Chứng minh
\(\dfrac{12}{1\cdot4\cdot7}+\dfrac{12}{4\cdot7\cdot10}+\dfrac{12}{7\cdot10\cdot13}+...+\dfrac{12}{54\cdot57\cdot60}< \dfrac{1}{2}\)
Những câu hỏi liên quan
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}+...
Đọc tiếp
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)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
\(=\frac{1}{2}.\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+\frac{11-9}{9.11}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{11}\right)\)
\(\frac{10}{22}\)
\(\dfrac{2}{1\cdot4}+\dfrac{2}{4\cdot7}+\dfrac{2}{7\cdot10}+\dfrac{2}{49\cdot50}\)
Đăt :
\(A=\dfrac{2}{1.4}+\dfrac{2}{4.7}+.........+\dfrac{2}{49.51}\)
\(\dfrac{3}{2}A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+..........+\dfrac{3}{49.51}\)
\(\dfrac{3}{2}A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+.........+\dfrac{1}{49}-\dfrac{1}{51}\)
\(\dfrac{3}{2}A=1-\dfrac{1}{51}\)
\(\dfrac{3}{2}A=\dfrac{50}{51}\)
\(\Rightarrow A=\dfrac{50}{51}:\dfrac{3}{2}=\dfrac{100}{153}\)
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Ta có công thức nha sau :
\(\dfrac{a}{b.c}=\dfrac{a}{c-b}.\left(\dfrac{1}{b}-\dfrac{1}{c}\right)\)
Ta gọi biểu thức phân số là A
Vậy \(\dfrac{2}{1.4}=\dfrac{2}{4-1}.\left(1-\dfrac{1}{4}\right)\)
\(\dfrac{2}{4.7}=\dfrac{2}{7-4}.\left(\dfrac{1}{4}-\dfrac{1}{7}\right)\)
\(\dfrac{2}{7.10}=\dfrac{2}{10-7}.\left(\dfrac{1}{7}-\dfrac{1}{10}\right)\)
Ta thấy 50 - 49 = 1 , không bằng những biểu thức kia bằng 3 nên ta tách những biểu thức đó ra.
A= \(\dfrac{2}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}\right)+\dfrac{2}{49.50}\)
\(A=\dfrac{2}{3}.\left(1-\dfrac{1}{10}\right)+2.\left(\dfrac{1}{49}-\dfrac{1}{50}\right)\)
\(A=\dfrac{18}{30}+\left(\dfrac{1}{1225}\right)=\dfrac{736}{1225}\)
mink chắc chắn, ủng hộ nha
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\(\dfrac{2}{1.4}+\dfrac{2}{4.7}+\dfrac{2}{7.10}+\dfrac{2}{49.50}\)
\(=\dfrac{2}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}\right)+\left(2.\dfrac{1}{49.50}\right)\)
\(=\dfrac{2}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}\right)+2\left(\dfrac{1}{49}-\dfrac{1}{50}\right)\)
\(=\dfrac{2}{3}\left(1-\dfrac{1}{10}\right)+2\left(\dfrac{50}{2450}-\dfrac{49}{2450}\right)\)
\(=\dfrac{2}{3}.\dfrac{9}{10}+2.\dfrac{1}{2450}\)
\(=\dfrac{3}{5}+\dfrac{1}{1225}\)
\(=\dfrac{735}{1225}+\dfrac{1}{1225}\)
\(=\dfrac{736}{1225}\)
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Tìm x
\(\dfrac{2}{1\cdot4}x+\dfrac{2}{4\cdot7}x+\dfrac{2}{7\cdot10}x+....+\dfrac{2}{31\cdot34}x=10\)
\(\dfrac{2}{1.4}x+\dfrac{2}{4.7}x+\dfrac{2}{7.10}x+...+\dfrac{2}{31.34}x=10\)
\(=>1,5.\left(\dfrac{2}{1.4}x+\dfrac{2}{4.7}x+\dfrac{2}{7.10}x+...+\dfrac{2}{31.34}x\right)=15\)
\(=>\dfrac{3}{1.4}x+\dfrac{3}{4.7}x+\dfrac{3}{7.10}x+...+\dfrac{3}{31.34}x=15\)
\(=>x\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{31.34}\right)=15\)
\(=>x\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{31}-\dfrac{1}{34}\right)=15\)
\(=>x\left(1-\dfrac{1}{34}\right)=15\)
\(=>\dfrac{33}{34}x=15\)
\(=>x=\dfrac{170}{11}\)
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Cho S=\(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{40\cdot43}+\dfrac{3}{43\cdot46}\)
Hãy chứng minh rằng S<1
Ta có :
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+..............+\dfrac{3}{40.43}+\dfrac{3}{43.46}\)
\(S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...............+\dfrac{1}{40}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{46}\)
\(S=1-\dfrac{1}{46}< 1\)
\(\Rightarrow S< 1\rightarrowđpcm\)
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\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{40.43}+\dfrac{3}{43.46}\)
\(S=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{40.43}+\dfrac{1}{43.46}\)
\(S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{40}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{46}\)
\(S=1-\dfrac{1}{46}=\dfrac{45}{46}\)
\(\dfrac{45}{46}< 1\)
=> \(S< 1\)
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3)Rút gọn
\(\dfrac{1\cdot3\cdot5+2\cdot6\cdot10+\text{4*12*20+7*21*35}}{1\cdot5\cdot7+2\cdot10\cdot14+4\cdot20\cdot28+7\cdot35\cdot49}\)
\(\dfrac{1.3.5+2.6.10+4.12.20+7.21.35}{1.5.7+2.10.14+4.20.28+7.35.49}\)
\(=\dfrac{1.3.5+2^3.1.3.5+2^6.1.3.5+7^3.1.3.5}{1.5.7+2^3.1.5.7+2^6.1.5.7+7^3.1.5.7}\)
\(=\dfrac{1.3.5\left(1+2^3+2^6+7^3\right)}{1.5.7\left(1+2^3+2^6+7^3\right)}\)
\(=\dfrac{1.3.5}{1.5.7}\)
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1)A=\(\dfrac{5}{1\cdot2}+\dfrac{5}{2\cdot3}+.....+\dfrac{5}{99\cdot100}\)
C=\(1\cdot2\cdot3+2\cdot3\cdot4++3\cdot4\cdot5+4\cdot5\cdot6+5\cdot6\cdot7+6\cdot7\cdot8+7\cdot8\cdot9+8\cdot9\cdot10\)
D=\(1^2+2^2+3^2+...+99^2+100^2\)
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.
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\(1)A=\dfrac{5}{1.2}+\dfrac{5}{2.3}+...+\dfrac{5}{99.100}\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{100}\right)\)
\(\Leftrightarrow A=5\cdot\dfrac{99}{100}\)
\(\Leftrightarrow A=\dfrac{99}{20}\)
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1)So sánh
\(\dfrac{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20+7\cdot21\cdot35}{1\cdot5\cdot7+2\cdot10\cdot14+4\cdot20\cdot28+7\cdot35\cdot45}\)với\(\dfrac{303}{708}\)
\(\dfrac{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20+7\cdot21\cdot35}{1\cdot5\cdot7+2\cdot10\cdot14+4\cdot20\cdot28+7\cdot35\cdot45}\)
=\(\dfrac{3+6+12+21\cdot35}{14+28+7\cdot45}\)
=\(\dfrac{450}{119}\)
Vì \(\dfrac{450}{119}>1\) mà \(1>\dfrac{303}{708}\)
\(\Rightarrow\)\(\dfrac{450}{119}>\dfrac{303}{708}\)
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23 tháng 8 2023 lúc 21:06
\(B=\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+\dfrac{2}{3\cdot4\cdot5}+\dfrac{2}{4\cdot5\cdot6}+\dfrac{2}{5\cdot6\cdot7}+\dfrac{2}{6\cdot7\cdot8}\)
\(B=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+\dfrac{2}{4.5.6}+\dfrac{2}{5.6.7}+\dfrac{2}{6.7.8}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{6.7}-\dfrac{1}{7.8}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{7.8}\)
\(=\dfrac{1}{2}-\dfrac{1}{56}=\dfrac{27}{56}\)
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\(\frac{5}{1\cdot4\cdot7}+\frac{5}{4\cdot7\cdot10}+\frac{5}{7\cdot10\cdot13}+.....+\frac{5}{31\cdot34\cdot37}\)Tính nhanh ( giải kiểu lớp 5 và dấu . là nhân)
THANKS nhiều