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a, A= \(5\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\)
\(A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(A=5\left(1-\dfrac{1}{100}\right)\)
\(A=5.\dfrac{99}{100}=\dfrac{99}{20}.\)
b, \(C=1.2.3+2.3.4+...+8.9.10\)
\(4C=1.2.3.4+2.3.4.\left(5-1\right)+...+8.9.10.\left(11-7\right)\)\(4C=1.2.3.4+2.3.4.5-1.2.3.4+...+8.9.10.11-7.8.9.10\)\(4C=8.9.10.11\)
\(C=\dfrac{8.9.10.11}{4}=1980.\)
c, https://hoc24.vn/hoi-dap/question/384591.html
Câu này bạn vào đây mình đã giải câu tương tự nhé.
\(1)A=\dfrac{5}{1.2}+\dfrac{5}{2.3}+...+\dfrac{5}{99.100}\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{100}\right)\)
\(\Leftrightarrow A=5\cdot\dfrac{99}{100}\)
\(\Leftrightarrow A=\dfrac{99}{20}\)
\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{27.28.29.30}\)
\(A=\frac{1}{4.6}+\frac{1}{10.12}+\frac{1}{18.20}+...+\frac{1}{810.812}\)
.......
~ Chúc học tốt ~
Ai ngang qua xin để lại 1 L - I - K - E
\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+.....+\frac{1}{27.28.29.30}\)
\(3A=3.\left(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+......+\frac{1}{27.28.29.30}\right)\)
\(3A=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+..........+\frac{3}{27.28.29.30}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+........+\frac{1}{27.28.29}-\frac{1}{28.29.30}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{28.29.30}\)
\(3A=\frac{1}{6}-\frac{1}{24360}\)
\(3A=\frac{1353}{8120}\)
\(A=\frac{1353}{8120}:3\)
\(A=\frac{451}{8120}\)
sách 6,7,8 có 2 bài này nè. mk k bt ghi ps nên mk ko gửi đc sorry nha. Hhh
a)\(A=\frac{10^{2014}+2016}{10^{2015}+2016}=>10A=\frac{10^{2015}+20160}{10^{2015}+2016}=1+\frac{18144}{10^{2015}+2016}\left(1\right)\)
\(B=\frac{10^{2015}+2016}{10^{2016}+2016}=>10B=\frac{10^{2016}+20160}{10^{2016}+2016}=1+\frac{18144}{10^{2016}+2106}\left(2\right)\)
từ 1 zà 2
=> 10A>10B
=>A>B
\(\dfrac{1\cdot2\cdot3+2\cdot4\cdot6+4\cdot8\cdot12}{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20}\\ =\dfrac{1\cdot2\cdot3+2\cdot1\cdot2\cdot2\cdot2\cdot3+4\cdot1\cdot4\cdot2\cdot4\cdot3}{1\cdot3\cdot5+2\cdot1\cdot2\cdot3\cdot2\cdot5+4\cdot1\cdot4\cdot3\cdot4\cdot5}\\ =\dfrac{1\cdot2\cdot3\cdot\left(1+2^3+4^3\right)}{1\cdot3\cdot5\cdot\left(1+2^3+4^3\right)}\\ =\dfrac{1\cdot2\cdot3}{1\cdot3\cdot5}\\ =\dfrac{6}{15}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{14.15.16}\)
\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{14.15.16}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{14.15}-\frac{1}{15.16}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{15.16}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{240}\right)\)
\(=\frac{1}{2}.\frac{119}{240}\)
\(=\frac{119}{480}\)
Bài làm:
Ta có:\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{14.15.16}\)
\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{14.15.16}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{14.15}-\frac{1}{15.16}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{15.16}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{240}\right)\)
\(=\frac{1}{2}.\frac{119}{240}=\frac{119}{480}\)
A = \(\frac{1}{1.2.3}\) + \(\frac{1}{2.3.4}\) + \(\frac{1}{3.4.5}\) + ...+ \(\frac{1}{10.11.12}\)
A = \(\frac13\).(\(\frac{3}{1.2.3}\) + \(\frac{3}{2.3.4}\) + \(\frac{3}{3.4.5}\) + ..+\(\frac{3}{10.11.12}\))
A = \(\frac13\).(\(\frac{1}{1.2}\) - \(\frac{1}{2.3}\) +\(\frac{1}{2.3}-\frac{1}{3.4}\) +..+\(\frac{1}{10.11}-\frac{1}{11.12}\))
A = \(\frac13\).(\(\frac{1}{1.2}\) - \(\frac{1}{11.12}\))
A = \(\frac13\).(\(\frac12\) - \(\frac{1}{132}\))
A = \(\frac13\).\(\frac{65}{132}\)
A = \(\frac{65}{396}\)
H=\(\frac{1\cdot2\cdot3+2\cdot4\cdot6+3\cdot6\cdot9+5\cdot10\cdot15}{1\cdot3\cdot6+2\cdot6\cdot12+3\cdot9\cdot18+5\cdot15\cdot30}=\frac{1.2.3+2^3.\left(1.2.3\right)+3^3.\left(1.2.3\right)+5^3.\left(1.2.3\right)}{1.3.6+2^3.\left(1.3.6\right)+3^3.\left(1.3.6\right)+5^3.\left(1.3.6\right)}=\frac{1.2.3.\left(1+2^3+3^3+5^3\right)}{1.3.6.\left(1+2^3+3^3+5^3\right)}=\frac{2}{6}=\frac{1}{3}\)
📌 Bài toán:
\(S = \frac{1}{1 \cdot 2 \cdot 3 \cdot 4} + \frac{1}{2 \cdot 3 \cdot 4 \cdot 5} + \hdots + \frac{1}{7 \cdot 8 \cdot 9 \cdot 10}\)
💡 Mẹo dễ hiểu:
Ta biến mỗi phân số thành hiệu của 2 phân số nhỏ hơn:
\(\frac{1}{1 \cdot 2 \cdot 3 \cdot 4} = \frac{1}{3} \left(\right. \frac{1}{1 \cdot 2 \cdot 3} - \frac{1}{2 \cdot 3 \cdot 4} \left.\right)\)
Tương tự:
\(\frac{1}{2 \cdot 3 \cdot 4 \cdot 5} = \frac{1}{3} \left(\right. \frac{1}{2 \cdot 3 \cdot 4} - \frac{1}{3 \cdot 4 \cdot 5} \left.\right)\)
👉 Cứ thế tiếp tục...
🔥 Khi cộng lại:
Các số ở giữa bị triệt tiêu hết (giống kiểu gạch đi):
\(S = \frac{1}{3} \left(\right. \frac{1}{1 \cdot 2 \cdot 3} - \frac{1}{8 \cdot 9 \cdot 10} \left.\right)\)
✏️ Tính:
\(\frac{1}{1 \cdot 2 \cdot 3} = \frac{1}{6}\) \(\frac{1}{8 \cdot 9 \cdot 10} = \frac{1}{720}\) \(S = \frac{1}{3} \left(\right. \frac{1}{6} - \frac{1}{720} \left.\right)\)
Quy đồng:
\(\frac{1}{6} = \frac{120}{720}\) \(\Rightarrow \frac{120 - 1}{720} = \frac{119}{720}\) \(S = \frac{119}{2160}\)
✅ Đáp án: hơi khó hiểu 1 chút mong bạn thông cảm
\(\boxed{\frac{119}{2160}}\)
S = \(\frac{1}{1.2.3.4}\) + \(\frac{1}{2.3.4.5}\) + \(\frac{1}{3.4.5.6}\) + ...+\(\frac{1}{7.8.9.10}\)
S = \(\frac13\).(\(\frac{3}{1.2.3.4}\) + \(\frac{3}{2.3.4.5}\) + \(\frac{3}{3.4.5.6}\) +...+\(\frac{3}{7.8.9.10}\))
S = \(\frac13\).(\(\frac{1}{1.2.3}-\frac{1}{2.3.4}\) + \(\frac{1}{2.3.4}-\frac{1}{3.4.5}\) +...+\(\frac{1}{7.8.9}-\frac{1}{8.9.10}\))
S = \(\frac13\).(\(\frac{1}{1.2.3}\) - ...- \(\frac{1}{8.9.10}\))
S = \(\frac13\).(\(\frac16\) - \(\frac{1}{720}\))
S = \(\frac13\).\(\frac{119}{720}\)
S = \(\frac{119}{2160}\)