M=1/2.2/3.3/4.4/5.......99/100 .Cmr 1/15<M<1/110
Những câu hỏi liên quan
A = 1+ 1/ 2.2 + 1 / 3.3 +1/ 4.4 + ....+1/99.99 + 1/ 100. 100
CMR: 1/2.2 +1/3.3+1/4.4+....+1/100.100<1
Mik cần gấp
Ta có : \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+....+\frac{1}{100.100}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}< 1\)(đpcm)
+)Ta thấy:\(\frac{1}{2.2}< \frac{1}{1.2}\)
\(\frac{1}{3.3}< \frac{1}{2.3}\)
............................
..............................
\(\frac{1}{100.100}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2.2}+\frac{1}{3.3}+...............+\frac{1}{100.100}< \frac{1}{1.2}+\frac{1}{2.3}+............+\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2.2}+\frac{1}{3.3}+...............+\frac{1}{100.100}< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..............+\frac{1}{99}-\frac{1}{100}< 1\)
\(\Rightarrow\frac{1}{2.2}+\frac{1}{3.3}+.............+\frac{1}{100.100}< 1\left(\text{Đ}PCM\right)\)
Chúc bạn học tốt
2/5<1/2.2+1/3.3+1/4.4+...+1/9.9<8/9
1/2.2 + 1/3.3 + 1/4.4 + ... + 1/9.9
> 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/9.10
> 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/9 - 1/10
> 1/2 - 1/10
> 5/10 - 1/10
> 2/5 (1)
1/2.2 + 1/3.3 + 1/4.4 + ... + 1/9.9
< 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/8.9
< 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/8 - 1/9
< 1 - 1/9
< 8/9 (2)
Từ (1) và (2) => 2/5 < 1/2.2 + 1/3.3 + 1/4.4 + ... + 1/9.9 < 8/9
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cho a=1/2.3/4.5/6. ... .99/100;b=2/3.4/5. ... .100/101;c=1/2.2/3.3/4. ... 98/99 a, so sánh a, b,c b,chứng minh a.c<a^2<1/100 c,chứng minh 1/15<a<1/10
A=1+2.2!+3.3!+4.4!+...+100.100!
1/2.2+1/3.3+1/4.4+....+1/100.100<1
1/2.2 < 1/1.2
1/3.3 < 1/2.3
..................
1/100.100 < 1/99.100
=> <
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Ta có: \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+....+\frac{1}{100.100}=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}\)
Vì \(\frac{1}{2^2}<\frac{1}{1.2}\)
\(\frac{1}{3^2}<\frac{1}{2.3}\)
\(\frac{1}{4^2}<\frac{1}{3.4}\)
.....
\(\frac{1}{100^2}<\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}<\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}<1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}<1\left(đpcm\right)\)
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1/2.2 < 1/1.2
1/3.3 < 1/2.3
..................
1/100.100 < 1/99.100
=> <
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tính : A= 1/2.2+1/3.3+1/4.4+...+1/9.9
B=2/3.3+2/5.5+2/7.7+...+2/2007.2007
C=1/4.4+1/6.6+1/8.8+...+1/2006.2006
Cho M=1/2.2/3.5/6......99/100 và N=2/3.3/4.5/6....100/101
M . N = ?
1/2.2 + 1/3.3 + 1/4.4 +...+ 1/99.99 + 1/100.100