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\(\frac{1}{3^2}<\frac{1}{3.4}\)
\(\frac{1}{4^2}<\frac{1}{4.5}\)
\(\frac{1}{5^2}<\frac{1}{5.6}\)
\(...\)
\(\frac{1}{100^2}<\frac{1}{100.101}\)
\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}<\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{100.101}\)
\(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}<\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{100}-\frac{1}{101}\)
\(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}<\frac{1}{3}-\frac{1}{101}\)
Mà \(\frac{1}{3}<\frac{1}{2}\) nên \(\frac{1}{3}-\frac{1}{101}<\frac{1}{2}\)
hay \(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}<\frac{1}{2}\)
Đặt A=1/3^2+1/4^2+1/5^2+...+1/100^2
Suy raA<1/2*3+1/3*4+1/4*5+..+1/99*100
A<1/2-1/100<1/2
Ta có điều phải chứng minh.
1) Tìm x
\(\frac{11}{2}.x+\frac{1}{3}.x=1\)
\(\Rightarrow x\left(\frac{11}{2}+\frac{1}{3}\right)=1\)
\(\Rightarrow x\left(\frac{33}{6}+\frac{2}{6}\right)=1\)
\(\Rightarrow x.\frac{35}{6}=1\)
\(\Rightarrow x=\frac{6}{35}\)
2) So sánh
\(\frac{59}{40}< \frac{50}{31}\)( cái này bạn quy đồng là ra, mik chỉ ghi kq, bạn tự tính )
3)\(\frac{1}{3}+\frac{4}{7}-\frac{5}{14}-\frac{1}{2}-\frac{2}{3}\)
\(=\left(\frac{1}{3}-\frac{2}{3}\right)+\left(\frac{4}{7}-\frac{5}{14}\right)-\frac{1}{2}\)
\(=-\frac{1}{3}+\frac{3}{14}-\frac{1}{2}\)
\(=-\frac{13}{21}\)
1)\(\frac{11}{2}.x+\frac{1}{3}.x=1\)
\(x.\left(\frac{11}{2}+\frac{1}{3}=1\right)\)
\(x.\frac{35}{6}=1\)
\(x=1:\frac{35}{6}\)
\(x=\frac{6}{35}\)
2) Ta có:
\(\frac{59}{40}=\frac{1829}{1240}\)
\(\frac{50}{31}=\frac{2000}{1240}\)
Vì \(2000>1829\Rightarrow\frac{2000}{1240}>\frac{1829}{1240}\Rightarrow\frac{50}{31}>\frac{59}{40}\)
3)\(\frac{1}{3}+\frac{4}{7}-\frac{5}{14}-\frac{1}{2}-\frac{2}{3}\)
\(=\left(\frac{1}{3}-\frac{2}{3}\right)+\left(\frac{4}{7}-\frac{5}{14}-\frac{1}{2}\right)\)
\(=-\frac{1}{3}+\left(\frac{8}{14}-\frac{5}{14}-\frac{7}{14}\right)\)
\(=\frac{-1}{3}+\frac{-4}{14}\)
\(=\frac{-1}{3}+\frac{-2}{7}\)
\(=\frac{-7}{21}+\frac{-6}{21}\)
\(=\frac{-13}{21}\)
\(S=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{20^2}\)
\(S=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{20^2}<\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)
\(S<\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)
\(S<\frac{1}{2}-\frac{1}{20}<\frac{1}{2}\)
Vậy \(S<\frac{1}{2}\)
câu 2:đặt B=1/1*2+1/2*3+...+1/2007*2008
ta có:\(A=3\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2008^2}\right)\)
\(\frac{A}{3}=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2008^2}
câu 2:đặt B=1/1*2+1/2*3+...+1/2007*2008
\(A=3\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2008^2}\right)\)
\(\frac{A}{3}=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2008^2}\)\( (1)
mà \(B=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2007.2008}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2007}-\frac{1}{2008}\)
\(=1-\frac{1}{2008}\)<1 (2)
mà 1<3 (3)
từ (1),(2) và (3)=> đpcm
\(a)\frac{2}{3}x-\frac{1}{2}=\frac{5}{12}\)
\(\Rightarrow\frac{2}{3}x=\frac{5}{12}+\frac{1}{2}=\frac{11}{12}\)
\(\Rightarrow x=\frac{11}{12}:\frac{2}{3}=\frac{11}{8}\)
\(b)\left(2\frac{4}{5}x-50\right):\frac{2}{3}=51\)
\(\Rightarrow\frac{14}{5}x-50=51.\frac{2}{3}=34\)
\(\Rightarrow\frac{14}{5}x=34+50=84\)
\(\Rightarrow x=84:\frac{14}{5}=30\)
a) 2/3.x - 1/2 = 5/12
2/3.x = 5/12 + 1/2
2/3.x = 11/12
x = 11/12 : 2/3
x = 11/8
b) \(\left(2\frac{4}{5}.x-50\right):\frac{2}{3}=51\)
\(\frac{14}{5}.x-50=51.\frac{2}{3}\)
\(\frac{14}{5}.x-50=34\)
\(\frac{14}{5}.x=34+50\)
\(\frac{14}{5}.x=84\)
\(x=84:\frac{14}{5}\)
\(x=30\)
Ta có: A = \(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}<\frac{1}{1^2}+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(\Rightarrow\) A < \(1+\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(\Rightarrow\) A < \(1+\left(1-\frac{1}{50}\right)\)
\(\Rightarrow\) A < 1 + 49/50
Mà 1+49/50 < 2 nên A < 1+49/50 < 2
\(\Rightarrow\) A < 2