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18 tháng 10 2016

C=.................................

\(\Rightarrow3C=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)

\(\Rightarrow3C-C=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)

\(\Rightarrow2C=1-\frac{1}{3^{99}}\)

\(\Rightarrow C=\frac{1-\frac{1}{3^{99}}}{2}\)

Dễ thấy \(1-\frac{1}{3^{99}}< 1\Leftrightarrow\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\Leftrightarrow C< \frac{1}{2}\)

22 tháng 6 2023

\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)

\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)

\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)

\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)

\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)

6 tháng 12 2017

3C=1+1/3+1/32+........+1/321

3C-C=2C=1+1/3+1/32+........+1/321-(1/3+1+32+1/33+...+1/322)

2C=1-1/322

C=1/2-1/322/2<1/2

Vậy C<1/2

2 tháng 8 2018

\(30A=\frac{30^{32}+30}{30^{32}+1}=\frac{30^{32}+1+29}{30^{32}+1}=1+\frac{29}{30^{32}+1}\)

\(30B=\frac{30^{33}+30}{30^{33}+1}=\frac{30^{33}+1+29}{30^{33}+1}=1+\frac{29}{30^{33}+1}\)

Vì \(\frac{29}{30^{32}+1}>\frac{29}{30^{33}+1}\) nên \(1+\frac{29}{30^{32}+1}>1+\frac{29}{30^{33}+1}\Rightarrow30A>30B\Rightarrow A>B\)

Vậy \(A>B.\)

Chúc bạn học tốt.

18 tháng 10 2016

\(\frac{C}{3}=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\)

\(\frac{2C}{3}=C-\frac{C}{3}=\frac{1}{3}-\frac{1}{3^{100}}\)

\(2C=1-\frac{1}{3^{99}}\Rightarrow C=\frac{1}{2}-\frac{1}{2.3^{99}}< \frac{1}{2}\)

14 tháng 10 2015

ta có: 2B=\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+..+\frac{1}{2^{97}}+\frac{1}{2^{98}}\)

B=\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+..+\frac{1}{2^{98}}+\frac{1}{2^{99}}\)

=>2B-B=\(1-\frac{1}{2^{99}}\)

mà 1/2^99>0 nên B<1 (đpcm)

31 tháng 10 2021

 Mk săpp thi rồi nên hơi nhiều bài mong mn giúp mk !!!

31 tháng 10 2021

\(1,\\ a,3^{2^3}=3^8>3^6=\left(3^2\right)^3\\ b,\left(-8\right)^9=\left(-2\right)^{27}< \left(-2\right)^{25}=\left(-32\right)^5\\ c,2^{21}=8^7< 9^7=3^{14}\\ 2,\)

\(a,\) Áp dụng tcdtsbn:

\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)

\(b,\) Sửa: \(\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow a=bk;c=dk\)

\(\Leftrightarrow\dfrac{ab}{cd}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2};\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\dfrac{b^2}{d^2}\\ \LeftrightarrowĐpcm\)