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21 tháng 10 2016
\(B=\left[\left(0,1\right)^2\right]^0+\left[\left(\frac{1}{7}\right)^{-1}\right]^2.\frac{1}{49}.\left[\left(2^2\right)^3.2^5\right]\)
\(=1+\left(\frac{1}{\frac{1}{7}}\right)^2.\frac{1}{49}.\left(2^6.2^5\right)\)
\(=1+7^2.\frac{1}{49}.2^{11}\)
\(\Rightarrow1+49.\frac{1}{49}.2^{11}\)
\(=1+2^{11}\)
Vậy \(B=1+2^{11}\)
LN
10 tháng 10 2017
Ta có :
\(0,0\left(8\right)=\dfrac{1}{10}.0,\left(8\right)=\dfrac{1}{10}.0,\left(1\right).8=\dfrac{1}{10}.\dfrac{1}{9}.8=\dfrac{4}{45}\)
\(0,1\left(2\right)=0,1+0,0\left(2\right)\)
\(=\dfrac{1}{10}+\dfrac{1}{10}.0,\left(2\right)=\dfrac{1}{10}+\dfrac{1}{10}.0,\left(1\right).2\)
\(=\dfrac{1}{10}+\dfrac{1}{10}.\dfrac{1}{9}.2=\dfrac{9}{90}+\dfrac{2}{90}=\dfrac{11}{90}\)
\(0,1\left(23\right)=0,1+0,0\left(23\right)=\dfrac{1}{10}+\dfrac{1}{10}.0,23\)
\(=\dfrac{1}{10}+\dfrac{1}{10}.0,\left(01\right).23\)
\(\dfrac{1}{10}+\dfrac{1}{10}.\dfrac{1}{99}.23=\dfrac{99}{990}+\dfrac{23}{990}=\dfrac{122}{990}=\dfrac{61}{495}\)
12 tháng 7 2019
a)\(\left(\frac{1}{3}\right)^{-1}-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^4.2^3=3-1+\frac{1}{16}.8=3-1+\frac{1}{2}=\frac{5}{2}\\ \)
b)\(2^2.2^3.\left(\frac{2}{3}\right)^{-2}=2^5.\frac{9}{4}=72\)
c)\(\left(\frac{4}{3}\right)^{-2}.\left(\frac{3}{4}\right)^3:\left(\frac{-2}{3}\right)^{-3}=\left(\frac{3}{4}\right)^2.\left(\frac{3}{4}\right)^3:\left(\frac{-2}{3}\right)^{-3}=\left(\frac{3}{4}\right)^5:\left(\frac{3}{2}\right)^3=\frac{9}{128}\)
2)
\(3^{x+1}=9^x\Leftrightarrow3^x.3=9^x\Rightarrow3=9^x:3^x\Rightarrow3=3^x\Rightarrow x=1\)
\(\left(x-0,1\right)^2=6,25\Leftrightarrow\left(x-0,1\right)^2=2,5^2\Rightarrow\left(x-0,1\right)=2,5\Rightarrow x=2,5+0,1=2,6\)
\(3^{2x-1}=243\Leftrightarrow3^{2x-1}=3^5\Rightarrow2x-1=5\Rightarrow2x=6\Rightarrow x=3\)
\(\left(4x-3\right)^4=\left(4x-3\right)^2\Rightarrow x=1\)
MV
4 tháng 7 2018
\(\left(\dfrac{1}{16}\right)^{10}=\left[\left(\dfrac{1}{2}\right)^4\right]^{10}=\left(\dfrac{1}{2}\right)^{40}< \left(\dfrac{1}{2}\right)^{50}\\ \left(\dfrac{1}{2}\right)^{300}=\left(\dfrac{1}{2}\right)^{3\cdot100}=\left[\left(\dfrac{1}{2}\right)^3\right]^{100}=\left(\dfrac{1}{8}\right)^{100}\\ \left(\dfrac{1}{3}\right)^{200}=\left(\dfrac{1}{3}\right)^{2\cdot100}=\left[\left(\dfrac{1}{3}\right)^2\right]^{100}=\left(\dfrac{1}{9}\right)^{100}\\ \dfrac{1}{8}>\dfrac{1}{9}\Rightarrow\left(\dfrac{1}{8}\right)^{100}>\left(\dfrac{1}{9}\right)^{100}\Rightarrow\left(\dfrac{1}{2}\right)^{300}>\left(\dfrac{1}{3}\right)^{200}\\ \left(0,3\right)^{20}=\left(0,3\right)^{2\cdot10}=\left[\left(0,3\right)^2\right]^{10}=\left(0,09\right)^{10}< \left(0,1\right)^{10}\)
LD
5 tháng 7 2018
a) \(\left[\left(\dfrac{1}{2}\right)^4\right]^{10}=\left(\dfrac{1}{2}\right)^{40}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{40}< \left(\dfrac{1}{2}\right)^{50}\)
Vì \(40< 50\)
b)\(\left[\left(\dfrac{1}{2}\right)^3\right]^{100}=\left(\dfrac{1}{8}\right)^{100}\)
\(\left[\left(\dfrac{1}{3}\right)^2\right]^{100}=\left(\dfrac{1}{9}\right)^{100}\)
\(\Rightarrow\text{}\text{}\left(\dfrac{1}{2}\right)^{300}>\left(\dfrac{1}{3}\right)^{200}\)
Vì \(\dfrac{1}{8}>\dfrac{1}{9}\)
c)\(\left[\left(0,3\right)^2\right]^{10}=\left(0,09\right)^{10}\)
\(\Rightarrow\left(0,1\right)^{10}>\left(0,3\right)^{20}\)
Vì \(0,1>0,09\)
16 tháng 7 2015
a/
(0.3)20 = [(0.3)2]10 = (0.09)10
b/
(\(\frac{1}{6}\))10 = [(\(\frac{1}{2}\))4]10 = (\(\frac{1}{2}\))40
c/
2400 = (22)200 = 4200
27 tháng 5 2022
a: \(\left(\dfrac{5}{6}\right)^6\cdot\left(\dfrac{6}{5}\right)^6\cdot\left(\dfrac{6}{5}\right)^2=\left(\dfrac{5}{6}\cdot\dfrac{6}{5}\right)^6\cdot\dfrac{36}{25}=\dfrac{36}{25}\)
b: \(=-\left(\dfrac{13}{8}\right)^3\cdot\left(\dfrac{32}{13}\right)^3\cdot\dfrac{32}{13}\)
\(=-\left(\dfrac{13}{8}\cdot\dfrac{32}{13}\right)^3\cdot\dfrac{32}{13}=-4^3\cdot\dfrac{32}{13}=\dfrac{-2048}{13}\)
c: \(=\left(0.1\right)^7\cdot10^{13}=\left(0.1\cdot10\right)^7\cdot10^6=10^6\)
\(0,1\left(2\right)=0,1+0,0\left(2\right)=\frac{1}{10}+\frac{0,\left(2\right)}{10}\)
\(=\frac{1}{10}+\frac{0,\left(1\right).2}{10}=\frac{1}{10}+\frac{\frac{1}{9}.2}{10}\)
\(=\frac{1}{10}+\frac{\frac{2}{9}}{10}=\frac{11}{9}.\frac{1}{10}=\frac{11}{90}\)
\(0,1\left(23\right)=0,1+0,0\left(23\right)=\frac{1}{10}+\frac{0,\left(23\right)}{10}\)
\(=\frac{1}{10}+\frac{0,\left(01\right).23}{10}=\frac{1}{10}+\frac{\frac{1}{99}.23}{10}\)
\(=\frac{1}{10}+\frac{\frac{23}{99}}{10}=\frac{122}{99}.\frac{1}{10}=\frac{61}{495}\)
Có đúng không ạ?