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Ta có :
\(B=3+3^2+3^3+.....+3^{2015}\)
\(\Leftrightarrow3B=3^2+3^3+.........+3^{2015}+3^{2016}\)
\(\Leftrightarrow3B-B=\left(3^2+3^3+.....+3^{2016}\right)-\left(3+3^2+......+3^{2015}\right)\)
\(\Leftrightarrow2B=3^{2016}-3\)
\(\Leftrightarrow2B+3=3^{2016}\)
Lại có : \(2B+3=3^x\)
\(\Leftrightarrow3^{2016}=3^x\Leftrightarrow x=2016\)
Vậy...
a/ \(8^5=\left(2^3\right)^5=2^{15}\)và \(32^3=\left(2^5\right)^3=2^{15}\Rightarrow8^5=32^3\)
b/ \(27^4=\left(3^3\right)^4=3^{12}\) và \(9^6=\left(3^2\right)^6=3^{12}\Rightarrow27^4=9^6\)
c/ \(23^{17}-23^{16}=23^{16}\left(23-1\right)=22.23^{16}\)
\(23^{16}-23^{15}=23^{15}\left(23-1\right)=22.23^{15}\)
\(\Rightarrow22.23^{16}>22.23^{15}\Rightarrow23^{17}-23^{16}>23^{16}-23^{15}\)
d/ \(\frac{3^{2015}+1}{3^{2016}}=\frac{1}{3}+\frac{1}{3^{2016}}\) và \(\frac{3^{2016}+1}{3^{2017}+1}=\frac{3^{2017}+3}{3\left(3^{2017}+1\right)}=\frac{3^{2017}+1+2}{3\left(3^{2017}+1\right)}=\frac{1}{3}+\frac{2}{3}.\frac{1}{3^{2017}+1}\)
\(\frac{1}{3^{2016}}>\frac{1}{3^{2017}}>\frac{1}{3^{2017}+1}>\frac{2}{3}.\frac{1}{3^{2017}+1}\)
\(\Rightarrow\frac{3^{2015}+1}{3^{2016}}>\frac{3^{2016}+1}{3^{2017}+1}\)
Câu cuối phân tích tương tự
\(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
\(< \frac{1}{1}+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=\frac{1}{1}+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{1}+\frac{1}{1}=2\)
\(\Rightarrow\)\(A< 2\left(đpcm\right)\)
chúc bạn học tốt!!!
Bài 6 :
2S = 6 + 3 + 3/2 + ... + 3/2^8
2S = 6 - 3/2^9 + S
S = 6 - 3/2^9
Vậy S = 6 - 3/2^9
Bài 7 :
Ta có :
A = 1/1 + 1/2^2 + 1/3^2 + ... + 1/50^2 < 1 + 1/(1x2) + 1/(2x3) + ... + 1/(49x50) = 1 + 1 - 1/50 < 1 + 1 = 2
=) A < 2
Vậy A < 2
Bài 8 :
Do A = 1 + 2/(2015^2014 - 1 ) và B = 1 + 2/(2015^2014 - 3 ) mà 2/(2015^2014 -1) < 2/(2015^2014 - 3 )
=) A < B
Vậy A < B
Bài 9:
Do 196/197 > 196/(197+198) và 197/198 > 197/(197+198)
=) A > B
Vậy A > B
Mấy bài dễ u tự giải quyết nha
3) \(\dfrac{2013}{2014}+\dfrac{2014}{2015}+\dfrac{2015}{2013}\)
\(=\left(1-\dfrac{1}{2014}\right)+\left(1-\dfrac{1}{2015}\right)+\left(1+\dfrac{2}{2013}\right)\)
\(=3+\dfrac{2}{2013}-\dfrac{1}{2014}-\dfrac{1}{2015}\)
\(=3+\left(\dfrac{1}{2013}-\dfrac{1}{2014}\right)+\left(\dfrac{1}{2013}-\dfrac{1}{2015}\right)>3\)
\(B=3+3^2+3^3+...+3^{2014}+3^{2015}\)
\(3B=3^2+3^3+3^4+...+3^{2015}+3^{2016}\)
\(3B-B=\left(3^2+3^3+...+3^{2016}\right)-\left(3+3^2+...+3^{2015}\right)\)
\(2B=3^{2016}-3\). \(\Leftrightarrow3^{2016}-1+3=3^x\Leftrightarrow3^{2016}=3^x\Leftrightarrow x=2016\)
Vậy \(x=2016\)
B = 3 + 32 + 33 + ... + 32015
=> 3B = 32 + 33 + 34 + ... + 32016
Lấy 3B trừ B theo vế ta có
3B - B = (32 + 33 + 34 + ... + 32016) - ( 3 + 32 + 33 + ... + 32015)
2B = 32016 - 3
Khi đó 2B + 3 = 3x
<=> 32016 - 3 + 3 = 3x
=> 3x = 32016
=> x = 2016