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VM
22 tháng 9 2019
3)
\(\left(\frac{1}{2}\right)^{x+1}=\frac{1}{16}\)
⇒ \(\left(\frac{1}{2}\right)^{x+1}=\left(\frac{1}{2}\right)^4\)
⇒ \(x+1=4\)
⇒ \(x=4-1\)
⇒ \(x=3\)
Vậy \(x=3.\)
Chúc bạn học tốt!
DT
1
VM
15 tháng 10 2019
Ta có:
\(\left(\frac{1}{16}\right)^{250}=\left(\frac{1}{16}\right)^{250}.\)
\(\left(\frac{1}{2}\right)^{1500}=\left[\left(\frac{1}{2}\right)^6\right]^{250}=\left(\frac{1}{64}\right)^{250}.\)
Vì \(\frac{1}{16}>\frac{1}{64}\) nên \(\left(\frac{1}{16}\right)^{250}>\left(\frac{1}{64}\right)^{250}.\)
\(\Rightarrow\left(\frac{1}{16}\right)^{250}>\left(\frac{1}{2}\right)^{1500}.\)
Chúc bạn học tốt!
15 tháng 2 2015
x=[(1/2)^3]^75 =>(1/8)^75
y=[(1/3)^2]^75 =>(1/9)^75
vì 1/8>1/9
=>(1/2)^225 > (1/3)^150
30 tháng 8 2016
Đặt \(100=n\) , ta có :
\(B=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)....\left(\frac{1}{n^2}-1\right)\)
\(=\frac{\left(-1\right).3}{2^2}.\frac{\left(-2\right).4}{3^2}.....\frac{\left(1-n\right)\left(1+n\right)}{n^2}\)
\(=\frac{\left(-1\right).\left(-2\right)....\left(1-n\right)}{2.3.....n}.\frac{3.4........\left(1+n\right)}{2.3.....n}\)
\(=\frac{\left(-1\right).2.3.....\left(n-1\right)}{2.3......n}.\frac{3.4.....\left(n+1\right)}{2.3.......n}\)
\(=\frac{\left(-1\right)}{n}.\frac{n+1}{2}=\frac{-1}{2}.\frac{n+1}{n}< \frac{-1}{2}\)
Vậy \(B< \frac{-1}{2}\)
25 tháng 9 2019
Bài 2:
Ta có: \(\frac{\left(3^3\right)^2.\left(2^3\right)^5}{\left(2.3\right)^6.\left(2^5\right)^3}\)\(=\frac{3^6.2^{15}}{2^6.3^6.2^{15}}\)\(\frac{1}{2^6}=\frac{1}{64}\)
Chúc hk tốt nha!!!
HH
14 tháng 6 2018
Giải:
\(C=\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)\left(1-\dfrac{2}{4.5}\right)...\left(1-\dfrac{2}{n\left(n+1\right)}\right)\)
Đk: \(n\ne0;n\ne-1\)
\(C=\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)\left(1-\dfrac{2}{4.5}\right)...\left(1-\dfrac{2}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\left(\dfrac{2.3-2}{2.3}\right)\left(\dfrac{3.4-2}{3.4}\right)\left(\dfrac{4.5-2}{4.5}\right)...\left(\dfrac{n\left(n-1\right)-2}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\dfrac{4}{2.3}.\dfrac{10}{3.4}.\dfrac{18}{4.5}...\left(\dfrac{n\left(n-1\right)-2}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}.\dfrac{3.6}{4.5}...\left(\dfrac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\dfrac{1.4.2.5.3.6...\left(n-1\right)\left(n+2\right)}{2.3.3.4.4.5.n\left(n+1\right)}\)
\(\Leftrightarrow C=\dfrac{\left[1.2.3...\left(n-1\right)\right]\left[4.5.6\left(n+2\right)\right]}{\left(2.3.4...n\right)\left[3.4.5....\left(n+1\right)\right]}\)
\(\Leftrightarrow C=\dfrac{n+2}{3n}\)
Vì \(\dfrac{n+2}{3n}< \dfrac{2n+2}{3n}\)
\(\Leftrightarrow C< \dfrac{2n+2}{3n}\)
Vậy ...
HH
14 tháng 6 2018
Giải:
\(C=\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)\left(1-\dfrac{2}{4.5}\right)...\left(1-\dfrac{2}{n\left(n+1\right)}\right)\)
Đk: \(n\ne0;n\ne-1\)
\(C=\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)\left(1-\dfrac{2}{4.5}\right)...\left(1-\dfrac{2}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\left(\dfrac{2.3-2}{2.3}\right)\left(\dfrac{3.4-2}{3.4}\right)\left(\dfrac{4.5-2}{4.5}\right)...\left(\dfrac{n\left(n-1\right)-2}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\dfrac{4}{2.3}.\dfrac{10}{3.4}.\dfrac{18}{4.5}...\left(\dfrac{n\left(n-1\right)-2}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}.\dfrac{3.6}{4.5}...\left(\dfrac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\dfrac{1.4.2.5.3.6...\left(n-1\right)\left(n+2\right)}{2.3.3.4.4.5.n\left(n+1\right)}\)
\(\Leftrightarrow C=\dfrac{\left[1.2.3...\left(n-1\right)\right]\left[4.5.6\left(n+2\right)\right]}{\left(2.3.4...n\right)\left[3.4.5....\left(n+1\right)\right]}\)
\(\Leftrightarrow C=\dfrac{n+2}{3n}\)
Vì \(\dfrac{n+2}{3n}< \dfrac{2n+2}{3n}\)
\(\Leftrightarrow C< \dfrac{2n+2}{3n}\)
Vậy ...
(1/2)1500=(1/26)250=(1/64)250
Do 1/16>1/64 =>(1/16)250>(1/64)250
Vậy (1/16)250>(1/2)1500
\(\left(\frac{1}{16}\right)^{250}\) và \(\left(\frac{1}{2}\right)^{1500}\)
=> \(\left(\frac{1}{16}\right)^{250}\) và \(\left(\left(\frac{1}{2}\right)^6\right)^{250}\)
=> \(\frac{1}{16}\) và \(\left(\frac{1}{2}\right)^6\)
=> \(\frac{1}{16}\) và \(\frac{1}{64}\)
=> \(\frac{1}{16}\) > \(\frac{1}{64}\) hay \(\left(\frac{1}{16}\right)^{250}\) > \(\left(\frac{1}{2}\right)^{1500}\)