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a)
Vì 3<5
\(\Rightarrow3^{30}< 5^{30}\)
\(\Rightarrow\left(-3\right)^{30}< \left(-5\right)^{30}\)
b)
Ta có
\(\left(\frac{1}{2}\right)^{50}=\left[\left(\frac{1}{2}\right)^4\right]^{10}.\left(\frac{1}{2}\right)^{10}\)
\(=\left(\frac{1}{16}\right)^{10}.\left(\frac{1}{2}\right)^{10}\)
Ta có
\(\left(\frac{1}{2}\right)^{10}< 1\)
\(\Leftrightarrow\left(\frac{1}{16}\right)^{10}.\left(\frac{1}{2}\right)^{10}< \left(\frac{1}{16}\right)^{10}\)
\(\Leftrightarrow\left(\frac{1}{2}\right)^{50}< \left(\frac{1}{16}\right)^{10}\)
Cách1:Ta có:\(\left(\frac{1}{2}\right)^{50}< \left(\frac{1}{2}\right)^{40}=\left[\left(\frac{1}{2}\right)^4\right]^{10}=\left(\frac{1}{16}\right)^{10}\)
Vậy..................
Cách 2:Ta có:\(\left(\frac{1}{16}\right)^{10}=\left[\left(\frac{1}{2}\right)^4\right]^{10}=\left(\frac{1}{2}\right)^{40}>\left(\frac{1}{2}\right)^{50}\)
Vậy......................
\(\left(\frac{1}{16}\right)^{10}=\left(\frac{1}{2^4}\right)^{10}=\frac{1^{10}}{2^{40}}=\frac{1}{2^{40}}\)
\(\left(\frac{1}{2}\right)^{50}=\frac{1^{50}}{2^{50}}=\frac{1}{2^{50}}\)
Do 250 > 240 => \(\frac{1}{2^{40}}>\frac{1}{2^{50}}\)
=> \(\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
a, Ta có :
\(\left(\frac{1}{2}\right)^{50}=\left(\left(\frac{1}{2}\right)^5\right)^{10}=\left(\frac{1}{32}\right)^{10}\)
bạn so sánh nha :)
b,
T/c : \(99^{20}=\left(\left(99\right)^2\right)^{10}=9801^{10}\)
tiếp đây thì bạn tự làm nha có gì k hiểu ibx mk
=(-1/2) : (-2/3) :( -3/4) :...: (-49/50)
= -1/2 . (-3/2) . (-4/3) . ... . (-50/49)
= -1/2.(-1/2) . (-50)
= - 1/100
Ta có:
\(\left(\frac{1}{16}\right)^{50}=\left[\left(\frac{1}{2}\right)^4\right]^{50}=\left(\frac{1}{2}\right)^{200}=\frac{1^{200}}{2^{200}}=\frac{1}{2^{200}}\)
\(\left(\frac{1}{2}\right)^{60}=\frac{1^{60}}{2^{60}}=\frac{1}{2^{60}}\)
Vì \(2^{200}>2^{60}\Rightarrow\frac{1}{2^{200}}< \frac{1}{2^{60}}\Rightarrow\left(\frac{1}{16}\right)^{50}< \left(\frac{1}{2}\right)^{60}\)
\(\left(\frac{1}{16}\right)^{10}\) và \(\left(\frac{1}{2}\right)^{50}\)
Ta có: \(\left(\frac{1}{2}\right)^{50}=\left[\left(\frac{1}{2}\right)^5\right]^{10}=\left(\frac{1}{32}\right)^{10}\)
Do \(\frac{1}{6}>\frac{1}{32}\Rightarrow\left(\frac{1}{6}\right)^{10}>\left(\frac{1}{32}\right)^{10}\)
Vậy \(\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
a) \(10^{20}\) và \(9^{10}\)
Vì 10 > 9 ; 20 > 10
nên \(10^{20}>9^{10}\)
Vậy \(10^{20}>9^{10}\)
b) \(\left(-5\right)^{30}\) và \(\left(-3\right)^{50}\)
Ta có: \(\left(-5\right)^{30}=5^{30}=\left(5^3\right)^{10}=125^{10}\)
\(\left(-3\right)^{50}=3^{50}=\left(3^5\right)^{10}=243^{10}\)
Vì 243 > 125 nên \(125^{10}< 243^{10}\)
Vậy \(\left(-5\right)^{30}< \left(-3\right)^{50}\)
c) \(64^8\) và \(16^{12}\)
Ta có: \(64^8=\left(4^3\right)^8=4^{24}\)
\(16^{12}=\left(4^2\right)^{12}=4^{24}\)
Vậy \(64^8=16^{12}\left(=4^{24}\right)\)
d) \(\left(\frac{1}{6}\right)^{10}\) và \(\left(\frac{1}{2}\right)^{50}\)
Ta có: \(\left(\frac{1}{6}\right)^{10}=\left[\left(\frac{1}{2}\right)^4\right]^{10}=\left(\frac{1}{2}\right)^{40}\)
Vì 40 < 50 nên \(\left(\frac{1}{2}\right)^{40}< \left(\frac{1}{2}\right)^{50}\)
Vậy \(\left(\frac{1}{16}\right)^{10}< \left(\frac{1}{2}\right)^{50}\)
\(P=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{16}\left(1+2+3+...+16\right)\)
\(=1+\frac{1}{2}.\frac{2.\left(2+1\right)}{2}+\frac{1}{3}.\frac{3.\left(3+1\right)}{2}+...+\frac{1}{16}.\frac{16.\left(16+1\right)}{2}\)
\(=1+\frac{2+1}{2}+\frac{3+1}{2}+...+\frac{16+1}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{17}{2}\)
\(=\frac{\left(17-2+1\right).\left(17+2\right)}{2}:2\)
\(=76\)
\(P=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{16}\left(1+2+3+...+16\right)\)
\(=1+\frac{1}{2}\left[\frac{\left(2+1\right)2}{2}\right]+\frac{1}{3}\left[\frac{\left(3+1\right)3}{3}\right]+...+\frac{1}{16}\left[\frac{\left(16+1\right)16}{2}\right]\)
\(=1+\frac{2+1}{2}+\frac{3+1}{2}+...+\frac{16+1}{2}\)
\(=\frac{2+2+1+3+1+...+16+1}{2}\)
\(=\frac{\left(1+1+1+..15cs.+1\right)+\left(2+3+...+16\right)+2}{2}\)
\(=\frac{15+135+2}{2}\)
\(=\frac{152}{2}\)\(=76\)
(\(\frac{1}{2}\))50=(\(\frac{1}{2^5}\))10=(\(\frac{1}{32}\))10
Do 1/6> 1/30 nên (\(\frac{1}{6}\))10>(\(\frac{1}{2}\))50
\(\left(\frac{1}{2}\right)^{50}=\left[\left(\frac{1}{2}\right)^5\right]^{10}=\left[\frac{1^5}{2^5}\right]^{10}=\left[\frac{1}{32}\right]^{10}\)
Vì 2 phân số này có cùng tử mà 6 < 30
=> \(\frac{1}{6}>\frac{1}{30}\)
=> \(\left(\frac{1}{6}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
Vì\(\left(\frac{1}{16}\right)^{10}\)= \(\left[\left(\frac{1}{2}\right)^4\right]^{10}\)= \(\left(\frac{1}{2}\right)^{40}\)
Mà 40<50 =>\(\left(\frac{1}{2}\right)^{40}\)< \(\left(\frac{1}{2}\right)^{50}\)hay \(\left(\frac{1}{16}\right)^{10}\)< \(\left(\frac{1}{2}\right)^{50}\)
Vậy \(\left(\frac{1}{16}\right)^{10}\)<\(\left(\frac{1}{2}\right)^{50}\)
Học giỏi!^^ (đúng thì k cho mik nhé,cảm ơn!)
\(\left(\frac{1}{2}\right)^{50}=\left(\left(\frac{1}{2}\right)^5\right)^{10}=\left(\frac{1}{32}\right)^{10}\)
Ta có\(\frac{1}{16}>\frac{1}{32}\)nên\(\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{32}\right)^{10}\)hay\(\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)