Thực hiện pháp tính : \(\dfrac{4}{5}+\dfrac{-3}{5.8}+\dfrac{-11}{8.19}+\dfrac{-12}{19.31}+\dfrac{-70}{31.101}+\dfrac{-99}{101.200}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(Q=\dfrac{3}{5.8}+\dfrac{11}{8.19}+\dfrac{12}{19.31}+\dfrac{70}{31.101}+\dfrac{90}{101.200}\)
\(=\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{101}+\dfrac{1}{101}-\dfrac{1}{200}\)
\(=\dfrac{1}{5}-\dfrac{1}{200}\)
\(=\dfrac{39}{200}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}+\frac{1}{31}-\frac{1}{101}+\frac{1}{101}-\frac{1}{200}\)
\(=\frac{1}{5}-\frac{1}{200}\)
\(=\frac{39}{200}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{3}{5.8}+\frac{11}{8.19}+\frac{12}{19.31}+\frac{70}{31.101}+\frac{99}{101.200}\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}+\frac{1}{31}-\frac{1}{101}+\frac{1}{101}-\frac{1}{200}\)
\(=\frac{1}{5}-\frac{1}{200}\)
\(=\frac{39}{200}\)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\frac{3}{5.8}+\frac{11}{8.19}+\frac{12}{19.31}+\frac{70}{31.101}+\frac{99}{101.200}\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}+\frac{1}{31}-\frac{1}{101}+\frac{1}{101}-\frac{1}{200}\)
\(=\frac{1}{5}-\frac{1}{200}\)
\(=\frac{39}{200}\)
\(\frac{x+2}{3}=\frac{x-2}{2}\)
=> \(\left(x+2\right)2=3\left(x-2\right)\)
2x + 4 = 3x - 6
2x - 3x = -6 - 4
-x = -10
x = 10
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Rightarrow A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}+\frac{1}{31}-\frac{1}{101}+\frac{1}{101}-\frac{1}{200}\)
\(\Rightarrow A=\frac{1}{5}-\frac{1}{200}\)
\(\Rightarrow A=\frac{39}{200}\)
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!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(D=\dfrac{2}{3}.\dfrac{5}{5.8}+\dfrac{11}{8.19}+\dfrac{13}{19.32}\)
\(D=\dfrac{2.5}{2.5.8}+\dfrac{1}{19}.\dfrac{11}{8}+\dfrac{1}{19}.\dfrac{13}{32}\)
\(D=\dfrac{1}{12}+\dfrac{1}{19}.\left(\dfrac{11}{8}+\dfrac{13}{32}\right)\)
\(D=\dfrac{1}{12}+\dfrac{1}{19}.\left(\dfrac{44}{32}+\dfrac{13}{32}\right)\)
\(D=\dfrac{1}{12}+\dfrac{1}{19}.\dfrac{57}{32}\)
\(D=\dfrac{1}{12}+\dfrac{3}{32}\)
\(D=\dfrac{8}{96}+\dfrac{9}{96}\)
\(D=\dfrac{17}{96}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(D=\dfrac{2}{3}.\dfrac{5}{5.8}+\dfrac{11}{8.19}+\dfrac{13}{19.32}\)
\(=\dfrac{2.5}{3.(5.8)}+\dfrac{11}{8.19}+\dfrac{13}{19.32}\)
\(=\dfrac{10}{3.5.8}+\dfrac{11}{8.19}+\dfrac{13}{19.32}\)
\(=\dfrac{1}{3.4}+\dfrac{11}{8.19}+\dfrac{13}{19.32}\)
\(=\dfrac{17}{96}\)
\(=\dfrac{4}{1\cdot5}-\left(\dfrac{3}{5\cdot8}+\dfrac{11}{8\cdot19}+\dfrac{12}{19\cdot31}+\dfrac{70}{31\cdot101}+\dfrac{99}{101\cdot200}\right)\)
\(=1-\dfrac{1}{5}-\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{101}+\dfrac{1}{101}-\dfrac{1}{200}\right)\)
\(=1-\dfrac{1}{5}-\dfrac{1}{5}+\dfrac{1}{200}\)
\(=\dfrac{201}{200}-\dfrac{2}{5}=\dfrac{201-80}{200}=\dfrac{121}{200}\)