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\(x+\frac{2}{5.9}+\frac{2}{9.13}+\frac{2}{13.17}+...+\frac{2}{41.45}=\frac{-37}{45}\)
\(\Leftrightarrow x+\left[\frac{2}{4}\cdot\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)\right]=\frac{-37}{45}\)
\(\Leftrightarrow x+\left[\frac{1}{2}\cdot\left(\frac{1}{5}-\frac{1}{45}\right)\right]=\frac{-37}{45}\)
\(\Leftrightarrow x+\left[\frac{1}{2}\cdot\frac{8}{45}\right]=\frac{-37}{45}\)
\(\Leftrightarrow x+\frac{4}{45}=\frac{-37}{45}\)
\(\Leftrightarrow x=-\frac{41}{45}\)
\(A=3.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{101}-\frac{1}{105}\right)\)
\(A=3.\left(1-\frac{1}{105}\right)\)
\(A=3.\frac{104}{105}\)
\(A=\frac{104}{35}\)
Em yêu cầu bác nhìn xuống dưới và bác sẽ biết cách làm
Bác thấy rồi mà còn đăng
Thay số mà làm nhé
:))
M = - ( 4/1.5 + 4/5.9 + ..................+ 4/(n-4).n
M = - (1-1/5 + 1/5 - 1/9 +..............+1/(n-4) - 1/n
M = -(1-1/n)
M = -1 + 1/n
M = -n + 1
Ta có : \(-\frac{4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-.....-\frac{4}{\left(n+4\right)n}\)
\(=-\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+......+\frac{4}{n\left(4+n\right)}\right)\)
\(=-\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+......+\frac{1}{n}-\frac{1}{n+4}\right)\)
\(=-\left(1-\frac{1}{n+4}\right)\)
\(=-\left(\frac{n+4}{n+4}-\frac{1}{n+4}\right)\)
\(=-\frac{n+3}{n+4}\)
x/1 - x/5 + x/5 - x/9 +x/9 - x/13 ..... + x/53 - x/57 = 56/57
x/1 - x/57 = 56/57
56x/57 = 56/57
56x = 56
=> X = 1
Tk mình với bạn ơi. Đúng rồi nhé!!
CHÚC BẠN HỌC TỐT ✓✓
\(\frac{x}{1.5}+\frac{x}{5.9}+\frac{x}{9.13}+...+\frac{x}{53.57}=\frac{56}{57}\)
\(\Leftrightarrow\frac{x}{1}-\frac{x}{5}+\frac{x}{5}-\frac{x}{7}+\frac{x}{9}-\frac{x}{13}+...+\frac{x}{53}-\frac{x}{57}=\frac{56}{57}\)
\(\Leftrightarrow\frac{x}{1}-\frac{x}{57}=\frac{56}{57}\)
\(\Leftrightarrow\frac{x.57}{57}-\frac{x}{57}=\frac{56}{57}\)
\(\Leftrightarrow\frac{x.57-x}{57}=\frac{56}{57}\)
\(\Leftrightarrow\frac{x.56}{57}=\frac{56}{57}\)
\(\Leftrightarrow x=1\)
\(S=\frac{-4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-...-\frac{4}{\left(n-4\right).n}\)
\(=-\left(\frac{1}{1}-\frac{1}{5}\right)-\left(\frac{1}{5}-\frac{1}{9}\right)-\left(\frac{1}{9}-\frac{1}{13}\right)-...-\left(\frac{1}{n-4}-\frac{1}{n}\right)\)
\(=-\frac{1}{1}+\frac{1}{5}-\frac{1}{5}+\frac{1}{9}-\frac{1}{9}+\frac{1}{13}-...-\frac{1}{n-4}+\frac{1}{n}\)
\(=-\frac{1}{1}+\frac{1}{n}=\frac{1}{n}+1\)
\(B=\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{41.45}\)
\(4B=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{41}-\frac{1}{45}\)
\(4B=\frac{44}{45}\)
\(B=\frac{11}{45}\)
\(B=\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{41.45}\)
\(=\frac{1}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{41.45}\right)\)
\(=\frac{1}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{41}-\frac{1}{45}\right)\)
\(=\frac{1}{4}.\left(1-\frac{1}{45}\right)\)
\(=\frac{1}{4}.\frac{44}{45}\)
\(=\frac{11}{45}\)
\(\frac{2}{1.5}+\frac{2}{5.9}+\frac{2}{9.13}+...+\frac{2}{95.99}\)
\(=\frac{1}{2}.\left(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{95.99}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{95}-\frac{1}{99}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{2}.\frac{98}{99}\)
\(=\frac{49}{99}\)
Chúc cậu học tốt !!!