Tính nhanh :1/3 x ( 3/2x5 + 3/5x8 + 3/8x11 + ...+ 3/92x95 + 3/95x98
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\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}+\frac{1}{95.98}\)
\(A=\frac{1}{3}\cdot\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{92.95}+\frac{3}{95.98}\right)\)
\(A=\frac{1}{3}\cdot\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{98}\right)\)
\(A=\frac{1}{3}\cdot\left(\frac{1}{2}-\frac{1}{98}\right)\)
\(A=\frac{1}{3}\cdot\frac{24}{49}\)
\(A=\frac{8}{49}\)
Đề sai rồi bạn nhé, đề là như thế này:
\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}+\frac{1}{95.98}\)
\(A=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{92.95}+\frac{3}{95.98}\right)\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}\right)\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{98}\right)\)
\(A=\frac{1}{3}.\frac{24}{49}\)
\(A=\frac{24}{147}=\frac{8}{49}\)
Đặt \(A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}\)
\(A=\dfrac{3}{2}-\dfrac{3}{5}+\dfrac{3}{5}-\dfrac{3}{8}+\dfrac{3}{8}-\dfrac{3}{11}+\dfrac{3}{11}-\dfrac{3}{14}\)
\(A=\dfrac{3}{2}-\dfrac{3}{14}\)
\(A=\dfrac{21}{14}-\dfrac{3}{14}\)
\(A=\dfrac{18}{14}\)
\(A=\dfrac{9}{7}\)
\(A=1\dfrac{2}{7}\)
\(A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}.\)
\(A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\)
\(A=\frac{1}{2}-\frac{1}{20}\)
\(A=\frac{10}{20}-\frac{1}{20}\)
\(A=\frac{9}{20}\)
Mình ra kết quả thứ nhất là 17/60 thứ 2 là 9/20 các bạn thấy cái nào đúng
\(A=\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+...+\frac{3}{93.96}+\frac{3}{96.99}\)
\(A=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{93}-\frac{1}{96}+\frac{1}{96}-\frac{1}{99}\)
\(A=1-\frac{1}{99}=\frac{98}{99}\)
Vậy A=\(\frac{98}{99}\)
\(B=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{95.98}\)
\(3B=\)\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{95.98}\)
\(3B=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{98}\)
\(3B=\frac{1}{2}-\frac{1}{98}=\frac{24}{49}\)
\(B=\frac{24}{49}:3=\frac{8}{49}\)
Vậy B=\(\frac{8}{49}\)
Dấu "." là dấu nhân.
_Học tốt_
\(B=\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{26\cdot29}\)
\(B=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{29}\)
\(B=\dfrac{1}{2}-\dfrac{1}{29}\)
\(B=\dfrac{27}{58}\)
B= 3/2x5 + 3/5x8+ 3/8x11 + ... + 3/26x29
B= 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/26 - 1/29
B= 1/2-1/29
B=27/58
Ta có : A=3/2x5+3/5x8+3/8x11+3/11x14+3/14x17+3/17x20
=> A=1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14+1/14-1/17+1/17-1/20
=> A=1/2-1/20
=> A=9/20
Vậy A=9/20
đó
Ta có: \(\frac{3x}{2\cdot5}+\frac{3x}{5\cdot8}+\frac{3x}{8\cdot11}+\frac{3x}{11\cdot14}=\frac{1}{21}\)
\(\Leftrightarrow x\cdot\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\right)=\frac{1}{21}\)
\(\Leftrightarrow x\cdot\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\right)=\frac{1}{21}\)
\(\Leftrightarrow x\cdot\left(\frac{1}{2}-\frac{1}{14}\right)=\frac{1}{21}\)
\(\Leftrightarrow x\cdot\frac{3}{7}=\frac{1}{21}\)
\(\Leftrightarrow x=\frac{1}{21}:\frac{3}{7}=\frac{1}{21}\cdot\frac{7}{3}=\frac{7}{63}=\frac{1}{9}\)
Vậy: \(x=\frac{1}{9}\)
A = \(\dfrac{1}{2}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{11}\) + \(\dfrac{1}{11}\) - \(\dfrac{1}{14}\) + \(\dfrac{1}{14}\) - \(\dfrac{1}{17}\) + \(\dfrac{1}{17}\) - \(\dfrac{1}{20}\)
= \(\dfrac{1}{2}\) - \(\dfrac{1}{20}\)
= \(\dfrac{9}{20}\)
`@` `\text {Ans}`
`\downarrow`
`A =`\(\dfrac{3}{2\times5}+\dfrac{3}{5\times8}+\dfrac{3}{8\times11}+...+\dfrac{3}{20\times23}+\dfrac{3}{23\times26}\)
`A=`\(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{23}-\dfrac{1}{26}\)
`A=`\(\dfrac{1}{2}-\dfrac{1}{26}\)
`A=`\(\dfrac{6}{13}\)
Vậy, `A=`\(\dfrac{6}{13}\).
\(\dfrac{1}{3}\times\left(\dfrac{3}{2\times5}+\dfrac{3}{5\times8}+\dfrac{3}{8\times11}+...+\dfrac{3}{92\times95}+\dfrac{3}{95\times98}\right)\\ =\dfrac{1}{3}\times\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}+\dfrac{1}{95}-\dfrac{1}{98}\right)\\ =\dfrac{1}{3}\times\left(\dfrac{1}{2}-\dfrac{1}{98}\right)\\ =\dfrac{1}{3}\times\left(\dfrac{49}{98}-\dfrac{1}{98}\right)\\ =\dfrac{1}{3}\times\dfrac{24}{49}\\ =\dfrac{8}{49}\)