so sánh
\(\left(\frac{1}{32}\right)^7\)với \(\left(\frac{1}{16}\right)^9\)
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Ta có :
\(M=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{99}{100}=\frac{3.8.15.....99}{4.9.16.....100}=\frac{1.3.2.4.3.5.....9.11}{2.2.3.3.4.4.....10.10}\)\(=\frac{1.2.3...9}{2.3...10}.\frac{3.4...11}{2.3...10}=\frac{1}{10}.\frac{11}{2}=\frac{11}{20}< \frac{11}{19}\)
ta có M = (1- 1/4) (1- 1/9)... ( 1- 1/100)
= 3/2^2.8/3^2 ... 99/10^2
= 1.3/2^2 . 2.4/3^2 ... 9.11/10^ 2
= 1.2.3...9/ 2.3.4...10 . 3.4.5... 11/ 2.3.4... 10
= 1/10 . 11/2 = 11/20 < 11/19
Vậy M < 11/19
\(\frac{2011.4023+2012}{2012.4023-2011}=\frac{2011.4023+2011+1}{2012.4023-2012-1}=\frac{2011.4023+2011.1+1}{2012.4023-2012.1-1}\)
\(=>\frac{2012.4023+2012.1+1}{2012.4023-2012.1-1}=\frac{2012.\left(4023+1\right)+1}{2012.\left(4023-1\right)-1}\)
\(=\frac{4023+1+1}{4023-1-1}=\frac{4023+2}{4023-2}=\frac{4025}{4021}\)
Vì 4025 > 4021 ( tử số lớn hơn mẫu số ) nên suy ra : 4025/4021 >1
<br class="Apple-interchange-newline"><div id="inner-editor"></div>=>2012.4023+2012.1+12012.4023−2012.1−1 =2012.(4023+1)+12012.(4023−1)−1
=4023+1+14023−1−1 =4023+24023−2 =40254021
Vì 4025 > 4021 ( tử số lớn hơn mẫu số ) nên suy ra : 4025/4021 >1
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
\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{400}-1\right)\)
\(-A=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{400}\right)\)
\(-A=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{399}{400}\)
\(-A=\frac{1\cdot3}{2\cdot2}\cdot\frac{2.4}{3.3}\cdot\frac{3.5}{4.4}\cdot...\cdot\frac{19.21}{20.20}\)
\(-A=\frac{1\cdot2\cdot3\cdot...\cdot19}{2\cdot3\cdot4\cdot...\cdot20}\cdot\frac{3\cdot4\cdot5\cdot...\cdot21}{2\cdot3\cdot4\cdot...\cdot20}\)
\(-A=\frac{1}{20}\cdot\frac{21}{2}=\frac{21}{40}>\frac{20}{40}=\frac{1}{2}\)
\(-A>\frac{1}{2}\Rightarrow A< \frac{1}{2}\)
\(B=\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{81}\right).\left(1-\frac{1}{100}\right)\)
\(B=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{80}{81}.\frac{99}{100}\)
\(B=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{8.10}{9.9}.\frac{9.11}{10.10}\)
\(B=\frac{1.2.3...8.9}{2.3.4...9.10}.\frac{3.4.5...10.11}{2.3.4...9.10}\)
\(B=\frac{1}{10}.\frac{11}{2}\)
\(B=\frac{11}{20}>\frac{11}{21}\)
\(B=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)......\left(1-\frac{1}{81}\right)\left(1-\frac{1}{100}\right)\)
= \(-\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.......\frac{80}{81}.\frac{99}{100}\)
=\(-\frac{1.3.2.4.3.5..............8.10.9.11}{2^2.3^2.4^2.......10^2}=-\frac{\left(1.2.3.....9\right)\left(3.4.5....11\right)}{2.3.4....10.2.3.4.....10}=-\frac{11}{20}\)
Ta có: (+) (1/32)^7 = [(1/2)^5]^7 =(1/2)^35
(+) (1/16)^9= [(1/2)^4]^9 =(1/2)^36
Vì 35 <36
=> (1/2)^35 > (1/2)^36
=> (1/32)^7 > (1/16)^9
Ta có :
\(\left(\frac{1}{32}\right)^7=\frac{1^7}{32^7}=\frac{1}{\left(2^5\right)^7}=\frac{1}{2^{5.7}}=\frac{1}{2^{35}}\)
\(\left(\frac{1}{16}\right)^9=\frac{1^9}{16^9}=\frac{1}{\left(2^4\right)^9}=\frac{1}{2^{4.9}}=\frac{1}{2^{36}}\)
Vì \(\frac{1}{2^{35}}>\frac{1}{2^{36}}\) ( cùng tử, mẫu nào bé hơn thì phân số đó lớn hơn ) nên \(\left(\frac{1}{32}\right)^7>\left(\frac{1}{16}\right)^9\)
Vậy \(\left(\frac{1}{32}\right)^7>\left(\frac{1}{16}\right)^9\)
Chúc bạn học tốt ~
Ta có : \(\left(\frac{1}{32}\right)^7=\left(\frac{1}{2^5}\right)^7=\frac{1}{2^{35}}\)
\(\left(\frac{1}{16}\right)^9=\left(\frac{1}{2^4}\right)^9=\frac{1}{2^{36}}\)
DO : \(\frac{1}{2^{35}}>\frac{1}{2^{36}}\)\(\Rightarrow\left(\frac{1}{32}\right)^7>\left(\frac{1}{16}\right)^9\)
Tk mk nha !!!