\(\dfrac{2}{1\cdot4}+\dfrac{2}{4\cdot7}+\dfrac{2}{7\cdot10}+\dfrac{2}{49\cdot50}\)
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ND
Tìm x
\(\dfrac{2}{1\cdot4}x+\dfrac{2}{4\cdot7}x+\dfrac{2}{7\cdot10}x+....+\dfrac{2}{31\cdot34}x=10\)
1
DA
28 tháng 4 2022
\(\dfrac{2}{1.4}x+\dfrac{2}{4.7}x+\dfrac{2}{7.10}x+...+\dfrac{2}{31.34}x=10\)
\(=>1,5.\left(\dfrac{2}{1.4}x+\dfrac{2}{4.7}x+\dfrac{2}{7.10}x+...+\dfrac{2}{31.34}x\right)=15\)
\(=>\dfrac{3}{1.4}x+\dfrac{3}{4.7}x+\dfrac{3}{7.10}x+...+\dfrac{3}{31.34}x=15\)
\(=>x\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{31.34}\right)=15\)
\(=>x\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{31}-\dfrac{1}{34}\right)=15\)
\(=>x\left(1-\dfrac{1}{34}\right)=15\)
\(=>\dfrac{33}{34}x=15\)
\(=>x=\dfrac{170}{11}\)
21 tháng 11 2017
\(\dfrac{12}{1.4.7}+\dfrac{12}{4.7.10}+\dfrac{12}{7.10.13}+...+\dfrac{12}{54.57.60}\)
\(=2\left(\dfrac{1}{1.4}-\dfrac{1}{4.7}+\dfrac{1}{4.7}-\dfrac{1}{7.10}+\dfrac{1}{7.10}-\dfrac{1}{10.13}+...+\dfrac{1}{54.57}-\dfrac{1}{57.60}\right)\)\(=2\left(\dfrac{1}{1.4}-\dfrac{1}{57.60}\right)\)
\(=2\left(\dfrac{1}{4}-\dfrac{1}{57.60}\right)=\dfrac{1}{2}-\dfrac{1}{2.57.60}< \dfrac{1}{2}\left(đpcm\right)\)
23 tháng 7 2017
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.
23 tháng 7 2017
\(1)A=\dfrac{5}{1.2}+\dfrac{5}{2.3}+...+\dfrac{5}{99.100}\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{100}\right)\)
\(\Leftrightarrow A=5\cdot\dfrac{99}{100}\)
\(\Leftrightarrow A=\dfrac{99}{20}\)
T
23 tháng 8 2023
\(B=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+\dfrac{2}{4.5.6}+\dfrac{2}{5.6.7}+\dfrac{2}{6.7.8}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{6.7}-\dfrac{1}{7.8}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{7.8}\)
\(=\dfrac{1}{2}-\dfrac{1}{56}=\dfrac{27}{56}\)
9 tháng 5 2017
Ta có :
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+..............+\dfrac{3}{40.43}+\dfrac{3}{43.46}\)
\(S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...............+\dfrac{1}{40}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{46}\)
\(S=1-\dfrac{1}{46}< 1\)
\(\Rightarrow S< 1\rightarrowđpcm\)
DA
9 tháng 5 2017
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{40.43}+\dfrac{3}{43.46}\)
\(S=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{40.43}+\dfrac{1}{43.46}\)
\(S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{40}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{46}\)
\(S=1-\dfrac{1}{46}=\dfrac{45}{46}\)
\(\dfrac{45}{46}< 1\)
=> \(S< 1\)
SN
8
Q
9 tháng 4 2017
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(A=1-\dfrac{1}{50}\)
\(A=\dfrac{49}{50}\)
9 tháng 4 2017
Ta có :
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...........+\dfrac{1}{49.50}\)
\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...........+\dfrac{1}{49}-\dfrac{1}{50}\)
\(A=\dfrac{1}{1}-\dfrac{1}{50}\)
\(A=\dfrac{49}{50}\)
~ Chúc bn học tốt ~
7 tháng 10 2021
\(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{49\cdot50}=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\\ =\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\\ =\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\\ =1+\dfrac{1}{2}+...+\dfrac{1}{50}-1-\dfrac{1}{2}-\dfrac{1}{3}-...-\dfrac{1}{25}\\ =\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)
7 tháng 10 2021
tại sao \(\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\\ =\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
T
5 tháng 8 2017
A = \(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{13.16}\)
\(A=1-\left(\dfrac{1}{4}+\dfrac{1}{4}\right)-\left(\dfrac{1}{7}+\dfrac{1}{7}\right)-\dfrac{1}{10}-\dfrac{1}{13}-\dfrac{1}{16}\)
\(A=1-\dfrac{1}{10}-\dfrac{1}{13}-\dfrac{1}{16}\)
(13 - 10 = 3 ; 16 - 13 = 3)
\(3A=1-\dfrac{1}{16}\)
\(=\dfrac{15}{16}\)
Vậy ... tự tìm a đi! Lười quá!
Bài 2: Dễ ; tự làm
Bài3: Áp dụng tính chất phép cộng ta có:
a + b = b + a
=> A và B có phép tính giống nhau chỉ đổi chỗ
Không mất công tính.
Ta có thể kết luận phép tính trên bằng nhau
Đăt :
\(A=\dfrac{2}{1.4}+\dfrac{2}{4.7}+.........+\dfrac{2}{49.51}\)
\(\dfrac{3}{2}A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+..........+\dfrac{3}{49.51}\)
\(\dfrac{3}{2}A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+.........+\dfrac{1}{49}-\dfrac{1}{51}\)
\(\dfrac{3}{2}A=1-\dfrac{1}{51}\)
\(\dfrac{3}{2}A=\dfrac{50}{51}\)
\(\Rightarrow A=\dfrac{50}{51}:\dfrac{3}{2}=\dfrac{100}{153}\)
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