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Gọi biểu thức là A, ta có:
A = \(\frac{12}{1.4.7}+\frac{12}{4.7.10}+\frac{12}{7.10.13}+...+\frac{12}{54.57.60}=2\left(\frac{6}{1.4.7}+\frac{6}{4.7.10}+\frac{6}{7.10.13}+...+\frac{6}{54.57.60}\right)\)
A = \(2\left(\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+\frac{1}{7.10}-\frac{1}{10.13}+...+\frac{1}{54.57}-\frac{1}{57.60}\right)\)
A = \(2\left(\frac{1}{1.4}-\frac{1}{57.60}\right)=2\left(\frac{427}{1710}\right)=\frac{427}{855}< \frac{427}{854}=\frac{1}{2}\)
Vậy A < \(\frac{1}{2}\)(điều cần chứng minh)
Câu hỏi của thục hà - Toán lớp 6 - Học toán với OnlineMath
Em tham khảo nhé!
Đề sai hả
\(P=\frac{12}{1.4.7}+\frac{12}{4.7.10}+...+\frac{12}{54.57.60}\)
\(\Rightarrow\frac{1}{2}P=\frac{6}{1.4.7}+\frac{6}{4.7.10}+...+\frac{6}{54.57.60}\)
\(\Rightarrow\frac{1}{2}P=\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+...+\frac{1}{54.57}-\frac{1}{57.60}\)
\(\Rightarrow\frac{1}{2}P=\frac{1}{1.4}-\frac{1}{57.60}< \frac{1}{4}\)
\(\Rightarrow P< \frac{1}{4}.2=\frac{1}{2}\)
Đặt \(\frac{12}{1.4.7}+\frac{12}{4.7.10}+...+\frac{12}{54.57.60}=A\)
\(\frac{A}{2}=\frac{6}{1.4.7}+\frac{6}{4.7.10}+...+\frac{6}{54.57.60}\)
\(\frac{A}{2}=\frac{7-1}{1.4.7}+\frac{10-4}{4.7.10}+...+\frac{60-54}{54.57.60}\)
\(\frac{A}{2}=\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+...+\frac{1}{54.57}-\frac{1}{57.60}=\frac{1}{1.4}-\frac{1}{57.60}\)
\(A=\frac{1}{2}-\frac{1}{30.57}< \frac{1}{2}\)
P = 2*[ 6/(1*4*7) + 6/(4*7*10) + ... + 6/(54*57*60) ]
= 2*[ 1/(1*4) - 1/(4*7) + 1/(4*7) - 1/(7*10) + ... + 1/(54*57) -1/(57*60) ]
= 2*[ 1/(1*4) - 1/(57*60) ]
= 2* (427/1710)
= 427/855 <1/2
S = 1+ 1/2^2 + 1/3^2 +... + 1/100^2
1/2^2 < 1/(1*2)
1/3^2 < 1/(2*3)
...
1/100^2 < 1/(99*100)
==> 1/2^2 +1/3^2 +.., +1/100^2 < 1/(1*2) + 1/(2*3) + ... + 1/(99*100) = 1 -1/2 +1/2 - 1/3 +1/3 -1/4 +... - 1/100
=1 - 1/100 <1
==> 1/2^2 + 1/3^2 +... + 1/100^2 < 1
==> 1 + 1/2^2 + 1/3^2 +... +1/100^2 <2
\(P=\dfrac{12}{1\cdot4\cdot7}+\dfrac{12}{4\cdot7\cdot10}+\dfrac{12}{7\cdot10\cdot13}+...+\dfrac{12}{54\cdot57\cdot60}\)
\(P=\dfrac{12}{6}\left(\dfrac{1}{1\cdot4}-\dfrac{1}{4\cdot7}+\dfrac{1}{4\cdot7}-\dfrac{1}{7\cdot10}+...+\dfrac{1}{54\cdot57}-\dfrac{1}{57\cdot60}\right)\)
\(P=2\left(\dfrac{1}{1\cdot4}-\dfrac{1}{57\cdot60}\right)\)
\(P=\dfrac{2}{4}-\dfrac{2}{57\cdot60}=\dfrac{1}{2}-\dfrac{1}{57\cdot30}\)
\(\Rightarrow P< \dfrac{1}{2}\)
\(\frac{2}{1.4.7}+\frac{2}{4.7.10}+...+\frac{2}{58.61.64}\)
\(=\frac{1}{3}.\left(\frac{6}{1.4.7}+\frac{6}{4.7.10}+...+\frac{6}{58.61.64}\right)\)
\(=\frac{1}{3}.\left(\frac{7 - 1}{1.4.7}+\frac{10 - 4}{4.7.10}+...+\frac{64 - 58}{58.61.64}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+...+\frac{1}{58.61}-\frac{1}{61.64}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{61.64}\right)=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{3904}\right)=\frac{1}{3}.\frac{975}{3904}=\frac{325}{3904}\)
\(\text{Giải :}\)
\(\frac{2}{1.4.7}+\frac{2}{4.7.10}+...+\frac{2}{58.61.64}=\frac{1}{3}.\left(\frac{6}{1.4.7}+\frac{6}{4.7.10}+...+\frac{6}{58.61.64}\right)\)
\(=\frac{1}{3}.\left(\frac{7-1}{1.4.7}+\frac{10-4}{4.7.10}+...+\frac{64-58}{58.61.64}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+...+\frac{1}{58.61}-\frac{1}{61.64}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{61.64}\right)=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{3904}\right)=\frac{1}{3}.\frac{975}{3904}=\frac{325}{3904}\)
\(\text{#Hok tốt!}\)
\(\frac{2}{1.4.7}+\frac{2}{4.7.10}+...+\frac{2}{58.61.64}\)
\(=\frac{1}{3}\left(\frac{6}{1.4.7}+\frac{6}{4.7.10}+...+\frac{6}{58.61.64}\right)\)
\(=\frac{1}{3}\left(\frac{7-1}{1.4.7}+\frac{10-4}{4.7.10}+...+\frac{64-58}{58.61.64}\right)\)
\(=\frac{1}{3}\left(\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+...+\frac{1}{58.61}-\frac{1}{61.64}\right)\)
\(=\frac{1}{3}\left(\frac{1}{1.4}-\frac{1}{61.64}\right)\)
\(=\frac{325}{3904}\)
a) \(\frac{1}{12}+\frac{3}{15}+\frac{11}{12}+\frac{1}{71}-\frac{12}{10}=\left(\frac{1}{12}+\frac{11}{12}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+\frac{1}{71}\)
\(=\frac{12}{12}+0+\frac{1}{71}=1+\frac{1}{71}=1\frac{1}{71}=\frac{72}{71}\)
b) \(\frac{2}{3}-4\left(\frac{1}{2}+\frac{3}{4}\right)=\frac{2}{3}-4.\frac{5}{4}=\frac{2}{3}-5=\frac{2}{3}-\frac{15}{3}=-\frac{13}{3}\)
c) \(\frac{-4}{13}.\frac{3}{17}+\frac{-12}{13}.\frac{4}{7}+\frac{4}{13}=\frac{4}{13}.\frac{-3}{17}+\frac{4}{13}.\frac{-12}{17}+\frac{4}{13}.1\)
\(=\frac{4}{13}\left(\frac{-3}{17}+\frac{-12}{17}+1\right)=\frac{4}{13}\left(\frac{-15}{17}+\frac{17}{17}\right)=\frac{4}{13}.\frac{2}{17}=\frac{8}{221}\)
d) \(\frac{10^3+2.5+5^3}{55}=\frac{1000+10+125}{55}=\frac{1135}{55}=\frac{227}{11}\)