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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.
\(1)A=\dfrac{5}{1.2}+\dfrac{5}{2.3}+...+\dfrac{5}{99.100}\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{100}\right)\)
\(\Leftrightarrow A=5\cdot\dfrac{99}{100}\)
\(\Leftrightarrow A=\dfrac{99}{20}\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(A=1-\dfrac{1}{50}\)
\(A=\dfrac{49}{50}\)
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 ~
Ta có :
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+..............+\dfrac{3}{40.43}+\dfrac{3}{43.46}\)
\(S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...............+\dfrac{1}{40}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{46}\)
\(S=1-\dfrac{1}{46}< 1\)
\(\Rightarrow S< 1\rightarrowđpcm\)
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{40.43}+\dfrac{3}{43.46}\)
\(S=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{40.43}+\dfrac{1}{43.46}\)
\(S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{40}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{46}\)
\(S=1-\dfrac{1}{46}=\dfrac{45}{46}\)
\(\dfrac{45}{46}< 1\)
=> \(S< 1\)
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
\(\dfrac{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20+7\cdot21\cdot35}{1\cdot5\cdot7+2\cdot10\cdot14+4\cdot20\cdot28+7\cdot35\cdot45}\)
=\(\dfrac{3+6+12+21\cdot35}{14+28+7\cdot45}\)
=\(\dfrac{450}{119}\)
Vì \(\dfrac{450}{119}>1\) mà \(1>\dfrac{303}{708}\)
\(\Rightarrow\)\(\dfrac{450}{119}>\dfrac{303}{708}\)
P=\(\dfrac{1}{1\cdot2}\)+\(\dfrac{2}{2\cdot4}\)+\(\dfrac{3}{4\cdot7}\)+...+\(\dfrac{10}{46\cdot56}\)
\(P=\dfrac{1}{1.2}+\dfrac{2}{2.4}+\dfrac{3}{4.7}+...+\dfrac{10}{46.56}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{46}-\dfrac{1}{56}\)
\(=1-\dfrac{1}{56}=\dfrac{55}{56}\)
Giải:
\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)
\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{3}-\dfrac{1}{10}\)
\(=\dfrac{7}{30}\)
Vậy ...
\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
=\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
=\(\dfrac{1}{3}-\dfrac{1}{10}\)
=\(\dfrac{7}{30}\)
Tìm x
\(\dfrac{2}{1\cdot4}x+\dfrac{2}{4\cdot7}x+\dfrac{2}{7\cdot10}x+....+\dfrac{2}{31\cdot34}x=10\)
\(\dfrac{2}{1.4}x+\dfrac{2}{4.7}x+\dfrac{2}{7.10}x+...+\dfrac{2}{31.34}x=10\)
\(=>1,5.\left(\dfrac{2}{1.4}x+\dfrac{2}{4.7}x+\dfrac{2}{7.10}x+...+\dfrac{2}{31.34}x\right)=15\)
\(=>\dfrac{3}{1.4}x+\dfrac{3}{4.7}x+\dfrac{3}{7.10}x+...+\dfrac{3}{31.34}x=15\)
\(=>x\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{31.34}\right)=15\)
\(=>x\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{31}-\dfrac{1}{34}\right)=15\)
\(=>x\left(1-\dfrac{1}{34}\right)=15\)
\(=>\dfrac{33}{34}x=15\)
\(=>x=\dfrac{170}{11}\)
1. \(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}=\dfrac{2}{4}=\dfrac{1}{2}\)
2. \(B=\dfrac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\dfrac{1}{5}\)
3.\(C=\dfrac{2^2.3^2.\text{4^2.5^2}.5^2}{1.2^2.3^2.4^2.5.6^2}=\dfrac{125}{36}\)
4.D=\(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right).\dfrac{4}{9}.\dfrac{1}{16}=\dfrac{19}{30}.\dfrac{1}{36}=\dfrac{19}{1080}\)
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