ChoM=1+1/2+1/3+...+1/[(2^100)-1]
Cm M<50
Những câu hỏi liên quan
\(ChoM+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+....+\frac{1}{1+2+3+4+...+59}.\)
Chứng minh rằng \(M< \frac{2}{3}\)
ChoM=1/2.3/4.5/6.......99/100
N=2/3.4/5.6/7......100/101
a)Chứng minh M<N
b)M.N=?
1.tim x nguyên đe 3.x chia het cho x-2 2 chom=1+1/2+1/3+......+1/2100-1a)chưng minh m< 100 b)chung minh m< 50. giup minh voi cac ban oi-khẩn cấp!
ChoM=-2/3x^2y^3+4-1/2xy và N=-7+2/3x^2y^3Tính M+N;M-N;N-M
\(M=-\dfrac{2}{3}x^2y^3+4-\dfrac{1}{2}xy=-\dfrac{2}{3}x^2y^3-\dfrac{1}{2}xy+4\)
\(N=-7+\dfrac{2}{3}x^2y^3=\dfrac{2}{3}x^2y^3-7\)
\(M+N=-\dfrac{2}{3}x^2y^3-\dfrac{1}{2}xy+4+\dfrac{2}{3}x^2y^3-7\)
\(=-\dfrac{1}{2}xy-3\)
\(M-N=-\dfrac{2}{3}x^2y^3-\dfrac{1}{2}xy+4-\left(\dfrac{2}{3}x^2y^3-7\right)\)
\(=-\dfrac{2}{3}x^2y^3-\dfrac{1}{2}xy+4-\dfrac{2}{3}x^2y^3+7\)
\(=-\dfrac{4}{3}x^2y^3-\dfrac{1}{2}xy+11\)
\(N-M=-\left(M-N\right)\)
\(=-\left(-\dfrac{4}{3}x^2y^3-\dfrac{1}{2}xy+11\right)\)
\(=\dfrac{4}{3}x^2y^3+\dfrac{1}{2}xy-11\)
\(Ayumu\)
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ChoM=1+3+32+33.....+319.Chứng minh rằng M không chia hết cho 13
\(ChoM=1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{999}\\ N=\frac{1}{1.999}+\frac{1}{3.997}+...+\frac{1}{997.3}+\frac{1}{999.1}\\ Tính\frac{M}{N}\)
\(N=\frac{1}{1.999}+\frac{1}{3.997}+...+\frac{1}{997.3}+\frac{1}{999.1}\)
\(1000N=1+\frac{1}{999}+\frac{1}{3}+\frac{1}{997}+...+\frac{1}{997}+\frac{1}{3}+1\)
\(1000N=2\left[1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right]\)
\(N=\frac{1}{50}\left[1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right]\)
\(\frac{M}{N}=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}}{\frac{1}{50}\left[1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right]}=\frac{1}{\frac{1}{50}}=50\)
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các bn nhanh lên nhé mk sẽ tick chom
1+2+3=???
![[A]ȵȟ•βê•ʠǔá♡](https://hoc24.vn/images/avt/avt1063100_256by256.jpg)
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cm: 1/2^2+1/3^2+1/4^2+...+1/100^2< 1
Vì \(\dfrac{1}{a}\left(a>1\right)< 1với\forall a\)
mà \(2^2;3^2;.....;100^2>1\)
\(=>\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< 1\)
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Đặt :
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+......+\dfrac{1}{100^2}\)
Ta có :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
.................
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(\Leftrightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+....+\dfrac{1}{100^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{99.100}\)
\(\Leftrightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Leftrightarrow A< 1-\dfrac{1}{100}< 1\left(đpcm\right)\)
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cm 1/2^2+1/3^2+1/4^2+...+1/100^2<1/3/4