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cho A= 1/1.4+1/3.8+1/5.12+...+1/99.200. chứng minh A<5/12
Cho A=1/1.4+1/3.8+1/5.12+...+1/99.200
CMR:A<5/12
Chứng minh : \(\frac{1}{1.4}+\frac{1}{3.8}+\frac{1}{5.12}+....+\frac{1}{99.200}<\frac{5}{12}\)
cho A=1/1.4+1/3.8+1/5.12+..+1/99.200 CMR A<5/12
chi tiết nha
Mọi người giải hộ mk nka, mình đang cần gấp. Thanks nhìu!
Cho A= 1/1.4+1/3.8+1/5.12+...+1/99.200
CMR: A<5/12
Ghi chi tiết hộ mk nha, thanks lần 2!
CMR:1/28+1/30+1/32+1/34+...+1/52=1/1.4+1/3.8+1/5.12+1/7.16+...+1/25.52
Chứng minh rằng:
A=3/1.4+3/2.6+3/3.8+..............+1/2012.1342<1,5
Chứng minh rằng: \(A=\dfrac{3}{1.4}+\dfrac{3}{2.6}+\dfrac{3}{3.8}+...+\dfrac{1}{2012.1342}< 1,5\)
\(A=\dfrac{3}{1\cdot4}+\dfrac{3}{2\cdot6}+\dfrac{3}{3\cdot8}+...+\dfrac{1}{2012\cdot1342}\\ =\dfrac{3}{1\cdot4}+\dfrac{3}{2\cdot6}+\dfrac{3}{3\cdot8}+...+\dfrac{3}{2012\cdot4026}\\ =\dfrac{6}{2\cdot4}+\dfrac{6}{4\cdot6}+\dfrac{6}{6\cdot8}+...+\dfrac{6}{4024\cdot4026}\\ =3\cdot\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{4024\cdot4026}\right)\\ =3\cdot\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{4024}-\dfrac{1}{4026}\right)\\ =3\cdot\left(\dfrac{1}{2}-\dfrac{1}{4026}\right)\\ =3\cdot\dfrac{1}{2}-3\cdot\dfrac{1}{4026}\\ =1,5-\dfrac{3}{4026}< 1,5\)
Vậy \(A< 1,5\left(đpcm\right)\)
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Chứng minh rằng \(\frac{3}{1.4}+\frac{3}{2.6}+\frac{3}{3.8}+...+\frac{1}{2012.1342}\)<1.5
\(\frac{3}{1.4}+\frac{3}{2.6}+\frac{3}{3.8}+...+\frac{1}{2012.1342}\)
\(=\frac{3}{1.4}+\frac{3}{2.6}+\frac{3}{3.8}+...+\frac{3}{2012.4026}\)
\(=\frac{6}{2.4}+\frac{6}{4.6}+\frac{6}{4.8}+...+\frac{6}{4024.4026}\)
\(=3\cdot\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{4024.4026}\right)\)
\(=3\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{4024}-\frac{1}{4026}\right)\)
\(=3\cdot\left(\frac{1}{2}-\frac{1}{4026}\right)\)
\(=3\cdot\frac{1}{2}-3\cdot\frac{1}{4026}\)
\(=1,5-\frac{3}{4026}< 1,5\)
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