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\(\left(-32\right)^9=-\left(2^5\right)^9=-\left(2^{45}\right)\)
\(\left(-16\right)^{13}=-\left(2^4\right)^{13}=-\left(2^{52}\right)\)
vì -2^45>-2^52hay -16^13>-32^9
Toán 6 ?
Ta có :
\(\left(-\frac{1}{16}\right)^{100}=\left(\frac{1}{16}\right)^{100}=\frac{1}{16^{100}}\)
\(\left(-\frac{1}{2}\right)^{500}=\left(\frac{1}{2}\right)^{500}=\frac{1}{2^{500}}=\frac{1}{\left(2^4\right)^{125}}=\frac{1}{16^{125}}\)
Do \(\frac{1}{16^{100}}>\frac{1}{16^{125}}\left(16^{100}< 16^{125}\right)\)
\(\Rightarrow\left(-\frac{1}{16}\right)^{100}>\left(-\frac{1}{.2}\right)^{500}\)
Vậy ...
a) \(\left(-\frac{1}{2}\right)^{500}=\left[\left(-\frac{1}{2}^5\right)^{100}\right]=\left(\frac{-1}{32}\right)^{100}\)
Vì \(\left(-\frac{1}{16}\right)^{100}\) > \(\left(\frac{-1}{32}\right)^{100}\) nên \(\left(-\frac{1}{16}\right)^{100}>\left(-\frac{1}{2}\right)^{500}\)
b) Câu này mk ko bt
Bạn thông cảm
\(\left(\frac{1}{16}\right)^{10}=\left[\left(\frac{1}{2}\right)^4\right]^{10}=\left(\frac{1}{2}\right)^{40}\)
vì 40<50 nên \(\left(\frac{1}{2}\right)^{40}<\left(\frac{1}{2}\right)^{50}\)
hay \(\left(\frac{1}{16}\right)^{10}<\left(\frac{1}{2}\right)^{50}\)
Ta có: \(\left(\frac{1}{16}\right)^{10}=\left(\frac{1}{2^4}\right)^{10}=\frac{1}{2^{40}}\)
\(\left(\frac{1}{2}\right)^{50}=\frac{1}{2^{50}}\)
Vì \(2^{40}< 2^{50}\Rightarrow\frac{1}{2^{40}}>\frac{1}{2^{50}}\)hay \(\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
Ta có: \(\left(0,3\right)^{20}=\left[\left(0,3\right)^2\right]^{10}=\left(0,09\right)^{10}\)
Vì \(0,09< 0,1\Rightarrow\left(0,09\right)^{10}< \left(0,1\right)^{100}\)
hay \(\left(0,3\right)^{20}< \left(0,1\right)^{10}\)
\(\left(\dfrac{7}{2}\right)^{50}=\left(\dfrac{16807}{32}\right)^{10}\)
mà 16807/32>1/16
nên \(\left(\dfrac{1}{16}\right)^{10}< \left(\dfrac{7}{2}\right)^{50}\)
Ta có 13x = \(\frac{13^{17}+13}{13^{17}+1}=1+\frac{12}{13^{17}+1}\)
13y = \(\frac{13^{16}+13}{13^{16}+1}=1+\frac{12}{13^{16}+1}\)
Vì 1317 + 1 > 1316 + 1
=> \(\frac{1}{13^{17}+1}< \frac{1}{13^{16}+1}\)
=> \(\frac{12}{13^{17}+1}< \frac{12}{13^{16}+1}\)
=> \(1+\frac{12}{13^{17}+1}< 1+\frac{12}{13^{16}+1}\)
=> 13x < 13y
=> x < y
Vậy x < y
\(\frac{x}{19}=\frac{19^{17}+1}{19^{17}+19}=1-\frac{18}{19^{17}+19}\)
\(\frac{y}{19}=\frac{19^{16}+1}{19^{16}+19}=1-\frac{18}{19^{16}+19}\)
Nhận thấy 1917 + 19 > 1916 + 19
=> \(\frac{18}{19^{17}+19}< \frac{18}{19^{16}+19}\)
=> \(-\frac{18}{19^{17}+19}>-\frac{18}{19^{16}+19}\)
=> \(1-\frac{18}{19^{17}+19}>1-\frac{18}{19^{16}+19}\)
=> \(\frac{x}{19}>\frac{y}{19}\)
=> x > y
Vậy x > y
Ta có : \(\frac{x}{19}=\frac{19^{17}+1}{19^{17}+19}=1-\frac{18}{19^{17}+19}\)
\(\frac{y}{19}=\frac{19^{16}+1}{19^{16}+19}=1-\frac{18}{19^{16}+19}\)
Vì\(\frac{18}{19^{17}+19}< \frac{18}{19^{16}+19}\)\(\Rightarrow\frac{x}{19}>\frac{y}{19}\)
mà \(x,y>0\)
\(\Rightarrow x>y\)
3300 = ( 33 )100 = 27100
2500 = ( 25 )100 = 32100
Vì 27 < 32 nên 27100 < 32100
Vậy : 3300 < 2500
Tick mình nha !
3^300=3^100x3=27^100
2^500=2^100x5=32^100
Vì 32>27
=>32^100>27^100
hay 2^500>3^300
Vậy 2^500>3^300
(-1/16)100 = (1/16)100
(-1/2)400 = (1/2)400 =[ (1/2)4 ]100= (1/16)100
---> (-1/16)100 =(-1/2)400