Tính:
A=\(\dfrac{1}{3\cdot8}+\dfrac{1}{6\cdot12}+\dfrac{1}{9\cdot16}+...+\dfrac{1}{1512\cdot2020}\)
Những câu hỏi liên quan
\(\dfrac{1\cdot2\cdot4+2\cdot4\cdot8+4\cdot8\cdot16+8\cdot16\cdot32}{1\cdot3\cdot4+2\cdot6\cdot8+4\cdot12\cdot16+8\cdot24\cdot32}\). dấu . là nhân .mọi ng cho m các phép tính cụ thể ạ.
\(\dfrac{1.2.4+2.4.8+4.8.16+8.16.32}{1.3.4+2.6.8+4.12.16+8.24.32}\)
\(=\dfrac{8.\left(1+8+4.16+16.32\right)}{12.\left(1+8+4.16+16.32\right)}\)
\(=\dfrac{8}{12}=\dfrac{2}{3}\)
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Bài 1 : Tính giá trị của phân số :
\(\dfrac{2\cdot4+2\cdot4\cdot8+4\cdot8\cdot16+8\cdot16\cdot32}{3\cdot4+2\cdot6\cdot8+4\cdot12\cdot16+8\cdot24\cdot32}\)
\(=\dfrac{8+8\cdot8+8\cdot64+8\cdot512}{12+12\cdot8+12\cdot64+12\cdot512}=\dfrac{8}{12}=\dfrac{2}{3}\)
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Rút gọn rồi so sánh
A=\(\dfrac{8056}{2012\cdot16-1982}\)
B=\(\dfrac{1\cdot2\cdot6+2\cdot4\cdot12+4\cdot8\cdot24+7\cdot14\cdot42}{1\cdot6\cdot9+2\cdot12\cdot18+4\cdot24\cdot36+7\cdot42\cdot63}\)
So sánh A và B
\(A=\frac{8056}{2012.16-1982}\)= \(\frac{2014.4}{2012.15+2012-1982}\)=\(\frac{2014.4}{2012.15+30}\)=\(\frac{2014.4}{2012.15+2.15}\)=\(\frac{2014.4}{15.\left(2012+2\right)}=\frac{2014.4}{15.2014}=\frac{4}{15}\)
B = \(\frac{1.2.6+2.4.12+4.8.24+7.14.42}{1.6.9+2.12.18+4.24.36+7.42.63}\)
= \(\frac{1.2.3.2+2.2.2.12+4.4.2.24+7.7.2.42}{1.2.3.9+2.12.2.9+4.24.4.9+7.42.7.9}\)
= \(\frac{2\left(1.2.3+2.2.12+4.4.24+7.7.42\right)}{9\left(1.2.3+2.2.12+4.4.24+7.7.42\right)}\)
= \(\frac{2}{9}\)
Ta có: \(\frac{4}{15}=\frac{4.3}{15.3}=\frac{12}{45};\frac{2}{9}=\frac{2.5}{9.5}=\frac{10}{45}\)
Vì \(\frac{12}{45}>\frac{10}{45}\Rightarrow\frac{4}{15}>\frac{2}{9}\Rightarrow A>B\)
Vậy A > B
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Rút gọn rồi so sánh
A=\(\dfrac{8056}{2012\cdot16-1982}\)
B=\(\dfrac{1\cdot2\cdot6+2\cdot4\cdot12+4\cdot8\cdot24+7\cdot14\cdot42}{1\cdot6\cdot9+2\cdot12\cdot18+4\cdot24\cdot36+7\cdot42\cdot63}\)
So sánh A và B
\(A=\dfrac{8056}{2012.16-1982}\)
\(A=\dfrac{8056}{32192-1982}\)
\(A=\dfrac{8056}{30210}=\dfrac{12}{45}\)
\(B=\dfrac{1.2.6+2.4.12+4.8.24+7.14.42}{1.6.9+2.12.18+4.24.36+7.42.63}\)
\(B=\dfrac{12+96+768+4116}{54+432+3456+18522}\)
\(B=\dfrac{4992}{22464}=\dfrac{10}{45}\)
Vậy: \(\dfrac{12}{45}>\dfrac{10}{45}\Rightarrow A>B\)
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Hãy tính các tổng sau:
a)dfrac{1}{1cdot3}+dfrac{1}{3cdot5}+dfrac{1}{5cdot7}+dfrac{1}{7cdot9}+dfrac{1}{9cdot11}
b)dfrac{1}{4cdot7}+dfrac{1}{7cdot10}+dfrac{1}{10cdot13}+dfrac{1}{13cdot16}
c)dfrac{1}{2cdot7}+dfrac{1}{7cdot12}+dfrac{1}{12cdot17}+...
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Hãy tính các tổng sau:
a)\(\dfrac{1}{1\cdot3}\)+\(\dfrac{1}{3\cdot5}\)+\(\dfrac{1}{5\cdot7}\)+\(\dfrac{1}{7\cdot9}\)+\(\dfrac{1}{9\cdot11}\)=
b)\(\dfrac{1}{4\cdot7}\)+\(\dfrac{1}{7\cdot10}\)+\(\dfrac{1}{10\cdot13}\)+\(\dfrac{1}{13\cdot16}\)=
c)\(\dfrac{1}{2\cdot7}\)+\(\dfrac{1}{7\cdot12}\)+\(\dfrac{1}{12\cdot17}\)+...=
1100444-88888=
a)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
\(=\frac{1}{2}.\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+\frac{11-9}{9.11}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{11}\right)\)
\(\frac{10}{22}\)
\(\dfrac{1\cdot2\cdot3+2\cdot4+6+4\cdot8\cdot12}{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20}\)
\(\dfrac{1\cdot2\cdot3+2\cdot4\cdot6+4\cdot8\cdot12}{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20}\\ =\dfrac{1\cdot2\cdot3+2\cdot1\cdot2\cdot2\cdot2\cdot3+4\cdot1\cdot4\cdot2\cdot4\cdot3}{1\cdot3\cdot5+2\cdot1\cdot2\cdot3\cdot2\cdot5+4\cdot1\cdot4\cdot3\cdot4\cdot5}\\ =\dfrac{1\cdot2\cdot3\cdot\left(1+2^3+4^3\right)}{1\cdot3\cdot5\cdot\left(1+2^3+4^3\right)}\\ =\dfrac{1\cdot2\cdot3}{1\cdot3\cdot5}\\ =\dfrac{6}{15}\)
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Bạn ghi đề sai rồi, mình sửa lại đề ở phần (*) rồi nhé! 
Ta có: \(\dfrac{1.2.3+2.4.6+4.8.12}{1.3.5+2.6.10+4.12.20}\) (*)
= \(\dfrac{1.2.3\left(1+2^3+4^3\right)}{1.3.5\left(1+2^3+4^3\right)}\) = \(\dfrac{1.2.3}{1.3.5}\) = \(\dfrac{2}{5}\)
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20 tháng 12 2021 lúc 19:20
tính :
\(\dfrac{5^2\cdot6^{11}\cdot16^2+6^2\cdot12^6\cdot15^2}{2\cdot6^{12}\cdot10^4-81^2\cdot960^3}\)
1. Thực hiện phép tính A3.dfrac{1}{1cdot2}- 5.dfrac{1}{2cdot3}+7.dfrac{1}{3cdot4}- ... +15dfrac{1}{7cdot8}-17dfrac{1}{8cdot9}
2.Tính tỉ số dfrac{A}{B} biết Adfrac{1}{1cdot300}+dfrac{1}{2cdot301}+dfrac{1}{3cdot302}+...+dfrac{1}{101cdot400} và Bdfrac{1}{1cdot102}+dfrac{1}{2cdot103}+dfrac{1}{3cdot104}+...+dfrac{1}{299cdot400}
Nhanh lên nhé, vội lắm rồi
Đọc tiếp
1. Thực hiện phép tính A=3.\(\dfrac{1}{1\cdot2}\)- 5.\(\dfrac{1}{2\cdot3}\)+7.\(\dfrac{1}{3\cdot4}\)- ... +15\(\dfrac{1}{7\cdot8}\)-17\(\dfrac{1}{8\cdot9}\)
2.Tính tỉ số \(\dfrac{A}{B}\) biết A=\(\dfrac{1}{1\cdot300}\)+\(\dfrac{1}{2\cdot301}\)+\(\dfrac{1}{3\cdot302}\)+...+\(\dfrac{1}{101\cdot400}\) và B=\(\dfrac{1}{1\cdot102}\)+\(\dfrac{1}{2\cdot103}\)+\(\dfrac{1}{3\cdot104}\)+...+\(\dfrac{1}{299\cdot400}\)
Nhanh lên nhé, vội lắm rồi
\(\dfrac{1}{2\cdot6}\)+\(\dfrac{1}{4\cdot9}\) +\(\dfrac{1}{6\cdot12}\) +....+\(\dfrac{1}{36\cdot57}\) +\(\dfrac{1}{38\cdot60}\) dấu * là dấu nhân . mọi ng cho mình các phép tính cụ thể với ạ.
\(=\dfrac{1}{2x1x3x2}+\dfrac{1}{2x2x3x3}+\dfrac{1}{2x3x3x4}+...+\dfrac{1}{2x18x3x19}+\dfrac{1}{2x19x3x20}=\)
\(=\dfrac{1}{2x3}x\left(\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+...+\dfrac{1}{18x19}+\dfrac{1}{19x20}\right)=\)
\(=\dfrac{1}{6}x\left(\dfrac{2-1}{1x2}+\dfrac{3-2}{2x3}+\dfrac{4-3}{3x4}+...+\dfrac{20-19}{19x20}\right)=\)
\(=\dfrac{1}{6}x\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)=\)
\(=\dfrac{1}{6}x\left(1-\dfrac{1}{20}\right)=\dfrac{1}{6}x\dfrac{19}{20}=\dfrac{19}{120}\)
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