\right)-k^2\)
= \(\left(k^2+\frac{1}{2}\right)^2-k^2\)= \(\left(k^2-k+\frac{1}{2}\right)\left(k^2+k+\frac{1}{2}\right)\)
Thay k từ 1 đến 2014 , ta được
M=
\(\frac{\left(2+\frac{1}{2}\right)\left(6+\frac{1.}{2}\right)...\left(4054182+\frac{1}{2}\right)\left(4058210+\frac{1}{2}\right)}{\frac{1}{2}\cdot\left(2+\frac{1}{2}\right)...\left(4050156+\frac{1}{2}\right)\left(4054182+\frac{1}{2}\right)}\)=\(\frac{4058210+\frac{1}{2}}{\frac{1}{2}}=8116421\)
ĐẶt A = 2,(4)
10A = 24 + 0,(4)
10 A - A = 24 + 0,(4) - 2,(4) = 24 + 0,(4) - 2 - 0,(4) = 22
9A = 22
=> A = 22/9
Đặt B = 2,(45)
100b = 245,(45)
100b- b = 245,(45) - 2,(45)
99b = 243
b = 243 /99
= 27/11
Thay vào bt ta có
\(3,5\cdot\frac{4}{49}-\left[\frac{22}{9}\cdot\frac{27}{11}\right]:\left(-8,4\right)\)
\(=\frac{7}{2}\cdot\frac{4}{49}-6:-\frac{42}{5}\)
\(=\frac{2}{7}-6\cdot-\frac{5}{42}=\frac{2}{7}+\frac{5}{7}=1\)