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\(\frac{3}{5.8}+\frac{11}{8.19}+\frac{12}{19.31}+\frac{70}{31.101}+\frac{99}{101.200}\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}+\frac{1}{31}-\frac{1}{101}+\frac{1}{101}-\frac{1}{200}\)
\(=\frac{1}{5}-\frac{1}{200}\)
\(=\frac{39}{200}\)
1/5-1/8+1/8-1/19+1/19-1/31+1/31-1/101+1/200=1/5-1/200=195/1000=39/200
\(\Rightarrow A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}+\frac{1}{31}-\frac{1}{101}+\frac{1}{101}-\frac{1}{200}\)
\(\Rightarrow A=\frac{1}{5}-\frac{1}{200}\)
\(\Rightarrow A=\frac{39}{200}\)
vì \(\frac{39}{200}< 1\) nên A < 1
\(A=\frac{3}{5.8}+\frac{11}{8.19}+\frac{12}{19.31}+\frac{70}{31.101}+\frac{99}{101.200}\)
Áp dụng công thức \(\frac{b-a}{a.b}=\frac{1}{a}-\frac{1}{b}\) với a < b và a khác b khác 0, ta có:
\(A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+...+\frac{1}{101}-\frac{1}{200}\\ =\frac{1}{5}-\frac{1}{200}\\ =\frac{40-1}{200}\\ =\frac{39}{200}\\ \frac{39}{200}< 1\\\Rightarrow A< 1\left(đpcm\right)\)
Chúc bạn học tốt!
\(Q=\dfrac{3}{5.8}+\dfrac{11}{8.19}+\dfrac{12}{19.31}+\dfrac{70}{31.101}+\dfrac{90}{101.200}\)
\(=\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{101}+\dfrac{1}{101}-\dfrac{1}{200}\)
\(=\dfrac{1}{5}-\dfrac{1}{200}\)
\(=\dfrac{39}{200}\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}+\frac{1}{31}-\frac{1}{101}+\frac{1}{101}-\frac{1}{200}\)
\(=\frac{1}{5}-\frac{1}{200}\)
\(=\frac{39}{200}\)
\(=2.\left(\frac{3}{5.8}+\frac{11}{8.19}+...+\frac{99}{101.200}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{200}\right)\)
\(=2.\frac{39}{200}=\frac{39}{100}\)
\(E=\frac{6}{5.8}+\frac{22}{8.19}+\frac{24}{19.31}+\frac{140}{31.101}+\frac{198}{101.200}\)
\(=2.\left(\frac{3}{5.8}+\frac{11}{8.19}+\frac{12}{19.31}+\frac{70}{31.101}+\frac{99}{101.200}\right)\)
\(=2\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}+\frac{1}{31}-\frac{1}{101}+\frac{1}{101}-\frac{1}{200}\right)\)
\(=2\left(\frac{1}{5}-\frac{1}{200}\right)\)
\(=2.\frac{39}{200}\)
\(=\frac{39}{100}\)
\(Cậu chép sai đề bài nha,ở ps 2 phải là 11/8*19\)
\(\frac{3}{5.8}+\frac{11}{8.19}+\)\(\frac{12}{19.31}+\frac{70}{31.101}\)\(+\frac{99}{101.200}\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}+\frac{1}{31}-\frac{1}{101}+\frac{1}{101}\)\(-\frac{1}{201}\)
\(=\frac{1}{5}\)\(-\frac{1}{201}\)
\(=\frac{196}{1005}\)
\(\frac{3}{5.8}+\frac{11}{8.19}+\frac{12}{19.31}+\frac{70}{31.101}+\frac{99}{101.200}\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}+\frac{1}{31}-\frac{1}{101}+\frac{1}{101}-\frac{1}{200}\)
\(=\frac{1}{5}-\frac{1}{200}=\frac{40-1}{200}=\frac{39}{200}\)