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\(B=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)
\(B=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(B=\frac{1}{3}.\left(1-\frac{1}{103}\right)\)
\(B=\frac{1}{3}.\frac{102}{103}\)
\(B=\frac{34}{103}\)
Bài 3: đổi ra phân số rồi tính, đổi:\(1,5=\frac{15}{10};2,5=\frac{25}{10};1\frac{3}{4}=\frac{7}{12}\)(cái này ko giải dùm, đổi ra như thek rồi tính nha)
\(B=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{100.103}\)
\(=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\right)\)
\(=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(=\frac{1}{3}.\left(1-\frac{1}{103}\right)\)
\(=\frac{1}{3}.\frac{102}{103}\)
\(=\frac{1}{1}.\frac{34}{103}=\frac{34}{103}\)
\(S=\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+...+\frac{1}{2002\cdot2005}\)
\(3S=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{2002\cdot2005}\)
\(3S=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{2002}-\frac{1}{2005}\)
\(3S=\frac{1}{1}-\frac{1}{2005}\)
\(3S=\frac{2004}{2005}\)
\(S=\frac{2004}{2005}\div3=\frac{668}{2005}\)
Ta có:
\(S=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2002.2005}\)
\(\Rightarrow S=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{2002.2005}\right)\)
\(\Rightarrow S=\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2002}-\frac{1}{2005}\right)\)
\(\Rightarrow S=\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{2005}\right)=\frac{1}{3}.\frac{2004}{2005}=\frac{668}{2005}\)
\(\frac{x}{1.4}+\frac{x}{4.7}+\frac{x}{7.10}+...+\frac{x}{36.39}=1\)
\(\frac{x}{3}.\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{36.39}\right)=1\)
\(\frac{x}{3}.[(\frac{1}{1}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{7})+(\frac{1}{7}-\frac{1}{10})+...+(\frac{1}{36}-\frac{1}{39})]=1\)
\(\frac{x}{3}.(\frac{1}{1}-\frac{1}{39})=1\)
\(\frac{x}{3}.\frac{38}{39}=1\)
\(\frac{x}{3}=1:\frac{38}{39}\)
\(\frac{x}{3}=\frac{39}{38}\)
\(\Rightarrow x=.....\)
Mình tính vội nên không tính kết quả đúng chưa, cậu kiểm tra lại nha, còn cách làm thế là chuẩn rồi! Học tốt!
\(\left(2\right)K=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(K=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\)
\(K=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(K=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
\(\left(3\right)L=\frac{5}{1\cdot4}+\frac{5}{4\cdot7}+\frac{5}{7\cdot10}+...+\frac{5}{100\cdot103}\)
\(L=\frac{5}{3}\cdot\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(L=\frac{5}{3}\cdot\left(\frac{1}{1}-\frac{1}{103}\right)=\frac{5}{3}\cdot\frac{102}{103}=\frac{510}{309}=\frac{170}{103}\)
Trả lời:
2,\(K=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(K=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(K=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(K=\frac{1}{2}-\frac{1}{100}\)
\(K=\frac{49}{100}\)
3,\(L=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)
\(L=\frac{5}{3}\times\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)
\(L=\frac{5}{3}\times\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(L=\frac{5}{3}\times\left(\frac{1}{1}-\frac{1}{103}\right)\)
\(L=\frac{5}{3}\times\frac{102}{103}\)
\(L=\frac{170}{103}\)
Học tốt
\(A=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(A=\frac{2}{3}.\left(1-\frac{1}{100}\right)\)
\(A=\frac{2}{3}.\frac{99}{100}\)
\(A=\frac{33}{50}\)
\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)
\(3B=5\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{100.103}\right)\)
\(3B=5\left(1-\frac{1}{103}\right)\)
\(3B=5.\frac{102}{103}\)
\(3B=\frac{510}{103}\)
\(\Rightarrow B=\frac{170}{103}\)
Ta có:
B=\(\frac{5}{1.4}\)+\(\frac{5}{4.7}+.....+\frac{5}{100.103}\)
B=\(\frac{5}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+....+\frac{3}{100.103}\right)\)
B=\(\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{100}-\frac{1}{103}\right)\)
B=\(\frac{5}{3}\left(1-\frac{1}{103}\right)\)
B=\(\frac{5}{3}.\frac{102}{103}\)
B=\(\frac{170}{103}\)
Vậy B=\(\frac{170}{103}\)
nhớ k
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
\(S=1-\frac{1}{46}\)
Đến đây ta suy được ra S<1
Ta có :
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
\(S=1-\frac{1}{46}\)
\(S=\frac{45}{46}< 1\)
Vậy \(S< 1\)
Võ Thiện Tuấn viết tổng quát kết quả hay phép đề bài hả bạn ?
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7} +....+\frac{1}{100}-\frac{1}{103}\)
\(=1-\frac{1}{103}\)
\(=\frac{102}{103}\)