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A = 1 / 31 + 1 / 32 + 1 / 33 + ... + 1 / 89 + 1 / 90 ... 5 / 6
A = 5 / 6 = 1 / 2 + 1 / 3
Ta đặt B = 1 / 31 + 1 / 32 + 1 / 33 + ... + 1 / 60 ( 30 phân số )
C = 1 / 61 + 1 / 62 + 1 / 63 + ... + 1 / 90 ( 30 phân số )
Ta có : B = 1 / 31 + 1 / 32 + 1 / 33 + ... + 1 / 60 > 1 / 60 + 1 / 60 + 1 / 60 + ... + 1 / 60 = 30 . 1 / 60 = 1 / 2
C = 1 / 61 + 1 / 62 + 1 / 63 + ... + 1 / 90 > 1 / 90 + 1 / 90 + 1 / 90 + ... + 1 / 90 = 30 . 1 / 90 = 1 / 3
Vì A = B + C > 1 / 2 + 1 / 3 = 5 / 6 nên 1 / 31 + 1 / 32 + ... + 1 / 89 + 1 / 90 > 5 / 6
GIẢI VẦY MỚI GỌI LÀ GIẢI CHI TIẾT
Ta sẽ lấy
\(1-\frac{1}{90}=\frac{89}{90}\)
Sau đó ta so sánh :
\(\frac{89}{90}>\frac{5}{6}\)
k mình nhé !!!
Ta nhận thấy mẫu số của các phân số có qui luật 1x3; 2x4; 3x5; 4x6...... => mẫu số của phân số thứ 98 là 98x100
\(\Rightarrow A=\frac{4}{3}x\frac{9}{8}x\frac{16}{15}x\frac{25}{24}x\frac{36}{35}x...x\frac{9801}{9800}\)
\(A=\frac{2x2x3x3x4x4x5x5x6x6x...x99x99}{1x2x3x3x4x4x5x5x...x96x96x97x97x98x98x99x100}=\frac{2x99}{100}=\frac{99}{50}=1\frac{49}{50}\)
= 2 x ( 1/2 x 5 + 1/ 5 x 8 + 1/ 8 x 11 + 1/ 11 x 14 + 1/ 14 x 17 )
= 2 x ( 1/2 - 1/5 + 1/5 - 1/8 + ....+1/14 - 1/17)
= 2 x (1/2 - 1/17)
= 2 x 15/34
= 15/17
ĐÚNG THÌ TÍCH CHO MÌNH NHA
CHÚC BẠN HỌC GIỎI
Đặt \(A=\frac{2}{2.5}+\frac{2}{5.8}+\frac{2}{8.11}+\frac{2}{11.14}+\frac{2}{14.17}\)
\(A=\frac{2}{3}\cdot\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{14}-\frac{1}{17}\right)\)
\(A=\frac{2}{3}\cdot\left(\frac{1}{2}-\frac{1}{17}\right)\)
\(A=\frac{2}{3}\cdot\frac{15}{34}=\frac{5}{17}\Rightarrow A< 1\)
a) \(\frac{18}{35};\frac{28}{35};\frac{31}{25};\frac{32}{25}\)
A=1/1.2+1/12.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8
A=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8
A=1/1-1/8
A=7/8
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(A=1-\frac{1}{8}\)
\(A=\frac{7}{8}\)
A,\(\frac{7}{13}+\frac{1}{7}+\frac{18}{30}=\frac{49}{91}+\frac{13}{91}+\frac{18}{30}=\frac{62}{91}+\frac{18}{30}=\frac{1860}{2730}+\frac{1638}{2730}=\frac{583}{455}\)
B,\(\left(\frac{20}{21}+\frac{3}{14}\right)-\frac{2}{3}=\left(\frac{280}{294}+\frac{63}{294}\right)-\frac{2}{3}=\frac{7}{6}-\frac{2}{3}=\frac{21}{18}-\frac{12}{18}=\frac{9}{18}=\frac{1}{2}\)
C,\(=\frac{17}{5}+\frac{31}{7}=\frac{119}{35}+\frac{155}{35}=\frac{234}{35}\)
Bài 3 :
\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+....+\frac{1}{99\times100}\)
Ta có : \(\frac{1}{1\times2}=\frac{2-1}{1\times2}=\frac{2}{1\times2}-\frac{1}{1\times2}=1-\frac{1}{2}\)
\(\frac{1}{2\times3}=\frac{3-2}{2\times3}=\frac{3}{2\times3}-\frac{2}{2\times3}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{99\times100}=\frac{100-99}{99\times100}=\frac{100}{99\times100}-\frac{99}{99\times100}=\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(B=\frac{1}{10\times11}+\frac{1}{11\times12}+...+\frac{1}{38\times39}\)
Ta có : \(\frac{1}{10\times11}=\frac{11-10}{10\times11}=\frac{11}{10\times11}-\frac{10}{10\times11}=\frac{1}{10}-\frac{1}{11}\)
\(\frac{1}{11\times12}=\frac{12-11}{11\times12}=\frac{12}{11\times12}-\frac{11}{11\times12}=\frac{1}{11}-\frac{1}{12}\)
\(\frac{1}{38\times39}=\frac{39-38}{38\times39}=\frac{39}{38\times39}-\frac{38}{38\times39}=\frac{1}{38}-\frac{1}{39}\)
\(\frac{1}{39\times40}=\frac{40-39}{39\times40}=\frac{40}{39\times40}-\frac{39}{39\times40}=\frac{1}{39}-\frac{1}{40}\)
\(\Rightarrow B=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{38}-\frac{1}{39}+\frac{1}{39}-\frac{1}{40}\)
\(B=\frac{1}{10}-\frac{1}{40}\)
\(B=\frac{3}{40}\)
3.
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(B=\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{38.39}+\frac{1}{39.40}\)
\(B=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{38}-\frac{1}{39}+\frac{1}{39}-\frac{1}{40}\)
\(B=\frac{1}{10}-\frac{1}{40}\)
\(B=\frac{3}{40}\)
????????????
Đặt S =\(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{1458}+\frac{1}{4374}\)
3S = \(3\times\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{1458}+\frac{1}{4374}\right)\)
3S \(=\frac{3}{2}+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{486}+\frac{1}{1458}\)
3S - S \(=\left(\frac{3}{2}+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{486}+\frac{1}{1458}\right)-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{1458}+\frac{1}{4374}\right)\)
2S = \(\frac{3}{2}+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{486}+\frac{1}{1458}-\frac{1}{2}-\frac{1}{6}-...-\frac{1}{1458}-\frac{1}{4374}\)
2S = \(\frac{3}{2}-\frac{1}{4374}\)
2S = \(\frac{3280}{2187}\)
\(\Rightarrow S=\frac{3280}{2187}:2=\frac{4373}{8748}\)