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\(B=\dfrac{1}{11}+\dfrac{1}{11^2}+\dfrac{1}{11^3}+...+\dfrac{1}{11^{99}}+\dfrac{1}{11^{100}}\\ 11B=1+\dfrac{1}{11}+\dfrac{1}{11^2}+...+\dfrac{1}{11^{98}}+\dfrac{1}{11^{99}}\\ 11B-B=1+\dfrac{1}{11}+\dfrac{1}{11^2}+...+\dfrac{1}{1^{99}0}-\dfrac{1}{11}-\dfrac{1}{11^2}-\dfrac{1}{11^3}-...-\dfrac{1}{11^{100}}\\ 10B=1-\dfrac{1}{11^{99}}\\ B=\dfrac{1-\dfrac{1}{11^{99}}}{10}\)
có : `1-1/(11^99)<1`
\(\Rightarrow\dfrac{1-\dfrac{1}{11^{99}}}{10}< \dfrac{1}{10}\)
hay `B<1/10`
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=-\dfrac{9}{10}\left(-\dfrac{10}{11}\right)\left(-\dfrac{11}{12}\right)...\left(-\dfrac{98}{99}\right)\left(-\dfrac{99}{100}\right)\)
Ta thấy C có \(\left(100-10\right):2+1=46\) thừa số nên số dấu âm là chẵn
Vậy \(C=\dfrac{9}{10}\cdot\dfrac{10}{11}\cdot\dfrac{11}{12}\cdot...\cdot\dfrac{99}{100}=\dfrac{9}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
mình ko đáng cái j linh tinh hết đây là các bài toán mà mình ko giải đc
b. (x-7)x+1-(x-7)x+11=0
(x-7)x+1.[1-(x-7)10]=0
=> (x-7)x+1=0 hoặc 1-(x-7)10=0
• (x-7)x+1= 0 => x-7=0 => x=7
• 1-(x-7)10=0=> (x-7)10=1=>x-7=1 hoặc x-7=-1 => x=8 hoặc x=6
Vậy x thuộc {6;7;8}
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\(A=\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\)
\(A=\frac{1}{2\cdot5}+\frac{1}{1\cdot11}+...+\frac{1}{10\cdot10}\)
\(A=\frac{1}{2}-\frac{1}{5}+\frac{1}{1}-\frac{1}{11}+...+\frac{1}{10}-\frac{1}{10}\)
\(A=\frac{1}{2}-\frac{1}{10}\)
\(A=\frac{2}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Xét C = \(\frac{1}{11}+\frac{1}{12}+....+\frac{1}{50}\)(40 số hạng)
=> C > \(\frac{1}{50}.40\)
=> C > \(\frac{4}{5}\)
Xét D = \(\frac{1}{51}+\frac{1}{52}+....+\frac{1}{100}\)(50 số hạng)
=> D > \(\frac{1}{100}.50\)
=> D > \(\frac{1}{2}\)
=> B = \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}>\frac{1}{10}+\frac{4}{5}+\frac{1}{2}\)
=> B > \(\frac{7}{5}\) > 1
=> B > 1 (Đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{9}{10}\times\frac{10}{11}\times\frac{11}{12}\times...\times\frac{99}{100}\)
\(A=\frac{9\times10\times11\times...99}{10\times11\times12\times...\times100}\)
\(\Rightarrow A=\frac{9}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(A=\left(\frac{1}{10}-1\right)\left(\frac{1}{11}-1\right)\left(\frac{1}{12}-1\right)...\left(\frac{1}{99}-1\right)\left(\frac{1}{100}-1\right)\)
\(A=\left(-\frac{9}{10}\right)\cdot\left(-\frac{10}{11}\right)\cdot\left(-\frac{11}{12}\right)\cdot....\cdot\left(-\frac{98}{99}\right)\left(-\frac{99}{100}\right)\)
\(A=\frac{\left(-9\right)\left(-10\right)\left(-11\right)...\left(-98\right)\left(-99\right)}{10\cdot11\cdot12\cdot....\cdot99\cdot100}\)
\(A=\frac{9\cdot10\cdot11\cdot....\cdot98\cdot99}{10\cdot11\cdot12\cdot...\cdot99\cdot100}=\frac{9}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\)
= \(\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\right)\)
Thấy : \(\frac{1}{11}>\frac{1}{100}\)
\(\frac{1}{12}>\frac{1}{100}\)
...
\(\frac{1}{99}>\frac{1}{100}\)
Cộng từng vế : \(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}>\frac{1}{100}+...+\frac{1}{100}\)( 90 SH 1/100)
\(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}>\frac{9}{10}\)
=> \(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}>\frac{9}{10}+\frac{1}{10}\)
=> Tổng trên > 1
\(=\frac{9}{10}.\frac{10}{11}...\frac{99}{100}=\frac{9}{100}\)