chung minh 1/5+1/14+1/28+1/44+1/61+1/85+1/97 < 1/2
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1/5+1/14+1/28+1/44+1/61+1/85+1/97 =0,3683
1/2 = 0,5
0,3683<0,5 nên 1/5+1/14+1/28+1/44+1/61+1/85+1/97 < 1/2
Sai đề. Sửa đề :v
Cmr: \(\dfrac{1}{5}+\dfrac{1}{14}+\dfrac{1}{28}+\dfrac{1}{44}+\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{97}< \dfrac{1}{2}\)
Giải:
Đặt \(A=\dfrac{1}{5}+\dfrac{1}{14}+\dfrac{1}{28}+\dfrac{1}{44}+\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{97}\)
Ta có:
\(A=\dfrac{1}{5}+\left(\dfrac{1}{14}+\dfrac{1}{28}+\dfrac{1}{44}\right)+\left(\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{97}\right)\)
\(A< \dfrac{1}{5}\left(\dfrac{1}{14.3}\right)+\left(\dfrac{1}{61.3}\right)\)
\(A< \dfrac{1}{5}+\dfrac{3}{14}+\dfrac{3}{61}\)
\(A< \dfrac{1}{5}+\dfrac{3}{12}+\dfrac{1}{20}\)
\(A< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}\)
\(\Rightarrow A< \dfrac{1}{2}\)
Vậy \(\dfrac{1}{5}+\dfrac{1}{14}+\dfrac{1}{28}+\dfrac{1}{44}+\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{97}< \dfrac{1}{2}\) \((đpcm)\)
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Ta có \(\frac{1}{5}=\frac{1}{5}\)
\(\frac{1}{14}< \frac{1}{10};\frac{1}{28}< \frac{1}{10}\)
\(\frac{1}{44}< \frac{1}{40};\frac{1}{61}< \frac{1}{40};\frac{1}{85}< \frac{1}{40};\frac{1}{97}< \frac{1}{40}\)
\(\Rightarrow\frac{1}{5}+\frac{1}{14}+\frac{1}{28}+\frac{1}{44}+\frac{1}{61}+\frac{1}{85}+\frac{1}{97}< \frac{1}{5}+\frac{1}{10}+\frac{1}{10}+\frac{1}{40}+\frac{1}{40}+\frac{1}{40}+\frac{1}{40}=\frac{1}{5}+\frac{1}{5}+\frac{1}{10}=\frac{5}{10}=\frac{1}{2}\)\(\Rightarrow A< \frac{1}{2}\)
Đặt \(\frac{1}{5}+\frac{1}{14}+\frac{1}{28}+\frac{1}{44}+\frac{1}{61}+\frac{1}{85}+\frac{1}{97}=A\)
Ta có : \(A=\frac{1}{5}+\left(\frac{1}{14}+\frac{1}{28}+\frac{1}{44}\right)+\left(\frac{1}{61}+\frac{1}{85}+\frac{1}{97}\right)\)
\(A< \frac{1}{5}\left(\frac{1}{14.3}\right)+\left(\frac{1}{61.3}\right)\)
\(A< \frac{1}{5}+\frac{3}{14}+\frac{3}{61}\)
\(A< \frac{1}{5}+\frac{3}{12}+\frac{1}{20}\)
\(A< \frac{1}{2}\left(ĐPCM\right).\)
Đặt 1/5 + 1/14 + 1/28 + 1/44 + 1/61 + 1/85 + 1/97 = A
A = 1/5 + (1/14 + 1/28 + 1/44) + (1/61 + 1/85 + 1/97)
A < 1/5 + (1/14 * 3) + (1/61 * 3)
A < 1/5 + 3/14 + 3/61
A < 1/5 + 3/12 + 3/60
A < 1/2 (đpcm)