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Ta có : \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)\)
\(=\frac{1}{2}.\frac{2}{3}....\frac{18}{19}.\frac{19}{20}\)
\(=\frac{1.2....18.19}{2.3...19.20}\)
\(=\frac{1}{20}>\frac{1}{21}\)
Vậy A > 1/21
Bài 1 và Bài 2 dễ, bn có thể tự làm được!
Bài 3:
a) ta có: 1020 = (102)10 = 10010
=> 10010>910
=> 1020>910
b) ta có: (-5)30 = 530 =( 53)10 = 12510 ( vì là lũy thừa bậc chẵn)
(-3)50 = 350 = (35)10= 24310
=> 12510 < 24310
=> (-5)30 < (-3)50
c) ta có: 648 = (26)8= 248
1612 = ( 24)12 = 248
=> 648 = 1612
d) ta có: \(\left(\frac{1}{16}\right)^{10}=\left(\frac{1}{2^4}\right)^{10}=\frac{1}{2^{40}}\)
\(\left(\frac{1}{2}\right)^{50}=\frac{1}{2^{50}}\)
\(\Rightarrow\frac{1}{2^{40}}>\frac{1}{2^{50}}\)
\(\Rightarrow\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
Ta có \(-A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{2014^2}\right)\)
\(=\left(\frac{2^2-1}{2^2}\right)\left(\frac{3^2-1}{3^2}\right)...\left(\frac{2014^2-1}{2014^2}\right)\)
\(=\frac{\left(2-1\right)\left(2+1\right)}{2^2}.\frac{\left(3-1\right)\left(3+1\right)}{3^2}...\frac{\left(2014-1\right)\left(2014+1\right)}{2014^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}...\frac{2013.2015}{2014.2014}\)
\(=\frac{1.2...2013}{2.3...2014}.\frac{3.4...2015}{2.3...2014}\)
\(=\frac{1}{2014}.\frac{2015}{2}\)
\(=\frac{2015}{2014.2}>\frac{1}{2}\)hay -A>1/2
=>\(A< \frac{-1}{2}\)hay A<B
\(a,3^{16}:3=3^{16-1}=3^{15}\)
\(b,3^6.3^4.3^2.3=3^{6+4+2+1}=3^{13}\)
\(c,\left(-\frac{1}{4}\right).\left(6\frac{2}{11}\right)+\left(3\frac{9}{11}\right).\left(-\frac{1}{4}\right)=\left(-\frac{1}{4}\right).\frac{68}{11}+\frac{42}{11}.\left(-\frac{1}{4}\right)\)
\(=\left(-\frac{1}{4}\right)\left(\frac{68}{11}+\frac{42}{11}\right)\)
\(=\left(-\frac{1}{4}\right).10\)
\(=-\frac{10}{4}=-\frac{5}{2}\)
\(d,\left(-\frac{1}{2}\right)^3+\frac{1}{2}:5=\left(-\frac{1}{2}\right)\left(\left(\frac{1}{2}\right)^2-\frac{1}{5}\right)\)
\(=-\frac{1}{2}.\left(\frac{1}{4}-\frac{1}{5}\right)\)
\(=-\frac{1}{2}.\frac{1}{20}\)
\(=-\frac{1}{40}\)
\(g,1\frac{1}{25}+\frac{2}{21}-\frac{1}{25}+\frac{19}{21}=\frac{26}{25}+\frac{2}{21}-\frac{1}{25}+\frac{19}{21}\)
\(=\left(\frac{26}{25}-\frac{1}{25}\right)+\left(\frac{2}{21}+\frac{19}{21}\right)\)
\(=1+1\)
\(=2\)
1) Ta có: \(\left|9y-1\right|+\left(2x+3\right)^2=0\)
Mà \(\hept{\begin{cases}\left|9y-1\right|\ge0\\\left(2x+3\right)^2\ge0\end{cases}}\left(\forall x,y\right)\)
=> \(\left|9y-1\right|+\left(2x+3\right)^2\ge0\left(\forall x,y\right)\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left|9y-1\right|=0\\\left(2x+3\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}9y-1=0\\2x+3=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{3}{2}\\y=\frac{1}{9}\end{cases}}\)
Vậy \(\hept{\begin{cases}x=-\frac{3}{2}\\y=\frac{1}{9}\end{cases}}\)
2)
a) Ta có: \(\left[\left(-\frac{1}{3}\right)^7\right]^4=\left(\frac{1}{3}\right)^{28}=\frac{1}{3^{28}}\)
và \(\left[\left(-\frac{1}{2}\right)^{14}\right]^2=\left(\frac{1}{2}\right)^{28}=\frac{1}{2^{28}}\)
Vì \(\frac{1}{3^{28}}< \frac{1}{2^{28}}\Rightarrow\left[\left(-\frac{1}{3}\right)^7\right]^4< \left[\left(-\frac{1}{2}\right)^{14}\right]^2\)
b) Ta có: \(\left(-\frac{2}{3}\right)^{12}=\left[\left(-\frac{2}{3}\right)^2\right]^6=\left(\frac{4}{9}\right)^6\)
Ta thấy \(0< \frac{4}{9}< 1\)\(\Rightarrow\left(\frac{4}{9}\right)^6>\left(\frac{4}{9}\right)^7\)
\(\Rightarrow\left(-\frac{2}{3}\right)^{12}>\left(\frac{4}{9}\right)^7\)
\(\text{a,}\frac{2}{13}.\frac{-5}{3}+\frac{11}{13}.\frac{-5}{3}=-\frac{5}{3}\left(\frac{2}{13}+\frac{11}{13}\right)\)
\(=\frac{-5}{3}.\frac{13}{13}\)
\(=-\frac{5}{3}\)
\(\text{b,}\left(-\frac{1}{3}\right)^2+\left(-\frac{1}{3}\right)^3.27+\left(\frac{-2017}{2018}\right)^0=\frac{1}{9}-\frac{1}{27}.27+1\)
\(=\frac{1}{9}-1+1\)
\(=\frac{1}{9}\)
\(\text{c,}1,2-\sqrt{\frac{1}{4}}:1\frac{1}{20}+\left|\frac{3}{4}-1,25\right|-\left(\frac{-3}{2}\right)^2=\frac{6}{5}-\frac{1}{2}:\frac{21}{20}+\left|\frac{3}{4}-\frac{5}{4}\right|-\frac{9}{4}\)
\(=\frac{6}{5}-\frac{10}{21}+\frac{1}{2}-\frac{9}{4}\)
\(=\frac{-431}{420}\)
Ta có : \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{20}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{19}{20}\)
\(=\frac{1.2.3.....19}{2.3.4.....20}\)
\(=\frac{1}{20}>\frac{1}{21}\)
SORRY EM MỚI LỚP 6