so sánh
(\(\frac{-1}{16}\))^100 va (\(\frac{-1}{2}\))^50
K
Khách
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9 tháng 7 2016
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: \(100^{99}+1>100^{68}+1\Rightarrow\frac{1}{100^{99}+1}< \frac{1}{100^{68}+1}\Rightarrow\frac{-99}{100^{99}+1}>\frac{-99}{100^{68}+1}\)
\(\Rightarrow100+\frac{-99}{100^{99}+1}>100+\frac{-99}{100^{68}+1}\Rightarrow\frac{100^{100}+1}{100^{99}+1}>\frac{100^{69}+1}{100^{68}+1}\)
MN
3
11 tháng 7 2017
Cách1:Ta có:\(\left(\frac{1}{2}\right)^{50}< \left(\frac{1}{2}\right)^{40}=\left[\left(\frac{1}{2}\right)^4\right]^{10}=\left(\frac{1}{16}\right)^{10}\)
Vậy..................
Cách 2:Ta có:\(\left(\frac{1}{16}\right)^{10}=\left[\left(\frac{1}{2}\right)^4\right]^{10}=\left(\frac{1}{2}\right)^{40}>\left(\frac{1}{2}\right)^{50}\)
Vậy......................
BT
11 tháng 7 2017
\(\left(\frac{1}{16}\right)^{10}=\left(\frac{1}{2^4}\right)^{10}=\frac{1^{10}}{2^{40}}=\frac{1}{2^{40}}\)
\(\left(\frac{1}{2}\right)^{50}=\frac{1^{50}}{2^{50}}=\frac{1}{2^{50}}\)
Do 250 > 240 => \(\frac{1}{2^{40}}>\frac{1}{2^{50}}\)
=> \(\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
19 tháng 5 2017
a, Ta có :
\(\left(\frac{1}{2}\right)^{50}=\left(\left(\frac{1}{2}\right)^5\right)^{10}=\left(\frac{1}{32}\right)^{10}\)
bạn so sánh nha :)
b,
T/c : \(99^{20}=\left(\left(99\right)^2\right)^{10}=9801^{10}\)
tiếp đây thì bạn tự làm nha có gì k hiểu ibx mk
NH
0
TH
1
10 tháng 4 2018
\(D=\frac{100^{15}+1}{100^{16}+1}\)
\(\Rightarrow D=\frac{100.\left(100^{15}+1\right)}{100.\left(100^{16}+1\right)}\)
\(\Rightarrow D=\frac{100^{16}+100}{100^{17}+100}\)
Vì \(\forall a;b\inℕ^∗;a< b;b\ne0\Rightarrow\frac{a}{b}< \frac{a+m}{b+m}\)
\(\Rightarrow C=\frac{100^{16}+1}{100^{17}+1}< \frac{100^{16}+1+99}{100^{17}+1+99}\)
\(\Rightarrow C< \frac{100^{16}+100}{100^{17}+100}=\frac{100^{15}+1}{100^{16}+1}\)
\(\Rightarrow C< D\)
1 tháng 5 2019
\(A=\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+\frac{1}{100}+\frac{1}{144}+\frac{1}{196}\)
\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\)
\(A=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{7^2}\right)\)
\(< \frac{1}{2^2}\left(1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{6}-\frac{1}{7}\right)\)
\(=\frac{1}{2^2}\left(1-\frac{1}{7}\right)\)
\(=\frac{1}{2^2}\cdot\frac{6}{7}\)
\(=\frac{3}{14}\)
\(< \frac{1}{2}\)
21 tháng 6 2017
\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+.....+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+....+\frac{1}{\sqrt{100}}\)
\(\Leftrightarrow\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+....+\frac{1}{\sqrt{100}}>100.\frac{1}{\sqrt{100}}=10.\)
S
7
TM
17 tháng 7 2016
Ta có:
\(\left(\frac{1}{16}\right)^{50}=\left[\left(\frac{1}{2}\right)^4\right]^{50}=\left(\frac{1}{2}\right)^{200}=\frac{1^{200}}{2^{200}}=\frac{1}{2^{200}}\)
\(\left(\frac{1}{2}\right)^{60}=\frac{1^{60}}{2^{60}}=\frac{1}{2^{60}}\)
Vì \(2^{200}>2^{60}\Rightarrow\frac{1}{2^{200}}< \frac{1}{2^{60}}\Rightarrow\left(\frac{1}{16}\right)^{50}< \left(\frac{1}{2}\right)^{60}\)
(-1/16)100= ((-1/2)4)100= (-1/2)4.100=(-1/2)400= (1/2)400
(-1/2)50=(1/2)50
Vì: (1/2)400> (1/2)50
=> (-1/16)100 > (-1/2)50
(-1/16)100= ((-1/2)4)100= (-1/2)4.100=(-1/2)400= (1/2)400
(-1/2)50=(1/2)50
Vì: (1/2)400> (1/2)50
=> (-1/16)100 > (-1/2)50