Haizzzzzzzz............. Vừa mới ik thi tháng về (=_=) . Có bài này khó
Các anh , các chị giúp em với ạ !
Cho \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\text{. Chứng minh rằng : }\frac{7}{12}< A< \frac{5}{6}\)
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\(B=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(B=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(B=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(B< \frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\)
\(B< \frac{50}{60}\Leftrightarrow B< \frac{5}{6}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.......+\frac{1}{99.100}\)
\(=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{100-99}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{49}{100}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}+\frac{1}{100}=\frac{49}{100}\)
\(a,\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{1}-\frac{1}{100}\)
\(=\frac{100}{100}-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(b,\frac{x}{y}=\frac{3}{5}\)
\(\Leftrightarrow\frac{x}{3}=\frac{y}{5}\)
\(\text{Áp dụng tính chất dãy tỉ số bằng nhau ta có :}\)
\(\frac{x}{3}=\frac{y}{5}=\frac{x+y}{3+5}=\frac{18}{8}=\frac{9}{4}\)
\(\Rightarrow\frac{x}{3}=\frac{9}{4}\Rightarrow x=\frac{27}{4}\)
\(\frac{y}{5}=\frac{9}{4}\Rightarrow y=\frac{45}{4}\)
\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{60}\)
\(=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+...+\frac{1}{50}+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)
2/ \(A=\frac{1}{2}+\frac{1}{12}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(A=\frac{7}{12}+\frac{1}{5.6}+\frac{1}{7.8}+...+\frac{1}{99.100}>\frac{7}{12}\)
Tương tự câu trên ta có: \(A=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(A=\frac{1}{51}+...+\frac{1}{60}+\frac{1}{61}+...+\frac{1}{70}+\frac{1}{71}+...+\frac{1}{80}+\frac{1}{81}+...+\frac{1}{90}+\frac{1}{91}+...+\frac{1}{100}\)
\(A< \frac{1}{50}+...+\frac{1}{50}+\frac{1}{60}+...+\frac{1}{60}+\frac{1}{70}+...+\frac{1}{70}+\frac{1}{80}+...+\frac{1}{80}+\frac{1}{90}+...+\frac{1}{90}\)
\(A< 10.\frac{1}{50}+10.\frac{1}{60}+10.\frac{1}{70}+10.\frac{1}{80}+10.\frac{1}{90}\)
\(A< \frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}< \frac{5}{6}\)
mình vừa mới trả lời xong đấy
Câu hỏi của Do Not Ask Why - Toán lớp 7 - Học toán với OnlineMath
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}.\text{ CMR : }\frac{7}{12}< A< \frac{5}{6}\)
Ta có :
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{99}+\frac{1}{100}-2.\frac{1}{2}-2.\frac{1}{4}-...-2.\frac{1}{98}\)
\(A=1+...+\frac{1}{100}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{49}\)
\(A=\frac{1}{51}+...+\frac{1}{100}\)
\(\Rightarrow A< \frac{1}{51.25}=\frac{25}{51}< \frac{25}{30}=\frac{5}{6}\) (đpcm)
Và \(A>25.\frac{1}{75}+25.\frac{1}{100}=\frac{7}{12}\)
Ta có : A = 1 / (1.2) + 1 / (3.4) + ... + 1 / (99.100) > 1 / (1.2) + 1 / (3.4) = 1 / 2 + 1 / 12 = 7 / 12 (1)
Lại có : A = 1 / (1.2) + 1 / (3.4) + ... + 1 / (99.100) = (1 - 1 / 2) + (1 / 3 - 1 / 4) + ... + (1 / 99 - 100)
= (1 - 1 / 2 + 1 / 3) - (1 / 4 - 1 / 5) - (1 / 6 - 1 / 7) - ... - (1 / 98 - 1 / 99) - 1 / 100 < 1 - 1 / 2 + 1 / 3 = 5 / 6 (2)
Từ (1) và (2) => 7 / 12 < A < 5 / 6