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a)
\(\Rightarrow A=\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{5\left(\frac{1}{11}-\frac{1}{13}-\frac{1}{17}\right)}+\frac{2\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}{7\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}\)
\(\Rightarrow A=\frac{1}{5}+\frac{2}{7}\)
\(\Rightarrow A=\frac{17}{35}\)
b)
\(\Rightarrow B=5\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+....+\frac{1}{56}-\frac{1}{61}\right)\)
\(\Rightarrow B=5\left(\frac{1}{11}-\frac{1}{61}\right)\)
\(\Rightarrow B=5.\frac{50}{671}=\frac{250}{671}\)
c)
\(\Rightarrow C=1-\left(\frac{1}{1.3}+\frac{1}{2.3}+\frac{1}{2.5}+....+\frac{1}{49.25}\right)\)
\(\Rightarrow C=1-2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{49.50}\right)\)
\(\Rightarrow C=1-2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{49}-\frac{1}{50}\right)\)
\(\Rightarrow C=1-1-\frac{1}{25}\)
\(\Rightarrow C=\frac{1}{25}\)
a) \(A=\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{\frac{5}{11}-\frac{5}{13}-\frac{5}{17}}+\frac{\frac{2}{3}-\frac{2}{9}-\frac{2}{27}+\frac{2}{81}}{\frac{7}{3}-\frac{7}{9}-\frac{7}{27}+\frac{7}{81}}\)
\(=\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{5\left(\frac{1}{11}-\frac{1}{13}-\frac{1}{17}\right)}+\frac{2\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}{7\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}\)
\(=\frac{1}{5}+\frac{2}{7}\)
\(=\frac{7}{35}+\frac{10}{35}\)
\(=\frac{17}{35}\)
Vậy \(A=\frac{17}{35}\)
b) \(B=\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}+\frac{5^2}{26.31}+...+\frac{5^2}{56.61}\)
\(=5.\left(\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}+...+\frac{5}{56.61}\right)\)
\(=5.\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}+...+\frac{1}{56}-\frac{1}{61}\right)\)
\(=5.\left(\frac{1}{11}-\frac{1}{61}\right)\)
\(=5.\left(\frac{61}{671}-\frac{11}{671}\right)\)
\(=5.\frac{50}{671}\)
\(=\frac{250}{671}\)
Vậy \(B=\frac{250}{671}\)
a) \(\frac{17}{9}-\frac{17}{9}:\left(\frac{7}{3}+\frac{1}{2}\right)\)
= \(\frac{17}{9}-\frac{17}{9}:\frac{17}{6}\)
= \(\frac{17}{9}-\frac{2}{3}\)
= \(\frac{11}{9}\)
b) \(\frac{4}{3}.\frac{2}{5}-\frac{3}{4}.\frac{2}{5}\)
= \(\frac{2}{5}.\left(\frac{4}{3}-\frac{3}{4}\right)\)
= \(\frac{2}{5}.\frac{7}{12}\)
= \(\frac{7}{30}\)
Mình lười làm quá, hay mình nói kết quả cho bn thôi nha
c) -6
d) 3
e) 3
g) 12
h) \(\frac{23}{18}\)
i) \(\frac{-69}{20}\)
k) \(\frac{-1}{2}\)
l) \(\frac{49}{5}\)
Phần 1)Đầu tiên bạn nhân B với 1 phần 4 rồi tính đến đoạn gần cuối sẽ ra 1/3 - 1/35 rồi quy đòng rồi tính sẽ ra kêt quả cuối là 32/105 nha
Mình lười lắm nên chỉ help 1 phần thui nha sr
Bài 1:
\(A=\frac{24.47-22}{24+47.23}.\frac{5+\frac{5}{7}+\frac{5}{11}-\frac{5}{13}+\frac{5}{1001}}{6+\frac{6}{7}+\frac{6}{11}-\frac{6}{13}+\frac{6}{1001}}\)\(=\frac{47.23+47-22}{24.47.23}.\frac{5\left(1+\frac{1}{7}+\frac{1}{11.}-\frac{1}{13}+\frac{1}{1001}\right)}{6\left(1+\frac{1}{7}+\frac{1}{11}-\frac{1}{13}+\frac{1}{1001}\right)}\)
\(=\frac{47.23+24}{24+47.23}.\frac{5}{6}\)
\(=1.\frac{5}{6}=\frac{5}{6}\)
Bài 2:
\(81^7-27^9-9^{13}=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}\)
\(=3^{28}-3^{27}-3^{26}\)
\(=3^{22}\left(3^6-3^5-3^4\right)\)
\(=3^{22}.405\) chia hết cho 405
=>đpcm
A = \(\frac{1}{2}-\frac{3}{4}+\frac{5}{6}-\frac{7}{12}\)
A = \(\left(-\frac{1}{4}\right)+\frac{5}{6}-\frac{7}{12}\)
A = \(\frac{7}{12}-\frac{7}{12}\)
A = \(0\).
Mình làm câu A thôi nhé.
Chúc bạn học tốt!