\(Q=\dfrac{3}{5.8}+\dfrac{11}{8.19}+\dfrac{12}{19.31}+\dfrac{70}{31.101}+\dfrac{90}{101.200}\)

\(=\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{101}+\dfrac{1}{101}-\dfrac{1}{200}\)

\(=\dfrac{1}{5}-\dfrac{1}{200}\)

\(=\dfrac{39}{200}\)

\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}+\frac{1}{31}-\frac{1}{101}+\frac{1}{101}-\frac{1}{200}\)

\(=\frac{1}{5}-\frac{1}{200}\)

\(=\frac{39}{200}\)

\(\frac{3}{5.8}+\frac{11}{8.19}+\frac{12}{19.31}+\frac{70}{31.101}+\frac{99}{101.200}\)

\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}+\frac{1}{31}-\frac{1}{101}+\frac{1}{101}-\frac{1}{200}\)

\(=\frac{1}{5}-\frac{1}{200}\)

\(=\frac{39}{200}\)

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!

<

\(=\dfrac{4}{1\cdot5}-\left(\dfrac{3}{5\cdot8}+\dfrac{11}{8\cdot19}+\dfrac{12}{19\cdot31}+\dfrac{70}{31\cdot101}+\dfrac{99}{101\cdot200}\right)\)

\(=1-\dfrac{1}{5}-\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{101}+\dfrac{1}{101}-\dfrac{1}{200}\right)\)

\(=1-\dfrac{1}{5}-\dfrac{1}{5}+\dfrac{1}{200}\)

\(=\dfrac{201}{200}-\dfrac{2}{5}=\dfrac{201-80}{200}=\dfrac{121}{200}\)

\(\Rightarrow A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}+\frac{1}{31}-\frac{1}{101}+\frac{1}{101}-\frac{1}{200}\)

\(\Rightarrow A=\frac{1}{5}-\frac{1}{200}\)

\(\Rightarrow A=\frac{39}{200}\)

Vì \(\frac{39}{200}<1\Rightarrow A<1\)

\(Cậu chép sai đề bài nha,ở ps 2 phải là 11/8*19\)

\(\frac{3}{5.8}+\frac{11}{8.19}+\)\(\frac{12}{19.31}+\frac{70}{31.101}\)\(+\frac{99}{101.200}\)

\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}+\frac{1}{31}-\frac{1}{101}+\frac{1}{101}\)\(-\frac{1}{201}\)

\(=\frac{1}{5}\)\(-\frac{1}{201}\)

\(=\frac{196}{1005}\)

\(\frac{3}{5.8}+\frac{11}{8.19}+\frac{12}{19.31}+\frac{70}{31.101}+\frac{99}{101.200}\)

\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}+\frac{1}{31}-\frac{1}{101}+\frac{1}{101}-\frac{1}{200}\)

\(=\frac{1}{5}-\frac{1}{200}=\frac{40-1}{200}=\frac{39}{200}\)

sr cậu

ở phần cuối mk làm nhầm. phải là:

\(\frac{1}{5}-\frac{1}{200}=\frac{39}{200}\)

haha