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Bài 1: \(\left(\frac{-1}{16}\right)^{100}=\frac{1}{\left(2^4\right)^{100}}=\frac{1}{2^{400}}>\frac{1}{2^{500}}=\left(\frac{-1}{2}\right)^{500}.\)
Bài 2: \(100^{99}+1>100^{68}+1\Rightarrow\frac{1}{100^{99}+1}< \frac{1}{100^{68}+1}\Rightarrow\frac{-99}{100^{99}+1}>\frac{-99}{100^{68}+1}\)
\(\Rightarrow100+\frac{-99}{100^{99}+1}>100+\frac{-99}{100^{68}+1}\Rightarrow\frac{100^{100}+1}{100^{99}+1}>\frac{100^{69}+1}{100^{68}+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
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\(\frac{1}{100^2}< \frac{1}{99.100}\)
Nên \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+......+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{99.100}\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+......+\frac{1}{100^2}^2< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{99}-\frac{1}{100}\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+......+\frac{1}{100^2}^2^2< 1-\frac{1}{100}=\frac{99}{100}\)
\(\frac{99}{100}\)> \(\frac{3}{4}\)thì sao mà so sánh được
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1
\(A=\frac{2019^{2019}+1}{2019^{2020}+1}< \frac{2019^{2019}+1+2018}{2019^{2020}+1+2018}=\frac{2019^{2019}+2019}{2019^{2020}+2019}=\frac{2019\left(2019^{2018}+1\right)}{2019\left(2019^{2019}+1\right)}\)
\(=\frac{2019^{2018}+1}{2019^{2019}+1}\)
2
\(M=\frac{100^{101}+1}{100^{100}+1}< \frac{100^{101}+1+99}{100^{100}+1+99}=\frac{100^{101}+100}{100^{100}+100}=\frac{100\left(100^{100}+1\right)}{100\left(100^{99}+1\right)}\)
\(=\frac{100^{100}+1}{100^{99}+1}=N\)
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ta có: \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\)
\(\Rightarrow\frac{1}{3}A=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{100}}+\frac{1}{3^{101}}\)
\(\Rightarrow A-\frac{1}{3}A=\frac{1}{3}-\frac{1}{3^{101}}< \frac{1}{3}\)
\(\Rightarrow\frac{2}{3}A< \frac{1}{3}\)
\(\Rightarrow A< \frac{1}{3}:\frac{2}{3}\)
\(\Rightarrow A< \frac{1}{2}\)
Cho \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}};B=\frac{1}{2}\).so sánh A và B
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Lời giải:
$A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}$
$3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}$
$\Rightarrow 3A-A=1-\frac{1}{3^{100}}$
$\Rightarrow 2A=1-\frac{1}{3^{100}}<1$
$\Rightarrow A< \frac{1}{2}$
$\Rightarrow A< B$
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Ta có :
\(\frac{1^{500}}{2}=\frac{1}{2}\)
\(\frac{1^{300}}{3}=\frac{1}{3}\)
Mà 3>2
\(\Rightarrow\frac{1}{2}>\frac{1}{3}\)
Hay \(\frac{1^{500}}{2}>\frac{1^{300}}{3}\)
ta có : (1/2)^500 = (1/2^5)^100 = (1/32)^100
(1/15)^100 > (1/32)^100 suy ra : (1/15)^100 > (1/2)^500
cho tui nha