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rinjin
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Nguyễn Lê Phước Thịnh
8 tháng 10 2023 lúc 13:56

a: \(A=\dfrac{3^3\cdot2^3+3^3\cdot2^2+3^3\cdot1}{-13}=\dfrac{27\left(2^3+2^2+1\right)}{-13}=-27\)

b: \(B=\dfrac{2\cdot2^{12}\cdot3^6+2^{11}\cdot3^9}{2^3\cdot2^7\cdot3^7+2^7\cdot2^3\cdot5\cdot3^8}\)

\(=\dfrac{2^{13}\cdot3^6+2^{11}\cdot3^9}{2^{10}\cdot3^7+2^{10}\cdot5\cdot3^8}\)

\(=\dfrac{2^{11}\cdot3^6\left(2^2+3^3\right)}{2^{10}\cdot3^7\left(1+5\cdot3\right)}=\dfrac{2}{3}\cdot\dfrac{4+27}{1+15}=\dfrac{2}{3}\cdot\dfrac{31}{16}=\dfrac{31}{24}\)

c: \(C=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{5\cdot2^{35}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)

\(=\dfrac{2^{29}\cdot3^{18}\left(5\cdot2-3^2\right)}{2^{29}\cdot3^{18}\left(5\cdot2^6-7\right)}=\dfrac{10-9}{5\cdot64-7}=\dfrac{1}{313}\)

Jin
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Jin
1 tháng 10 2021 lúc 20:44

mn giúp e vs ạT^T

Nguyễn Hoàng Minh
1 tháng 10 2021 lúc 20:51

\(a,=\dfrac{3}{2}-\dfrac{5}{6}:\dfrac{1}{4}+\sqrt{\dfrac{1}{4}-\dfrac{1}{2}}=\dfrac{3}{2}-\dfrac{10}{3}+\sqrt{\dfrac{1}{2}}=-\dfrac{11}{6}+\dfrac{\sqrt{2}}{2}=\dfrac{-33+3\sqrt{2}}{6}\)

\(b,=-\dfrac{4}{3}\cdot\dfrac{9}{2}+\dfrac{13}{12}\cdot\left(-\dfrac{8}{13}\right)=6-\dfrac{2}{3}=\dfrac{16}{3}\\ c,=\dfrac{1}{4}-\left(-\dfrac{1}{6}:4-8\cdot\dfrac{1}{16}\right)=\dfrac{1}{4}-\left(-\dfrac{1}{24}-\dfrac{1}{2}\right)\\ =\dfrac{1}{4}-\dfrac{13}{24}=-\dfrac{7}{24}\\ d,=\dfrac{3^{11}\cdot5^{11}\cdot5^7\cdot3^4}{5^{18}\cdot3^{18}}=\dfrac{1}{3^3}=\dfrac{1}{27}\)

Nguyễn Gia Hân
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Nguyễn Kiến Hoàng
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Giang シ)
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Lấp La Lấp Lánh
19 tháng 9 2021 lúc 9:57

\(\dfrac{4}{3.5}+\dfrac{4}{5.7}+\dfrac{4}{7.9}+...+\dfrac{4}{97.95}\)

\(=2\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{95.97}\right)\)

\(=2\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)

\(=2\left(\dfrac{1}{3}-\dfrac{1}{97}\right)\)

\(=2.\dfrac{94}{291}=\dfrac{188}{291}\)

Nguyễn Hoàng Minh
19 tháng 9 2021 lúc 9:57

\(=2\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{95\cdot97}\right)\\ =2\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\\ =2\left(\dfrac{1}{3}-\dfrac{1}{97}\right)=2\cdot\dfrac{94}{291}=\dfrac{188}{291}\)

Lê Ngọc Anh
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Vô danh
20 tháng 3 2022 lúc 8:37

\(j,\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{53.55}=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{53}-\dfrac{1}{55}=\dfrac{1}{5}-\dfrac{1}{55}=\dfrac{11}{55}-\dfrac{1}{55}=\dfrac{10}{55}=\dfrac{2}{11}\\ k,\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}...\dfrac{99}{100}=\dfrac{1}{100}.\dfrac{2}{2}.\dfrac{3}{3}...\dfrac{99}{99}=\dfrac{1}{100}.1.1...1=\dfrac{1}{100}\)

Lê Ngọc Linh
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Đào Trí Bình
25 tháng 8 2023 lúc 10:12

232/301

Nguyễn Minh Dương
25 tháng 8 2023 lúc 10:12

Gợi ý cho bn: Tách chúng ra nhé, rồi rút gọn số có ở trên cả tử và mẫu.

Nguyễn Ngọc Anh Minh
25 tháng 8 2023 lúc 10:20

\(=\dfrac{2^{12}.3^{15}.7^{30}-31.2^{12}.3^{12}.3^2.7^{28}}{2^{11}.3^{14}.7^{14}.7^{15}+2^{12}.3^{15}.7^{30}}=\)

\(=\dfrac{2^{12}.3^{15}.7^{30}-31.2^{12}.3^{14}.7^{28}}{2^{11}.3^{14}.7^{29}+2^{12}.3^{15}.7^{30}}=\)

\(=\dfrac{2^{11}.3^{14}.7^{28}\left(2.3.7^2-31.2\right)}{2^{11}.3^{14}.7^{28}\left(7+2.3\right)}=\dfrac{2.3.7^2-31.2}{7+2.3}=\dfrac{232}{13}\)

Nguyễn Thị Trà
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Phạm Ngân Hà
17 tháng 12 2017 lúc 20:19

a) \(\dfrac{1}{2}.\dfrac{1}{-3}+\dfrac{1}{-3}.\dfrac{1}{4}+\dfrac{1}{4}.\dfrac{1}{-5}+\dfrac{1}{-5}.\dfrac{1}{6}\)

\(=\dfrac{1}{-3}\left(\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{1}{-5}\left(\dfrac{1}{4}+\dfrac{1}{6}\right)\)

\(=\dfrac{1}{-3}.\dfrac{3}{4}+\dfrac{1}{-5}.\dfrac{5}{12}\)

\(=\left(-\dfrac{1}{4}\right)+\left(-\dfrac{1}{12}\right)\)

\(=-\dfrac{1}{3}\)

b) \(A=\dfrac{81^4.3^{10}.27^5.3^{12}}{3^{18}.9^3.243^2}\)

\(=\dfrac{9^8.9^8.9^{13}.9^{10}}{9^{16}.9^3.9^3}\)

\(=\dfrac{9^{39}}{9^{22}}\)

\(=9^{17}\)

Nguyễn Thị Thu
20 tháng 12 2017 lúc 21:38

\(A=\dfrac{81^4\cdot3^{10}\cdot27^5\cdot3^{12}}{3^{18}\cdot9^3\cdot243^2}=\dfrac{3^{16}\cdot3^{10}\cdot3^{15}\cdot3^{12}}{3^{18}\cdot3^6\cdot3^{10}}=\dfrac{3^{53}}{3^{34}}=3^{19}\)

Vậy A = 319

Ngân Hà làm đúng phần a) nhưng làm sai phần b) nên mk chỉ làm phần b) thôi

títtt
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a: \(A=\dfrac{25^6}{5^3}=\dfrac{\left(5^2\right)^6}{5^3}=\dfrac{5^{12}}{5^3}=5^9\)

b: \(B=32\cdot\left(\dfrac{3}{2}\right)^5=32\cdot\dfrac{3^5}{2^5}=32\cdot\dfrac{243}{32}=243\)

c: \(C=\left(\dfrac{1}{3}\right)^4\cdot3^{-3}=3^{-4}\cdot3^{-3}=3^{-4-3}=3^{-7}\)

d: \(D=4^{-2}\cdot\left(\dfrac{2}{5}\right)^5\cdot5^4\)

\(=\dfrac{1}{4^2}\cdot\dfrac{2^5}{5^5}\cdot5^4\)

\(=\dfrac{1}{16}\cdot\dfrac{32}{5}=\dfrac{2}{5}\)

e: \(E=9^{-5}:\left(\dfrac{5}{3}\right)^4\cdot25^2\)

\(=\dfrac{1}{9^5}:\dfrac{5^4}{3^4}\cdot\left(5^2\right)^2\)

\(=\dfrac{1}{3^{10}}\cdot\dfrac{3^4}{5^4}\cdot5^4=\dfrac{1}{3^6}\)

f: \(F=\left(\dfrac{5}{8}\right)^{-2}:4^2\)

\(=\left(1:\dfrac{5}{8}\right)^2:4^2\)

\(=\left(\dfrac{8}{5}\right)^2\cdot\dfrac{1}{16}=\dfrac{64}{25}\cdot\dfrac{1}{16}=\dfrac{4}{25}\)

g: \(G=\left(\dfrac{5}{3}\right)^3\cdot\left(\dfrac{9}{2}\right)^2:\left(\sqrt{3}\right)^4\)

\(=\dfrac{5^3}{3^3}\cdot\dfrac{9^2}{2^2}:9\)

\(=\dfrac{5^3\cdot3^4}{3^3\cdot2^2}\cdot\dfrac{1}{3^2}\)

\(=\dfrac{125}{2^2\cdot3}=\dfrac{125}{3\cdot4}=\dfrac{125}{12}\)

\(A=\dfrac{\left(5^2\right)^6}{5^3}=\dfrac{5^{12}}{5^3}=5^9\)

\(B=32.\left(\dfrac{3}{2}\right)^5=\dfrac{2^5.3^5}{2^5}=2^5\)

\(C=\left(\dfrac{1}{3}\right)^4.3^{-3}=\dfrac{1}{3^4.3^3}=\dfrac{1}{3^7}\)

\(D=4^{-2}.\left(\dfrac{2}{5}\right)^5.5^4=\dfrac{1}{\left(2^2\right)^2}.\dfrac{2^5}{5^5}.5^4=\dfrac{2}{5}\)

\(E=\dfrac{1}{9^5}.\dfrac{3^4}{5^4}.\left(5^2\right)^2=\dfrac{1}{3^{10}}.\dfrac{3^4}{5^4}.5^4=\dfrac{1}{3^6}\)

\(F=\dfrac{8^2}{5^2}:\left(2^2\right)^2=\dfrac{\left(2^3\right)^2}{5^2.2^4}=\dfrac{2^6}{5^2.2^4}=\dfrac{2^2}{5^2}\)

\(G=\dfrac{5^3}{3^3}.\dfrac{\left(3^2\right)^2}{2^2}:3^2=\dfrac{5^3}{3^3}.\dfrac{3^4}{2^2}.\dfrac{1}{3^2}=\dfrac{5^3}{3.2^2}\)