Tinh nhanh:
\(\frac{6}{2.5}+\frac{6}{5.8}+\frac{6}{8.11}+......+\frac{6}{299.302};\)
So sanh cac phan so sau ma khong can thuc hien cac phep tinh o mau
\(A=\frac{54.107-53}{53.107+54}\) \(B=\frac{135.269-133}{134.269+135}\)
K
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LN
31 tháng 1 2019
\(S=\frac{6}{2.5}+\frac{6}{5.8}+.......+\frac{6}{29.32}\)
\(S=2\left(\frac{3}{2.5}+\frac{3}{5.8}+......+\frac{3}{29.32}\right)\)
\(S=2\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+......+\frac{1}{29}-\frac{1}{32}\right)\)
\(S=2\left(\frac{1}{2}-\frac{1}{32}\right)\)
\(S=2.\frac{15}{32}\)
\(S=\frac{15}{16}< 1\RightarrowĐPCM\)
Vậy \(S=\frac{15}{16}\)
22 tháng 4 2017
a) Ý bạn là: \(S_1=\frac{3}{4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{40\cdot43}\)đúng không?
\(S_1=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)
\(S_1=1-\frac{1}{43}< 1\left(đpcm\right)\)
b) \(S_2=\frac{6}{2\cdot5}+\frac{6}{5.8}+\frac{6}{8\cdot11}+...+\frac{6}{29\cdot32}\)
=>\(\frac{S_2}{2}=\frac{3}{2\cdot5}+\frac{3}{5.8}+\frac{3}{8\cdot11}+...+\frac{3}{29\cdot32}\)
\(\frac{S_2}{2}=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{29}-\frac{1}{32}\)
\(\frac{S_2}{2}=\frac{1}{2}-\frac{1}{32}=\frac{16}{32}-\frac{1}{32}=\frac{15}{32}\)
=>\(S_2=\frac{15}{32}\cdot2=\frac{15}{16}< 1\left(đpcm\right)\)
NH
20 tháng 7 2016
\(a,A=\frac{3}{2}+\frac{3}{6}+\frac{3}{12}+\frac{3}{20}+...+\frac{3}{90}\)
\(A=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
\(A=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=3.\left(1-\frac{1}{10}\right)\)
\(A=3.\frac{9}{10}=\frac{27}{10}\)
\(b,B=\frac{2}{2.5}+\frac{2}{5.8}+\frac{2}{8.11}+\frac{2}{11.14}+\frac{2}{14.17}\)
\(B.\frac{3}{2}=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}\)
\(B.\frac{3}{2}=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)
\(B.\frac{3}{2}=\frac{1}{2}-\frac{1}{17}\)
\(B=\frac{15}{34}:\frac{3}{2}=\frac{5}{17}\)
10 tháng 5 2019
\(2x.\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}\right)=\frac{1}{21}\)
\(2x.\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}\right)=\frac{1}{21}\)
\(2x.\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{14}\right)=\frac{1}{21}\)
\(2x.\frac{1}{3}.\frac{3}{7}=\frac{1}{21}\)
\(2x.\frac{1}{7}=\frac{1}{21}\)
\(2x=\frac{1}{3}\)
\(x=\frac{1}{6}\)
vậy \(x=\frac{1}{6}\)
9 tháng 8 2018
a) 100 - 7 . (x - 5) = 58
7. (x - 5) = 100 - 58
7. (x - 5) = 42
x - 5 = 42 : 7
x - 5 = 6
x = 6 + 5
x = 11
b)\(x+\frac{1}{3}=\frac{7}{26}.\frac{13}{6}\)
\(x+\frac{1}{3}=\frac{7}{12}\)
\(x=\frac{7}{12}-\frac{1}{3}\)
\(x=\frac{3}{12}=\frac{1}{4}\)
SK
2
T
8 tháng 6 2019
#)Giải :
\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{98.101}\)
\(\Rightarrow3A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{99.101}\)
\(\Rightarrow3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{99}-\frac{1}{101}\)
\(\Rightarrow3A=\frac{1}{2}-\frac{1}{101}\)
\(\Rightarrow3A=\frac{99}{202}\)
\(\Leftrightarrow A=\frac{33}{202}\)
8 tháng 6 2019
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{101}\right)\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(A=\frac{1}{3}.\frac{99}{202}=\frac{33}{202}\)
PT
2
MN
0
TN
4
19 tháng 8 2020
\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{x\left(x+3\right)}=\frac{1}{6}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{1}{6}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{1}{6}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{1}{6}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{x+3}=-\frac{3}{10}\)
\(\Leftrightarrow1\cdot10=-3\left(x+3\right)\)
\(\Leftrightarrow10=-3x-9\)
\(\Leftrightarrow10+9=-3x\)
\(\Leftrightarrow19=-3x\)
\(\Leftrightarrow x=-\frac{19}{3}\)
Đề sai à -.-
19 tháng 8 2020
\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{x\left(x+3\right)}=\frac{1}{6}\)
=> \(\frac{1}{3}\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{1}{6}\)
=> \(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{1}{6}:\frac{1}{3}\)
=> \(\frac{1}{5}-\frac{1}{x+3}=\frac{1}{6}\cdot3=\frac{1}{2}\)
=> \(\frac{1}{x+3}=\frac{1}{5}-\frac{1}{2}=-\frac{3}{10}\)
=> \(10=-3\left(x+3\right)\)
=> 10 = -9x - 9
=> 10 + 9x + 9 = 0
=> 19 + 9x = 0
=> 9x = -19
=> x = -19/9
\(\frac{6}{2.5}+\frac{6}{5.8}+\frac{6}{8.11}+...+\frac{6}{299.302}\)
\(=2\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+..+\frac{3}{299.302}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{299}-\frac{1}{302}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{302}\right)=2.\frac{75}{151}=\frac{150}{151}\)
\(A=\frac{54.107-53}{53.107+54}=\frac{\left(53+1\right).107-53}{53.107+54}\)
\(=\frac{53.107+107-53}{53.107+54}=\frac{53.107+54}{53.107+54}=1\)
\(B=\frac{135.269-133}{134.269+135}=\frac{\left(134+1\right)269-133}{134.269+135}\)
\(=\frac{134.269+269-133}{134.269+135}=\frac{134.269+136}{134.269+135}>1\)
\(\Rightarrow B>A\)