so sánh A=1/5^1+1/5^2+....+1/5^2016+1/5^2017 với 1/4
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TP
0
26 tháng 11 2021
a.
\(x=9-\dfrac{1}{\sqrt{\dfrac{9-4\sqrt{5}}{4}}}+\dfrac{1}{\sqrt{\dfrac{9+4\sqrt{5}}{4}}}\\ x=9-\dfrac{1}{\dfrac{\sqrt{5}-2}{2}}+\dfrac{1}{\dfrac{\sqrt{5}+2}{2}}\\ x=9-\left(\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{\sqrt{5}+2}\right)=9-8=1\\ \Rightarrow f\left(x\right)=f\left(1\right)=\left(1-1+1\right)^{2016}=1\)
26 tháng 11 2021
c.
\(=\sin x\cdot\cos x+\dfrac{\sin^2x}{1+\dfrac{\cos x}{\sin x}}+\dfrac{\cos^2x}{1+\dfrac{\sin x}{\cos x}}\\ =\sin x\cdot\cos x+\dfrac{\sin^2x}{\dfrac{\sin x+\cos x}{\sin x}}+\dfrac{\cos^2x}{\dfrac{\sin x+\cos x}{\cos x}}\\ =\sin x\cdot\cos x+\dfrac{\sin^3x}{\sin x+\cos x}+\dfrac{\cos^3x}{\sin x+\cos x}\\ =\sin x\cdot\cos x+\dfrac{\left(\sin x+\cos x\right)\left(\sin^2x-\sin x\cdot\cos x+\cos^2x\right)}{\sin x+\cos x}\\ =\sin x\cdot\cos x-\sin x\cdot\cos x+\sin^2x+\cos^2x\\ =1\)
N
2
LQ
1
VC
1
KN
20 tháng 4 2019
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{2015}{2016}\)
\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2013}{2014}\)
\(\Rightarrow A>\frac{1.2.3...2013}{2.3.4...2014}\)
\(\Rightarrow A>\frac{1}{2014}>\frac{1}{2017}\)
Vậy \(A>\frac{1}{2017}\left(đpcm\right)\)
LN
1
19 tháng 7 2018
\(A=\frac{5^{2016}+1}{5^{2017}+1}\)
\(\Rightarrow5A=\frac{5^{2017}+5}{5^{2017}+1}=1+\frac{4}{5^{2017}+1}\)
\(B=\frac{5^{2017}+1}{5^{2018}+1}\)
\(\Rightarrow5B=\frac{5^{2018}+5}{5^{2018}+1}=1+\frac{4}{5^{2018}+1}\)
Do \(\frac{4}{5^{2018}+1}< \frac{4}{5^{2017}+1}\)
\(\Rightarrow5A>5B\Leftrightarrow A>B\)
LN
0
DL
4 tháng 5 2018
1) Đặt dãy trên là \(A\)
Theo bài ra ta có :
\(A=\frac{1}{3.3}+\frac{1}{4.4}+\frac{1}{5.5}+\frac{1}{6.6}+...+\frac{1}{100.100}\)
\(\Rightarrow A< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{100}< \frac{1}{2}\left(đpcm\right)\)
2) \(A=\frac{5^{2018}-2017+1}{5^{2018}-2017}=\frac{5^{2018}-2017}{5^{2018}-2017}+\frac{1}{5^{2018}-2017}=1+\frac{1}{5^{2018}-2017}\)( 1 )
\(B=\frac{5^{2018}-2019+1}{5^{2018}-2019}=\frac{5^{2018}-2019}{5^{2018}-2019}+\frac{1}{5^{2018}-2019}=1+\frac{1}{5^{2018}-2019}\)( 2 )
Từ ( 1 ) và ( 2 ) \(\Rightarrow\)\(A=1+\frac{1}{5^{2018}-2017}< 1+\frac{1}{5^{2018}-2019}=B\)
\(\Rightarrow A< B\)
Vậy \(A< B.\)
4 tháng 5 2018
1) Ta có B =
\(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\) < \(\frac{1}{1.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)= \(\frac{99}{100}\)
=> B < 1 ( chứ không phải \(\frac{1}{2}\) bạn nhé)
Sai thì thôi chứ mk chỉ làm rờ thôi
\(A=\frac{1}{5}+\frac{1}{5^2}+...............+\frac{1}{5^{2016}}+\frac{1}{5^{2017}}\)
\(\Rightarrow5A=1+\frac{1}{5}+...................+\frac{1}{5^{2015}}+\frac{1}{5^{2016}}\)
\(\Rightarrow5A-A=\left(1+\frac{1}{5}+...........+\frac{1}{5^{2015}}+\frac{1}{5^{2016}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+.............+\frac{1}{5^{2016}}+\frac{1}{5^{2017}}\right)\)
\(\Rightarrow4A=1-\frac{1}{5^{2017}}\)
\(\Rightarrow A=\left(1-\frac{1}{5^{2017}}\right):4=\left(1-\frac{1}{5^{2017}}\right).\frac{1}{4}=\frac{1}{4}-\frac{1}{5^{2017}.4}< \frac{1}{4}\)
\(\Rightarrow A< \frac{1}{4}\)
Vậy \(A< \frac{1}{4}\)
Chúc bạn học tốt
\(A=\frac{1}{5}+\frac{1}{5^2}+....+\frac{1}{5^{2016}}+\frac{1}{5^{2017}}\)
=>5A = \(1+\frac{1}{5}+...+\frac{1}{5^{2015}}+\frac{1}{5^{2016}}\)
=> 5A -A = \(\left(1+\frac{1}{5}+...+\frac{1}{5^{2015}}+\frac{1}{5^{2016}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2016}}+\frac{1}{5^{2017}}\right)\)
=> 4A = \(1-\frac{1}{5^{2017}}\)
=> \(A=\frac{1-\frac{1}{5^{2017}}}{4}=\frac{1}{4}-\frac{1}{5^{2017}}< \frac{1}{4}\)