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a) \(\left(\frac{2}{3}\right)^x=\left(\frac{4}{9}\right)^{50}\)
\(\Rightarrow\left(\frac{2}{3}\right)^x=\left(\frac{2^2}{3^2}\right)^{50}\)
\(\Rightarrow\left(\frac{2}{3}\right)^x=\left(\frac{2}{3}\right)^{100}\)
\(\Rightarrow x=100\)
Vậy x = 100
b) \(\left(\frac{2}{3}-x\right)^2=\frac{1}{36}\)
\(\Rightarrow\left(\frac{2}{3}-x\right)^2=\left(\frac{1}{6}\right)^2\)
\(\Rightarrow\frac{2}{3}-x=\frac{1}{6}\)
\(\Rightarrow x=\frac{2}{3}-\frac{1}{6}\)
\(\Rightarrow x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
2)
Ta có:
\(74^{m+1}+74^m=74^m.74^1+74^m=74^m.\left(74+1\right)=74^m.75⋮25\)
( vì \(75⋮25\) )
\(\Rightarrowđpcm\)
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\(\dfrac{9}{10}-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)
\(=\dfrac{9}{10}-\left(\dfrac{1}{90}+\dfrac{1}{72}+\dfrac{1}{56}+\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\right)\)
\(=\dfrac{9}{10}-\left(\dfrac{1}{9\cdot10}+\dfrac{1}{9\cdot8}+\dfrac{1}{7\cdot8}+\dfrac{1}{7\cdot6}+\dfrac{1}{5\cdot6}-\dfrac{1}{5\cdot4}-\dfrac{1}{3\cdot4}-\dfrac{1}{3\cdot2}-\dfrac{1}{1\cdot2}\right)\)
\(=\dfrac{9}{10}-\left(\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{8}-\dfrac{1}{9}+...+1-\dfrac{1}{2}\right)\)
\(=\dfrac{9}{10}-\left(1-\dfrac{1}{10}\right)\)
\(=\dfrac{9}{10}-\dfrac{9}{10}\)
\(=0\)
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Lời giải:
Ta có:
\(\frac{1}{3^2}=\frac{1}{3.3}< \frac{1}{2.3}\)
..........
\(\frac{1}{50^2}< \frac{1}{49.50}\)
Cộng theo vế:
\(B< \frac{1}{2^2}+\frac{1}{2.3}+...+\frac{1}{49.50}=\frac{1}{4}+\frac{3-2}{2.3}+....+\frac{50-49}{49.50}=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{50}< \frac{3}{4}\)
Ta có đpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
a -35/50 = -7/10
b 510/2805 = 2/11
c 119/126
B2
-2/3= -8/12 , -1/4= -3/12
-8/12<-3/12 nên -2/3<-1/4
b 2/3 5/6
12/18 và 15/18
12/18<15/18
nên 14/21<60/72
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(S=\dfrac{1}{7^2}-\dfrac{1}{7^4}+...+\dfrac{1}{7^{4n-2}}-\dfrac{1}{7^{4n}}+...+\dfrac{1}{7^{98}}-\dfrac{1}{7^{100}}\)
\(\Rightarrow\dfrac{S}{7^2}=\dfrac{1}{7^4}-\dfrac{1}{7^6}+...+\dfrac{1}{7^{100}}-\dfrac{1}{7^{102}}\)
\(\Rightarrow S+\dfrac{S}{7^2}=\left(\dfrac{1}{7^2}-\dfrac{1}{7^4}+...+\dfrac{1}{7^{98}}-\dfrac{1}{7^{100}}\right)+\left(\dfrac{1}{7^4}-\dfrac{1}{7^6}+...+\dfrac{1}{7^{100}}-\dfrac{1}{7^{102}}\right)\)
\(\Leftrightarrow\dfrac{50S}{49}=\dfrac{1}{7^2}-\dfrac{1}{7^{102}}< \dfrac{1}{7^2}=\dfrac{1}{49}< \dfrac{1}{50}\)
\(\Leftrightarrow S< \dfrac{1}{50}\)
Vậy \(\dfrac{1}{7^2}-\dfrac{1}{7^4}+...+\dfrac{1}{7^{98}}-\dfrac{1}{7^{100}}< \dfrac{1}{50}\) (Đpcm)
GIUP MINH LAM BAI NAY VOI
Gọi A=\(\frac{1}{7^2}-\frac{1}{7^4}+\frac{1}{7^6}-\frac{1}{7^8}+...+\frac{1}{7^{98}}-\frac{1}{7^{100}}\)
Nhân \(\frac{1}{7^2}\)vào A ta được
\(\frac{1}{7^2}\).A= \(\frac{1}{7^4}-\frac{1}{7^6}+\frac{1}{7^8}-...-\frac{1}{7^{98}}+\frac{1}{7^{100}}+\frac{1}{7^{102}}\)
A=\(\frac{1}{7^2}-\frac{1}{7^4}+\frac{1}{7^6}-\frac{1}{7^8}+....+\frac{1}{7^{98}}-\frac{1}{7^{100}}\)
Cộng \(\frac{1}{7^2}A\)+\(A\)=\(\frac{1}{49}-\frac{1}{7^{102}}\)\(\Rightarrow\frac{50}{49}A=\frac{1}{49}-\frac{1}{7^{102}}\Rightarrow A=\left(\frac{1}{49}-\frac{1}{7^{102}}\right).\frac{49}{50}\)
\(A=\frac{1}{50}-\frac{1}{7^{102}}.\frac{49}{50}