Tính nhanh:
\(A=\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{92\cdot95}+\dfrac{1}{95\cdot98}\)
(cảm ơn nhiều ạa)
Tính nhanh:
\(\dfrac{3}{2\cdot5}\)+\(\dfrac{3}{5\cdot8}\)+\(\dfrac{3}{8\cdot11}\)+\(\dfrac{3}{11\cdot14}\)
`# \text {Ryo}`
\(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}\\ =\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}\\ =\dfrac{1}{2}-\left(\dfrac{1}{5}-\dfrac{1}{5}\right)-\left(\dfrac{1}{8}-\dfrac{1}{8}\right)-\left(\dfrac{1}{11}-\dfrac{1}{11}\right)-\dfrac{1}{14}\\ =\dfrac{1}{2}-\dfrac{1}{14}\\ =\dfrac{7}{14}-\dfrac{1}{14}\\ =\dfrac{6}{14}\\ =\dfrac{3}{7}\)
Tính các tổng:
M = \(\dfrac{6}{2\cdot5}+\dfrac{6}{5\cdot8}+\dfrac{6}{8\cdot11}+...+\dfrac{6}{47\cdot50}\)
K = \(\dfrac{1}{9\cdot11}+\dfrac{1}{11\cdot13}+\dfrac{1}{13\cdot15}+...+\dfrac{1}{43\cdot45}\)
\(M=\dfrac{6}{2.5}+\dfrac{6}{5.8}+\dfrac{6}{8.11}+...+\dfrac{6}{47.50}\)
\(\Rightarrow\dfrac{M}{2}=\dfrac{6:2}{2.5}+...+\dfrac{6:2}{47.50}\)
\(=\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{47.50}\)
\(=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{47}-\dfrac{1}{50}\)
\(=\dfrac{1}{2}-\dfrac{1}{50}\)
\(=\dfrac{12}{25}\)
\(\Rightarrow M=\dfrac{12}{25}.2=\dfrac{24}{25}\)
\(K=\dfrac{1}{9.11}+\dfrac{1}{11.13}+\dfrac{1}{13.15}+...+\dfrac{1}{43.45}\)
\(\Rightarrow2K=\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}+...+\dfrac{2}{43.45}\)
\(=\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{43}-\dfrac{1}{45}\)
\(=\dfrac{1}{9}-\dfrac{1}{45}\)
\(=\dfrac{4}{45}\)
\(\Rightarrow K=\dfrac{4}{45}:2=\dfrac{2}{45}\)
\(M=\dfrac{6}{2.5}+\dfrac{6}{5.8}+\dfrac{6}{8.11}+...+\dfrac{6}{47.50}\)
\(M=\dfrac{6}{3}.\left(\dfrac{6}{2}-\dfrac{6}{5}+\dfrac{6}{5}-\dfrac{6}{8}+\dfrac{6}{8}-\dfrac{6}{11}+...+\dfrac{6}{47}-\dfrac{6}{50}\right)\)
\(M=\dfrac{6}{3}.\left(\dfrac{6}{2}-\dfrac{6}{50}\right)\)
\(M=\dfrac{6}{3}.\left(\dfrac{150}{50}-\dfrac{6}{50}\right)\)
\(M=\dfrac{6}{3}.\dfrac{144}{50}\)
\(M=\dfrac{144}{25}\)
\(K=\dfrac{1}{9.11}+\dfrac{1}{11.13}+\dfrac{1}{13.15}+...+\dfrac{1}{43.45}\)
\(K=\dfrac{1}{2}.\left(\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{43}-\dfrac{1}{45}\right)\)
\(K=\dfrac{1}{2}.\left(\dfrac{1}{9}-\dfrac{1}{45}\right)\)
\(K=\dfrac{1}{2}.\left(\dfrac{5}{45}-\dfrac{1}{45}\right)\)
\(K=\dfrac{1}{2}.\dfrac{4}{45}\)
\(K=\dfrac{2}{45}\)
Câu hỏi today là a, \(\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{x+\left(x+1\right)}=\dfrac{101}{1540}\)
b,\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{x+\left(x+1\right)}=\dfrac{2020}{2021}\)
Chúc các bn học tốt ❤
Giải:
a)1/5.8+1/8.11+...+1/x.(x+1)=101/1540
1/3.(3/5.8+3/8.11+...+3/x.(x+1))=101/1540
1/3.(1/5-1/8+1/8-1/11+...+1/x-1/x+1)=101/1540
1/3.(1/5-1/x+1)=101/1540
1/5-1/x+1=101/1540:1/3
1/5-1/x+1=303/1540
1/x+1=1/5-303/1540
1/x+1=1/308
⇒x+1=308
x=308-1
x=307
b)1/1.2+1/2.3+1/3.4+...+1/x.(x+1)=2020/2021
1/1-1/2+1/2-1/3+1/3-1/4+...+1/x-1/x+1=2020/2021
1/1-1/x+1=2020/2021
1/x+1=1/1-2020/2021
1/x+1=1/2021
⇒x+1=2021
x=2021-1
x=2020
Mk thấy đề bài hơi sai là:
1/x+(x+1) ➜ 1/x.(x+1)
mới ra đc kết quả!
Bài 1tính
A=\(\dfrac{3^2}{2\cdot5}+\dfrac{3^2}{5\cdot8}+\dfrac{3^2}{8\cdot11}+...+\dfrac{3^2}{98\cdot101}\)
giúp mình với mai thầy mình kiểm tra rồi!
A=\(\dfrac{3}{1}\).(\(\dfrac{3}{2.5}\)+\(\dfrac{3}{5.8}\)+...+\(\dfrac{3}{98.101}\))
A=3.(\(\dfrac{1}{2}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{8}\)+...+\(\dfrac{1}{98}\)-\(\dfrac{1}{101}\))
A=3.(\(\dfrac{1}{2}\)-\(\dfrac{1}{101}\))
A=3.\(\dfrac{98}{202}\)
A=\(\dfrac{294}{202}\)=\(\dfrac{147}{101}\)
1. Chứng minh rằng với \(\forall N\ne0̸\) ta đều có :
a, \(\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+\dfrac{1}{\left(3n-1\right)\cdot\left(3n+1\right)}=\dfrac{n}{6n+4}\).
2. Tìm GTLN hoặc GTNN của biểu thức \(A=\dfrac{\left|2-x\right|-3}{\left|2-x\right|+11}\).
tìm x
\(\dfrac{1}{5\cdot8}\)+\(\dfrac{1}{8\cdot11}\)+\(\dfrac{1}{11\cdot13}\)+...+\(\dfrac{1}{x\left(x+3\right)}\)=\(\dfrac{101}{1540}\)
sửa đề: phải là 14 chứ sao lại là 13 nhỉ?=))
\(\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+\dfrac{1}{11\cdot14}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\) \(\left(x\ne0;x\ne-3\right)\)
\(\left(\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+\dfrac{1}{11\cdot14}+...+\dfrac{1}{x\left(x+3\right)}\right)\cdot3=\dfrac{101}{1540}\cdot3\)
\(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{303}{1540}\)
\(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
\(\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
\(\dfrac{308\left(x+3\right)}{1540\left(x+3\right)}-\dfrac{1540}{1540\left(x+3\right)}=\dfrac{303\left(x+3\right)}{1540\left(x+3\right)}\)
suyy ra
`308x+924-1540=303x+909`
`5x=1525`
`x=305(tm)`
Tính:
\(\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}+\dfrac{1}{9\cdot11}+\dfrac{1}{11\cdot13}+\dfrac{1}{13\cdot15}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{13\cdot15}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{4}{15}=\dfrac{2}{15}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{1}{15}=\dfrac{5}{15}-\dfrac{1}{15}=\dfrac{4}{15}\)
= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{11}-\dfrac{1}{13}\)
= \(\dfrac{1}{3}-\dfrac{1}{13}\)
=\(\dfrac{10}{26}=\dfrac{5}{13}\)
Tính:
\(\dfrac{1}{2\text{×}5}+\dfrac{1}{5\text{×}8}+\dfrac{1}{8\text{×}11}+...+\dfrac{1}{92\text{×}95}+\dfrac{1}{95\text{×}97}\)
đặt
\(A=\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+..+\dfrac{1}{92\cdot95}+\dfrac{1}{95\cdot97}\)
\(3A=\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{92\cdot95}+\dfrac{3}{95\cdot97}\)
\(3A=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}+\dfrac{1}{95}-\dfrac{1}{97}\)
\(3A=\dfrac{1}{2}-\dfrac{1}{97}\\ 3A=\dfrac{95}{194}\\ A=\dfrac{95}{582}\)
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)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
\(=\frac{1}{2}.\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+\frac{11-9}{9.11}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{11}\right)\)
\(\frac{10}{22}\)