So sánh các lũy thừa sau
a, \(\left(\frac{1}{16}\right)^{10}va\left(\frac{1}{2}\right)^{50}\)
b, 9920 và 999910
1) so sánh
\(\left(\frac{1}{16}\right)^{10}\)và \(\left(\frac{1}{2}\right)^{50}\)
2) viết các biểu thức sau dưới dạng lũy thừa của một hữu tỉ
a) \(25.5^3.\frac{1}{625}.5^2\)
b) \(4.32:\left(2^3.\frac{1}{16}\right)\)
c) \(5^2.3^5.\left(\frac{3}{5}\right)^2\)
d)\(\left(\frac{1}{7}\right)^2.\frac{1}{7}.49^2\)
So sánh
\(\left(\frac{1}{16}\right)^{10}va\left(\frac{1}{2}\right)^{50}\)
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 => \(\frac{1}{2^{40}}>\frac{1}{2^{50}}\)
=> \(\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
Ta có: \(\left(\frac{1}{16}\right)^{10}=\left(\left(\frac{1}{2}\right)^4\right)^{10}=\left(\frac{1}{2}\right)^{40}\)
Mà \(\left(\frac{1}{2}\right)^{40}>\left(\frac{1}{2}\right)^{50}\)Vì \(2^{40}< 2^{50}\)
\(\Rightarrow\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
So sánh các lũy thừa sau :
1) \(\left(-2\right)^{240}\)và \(\left(-3\right)^{160}\)
2) \(\left(-84\right)^{11}\)và \(\left(-9\right)^{21}\)
3)\(\left(\frac{-1}{8}\right)^7\)và \(\left(\frac{-1}{16}\right)^5\)
So sánh
\(a,\left(-5\right)^{30}\&\left(-3\right)^{50}\)
\(b,\left(\frac{1}{16}\right)^{10}\&\left(\frac{1}{2}\right)^{50}\)
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
\(\left(\frac{1}{2}\right)^{10}< 1\)
\(\Leftrightarrow\left(\frac{1}{16}\right)^{10}.\left(\frac{1}{2}\right)^{10}< \left(\frac{1}{16}\right)^{10}\)
\(\Leftrightarrow\left(\frac{1}{2}\right)^{50}< \left(\frac{1}{16}\right)^{10}\)
ta có :\(\left(-5\right)^{30}\) và \(\left(-3\right)^{50}\)là 2 lũy thừa bậc chẵn nên :\(\left(-5\right)^{30}=5^{30}=\left(5^3\right)^{10}=125^{10}\)
\(\left(-3\right)^{50}=3^{50}=\left(3^5\right)^{10}=243^{10}\)
từ trên suy ra (-5)^30<(-3)^50
b) Ta có:\(\left(\frac{1}{2}\right)^{50}=\left(\frac{1}{2^5}\right)^{10}=\left(\frac{1}{32}\right)^{10}\)
\(\Rightarrow\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
ta có :\(\left(-5\right)^{30}\) và \(\left(-3\right)^{50}\)là 2 lũy thừa bậc chẵn nên :\(\left(-5\right)^{30}=5^{30}=\left(5^3\right)^{10}=125^{10}\)
\(\left(-3\right)^{50}=3^{50}=\left(3^5\right)^{10}=243^{10}\)
từ trên suy ra (-5)^30<(-3)^50
b) Ta có:\(\left(\frac{1}{2}\right)^{50}=\left(\frac{1}{2^5}\right)^{10}=\left(\frac{1}{32}\right)^{10}\)
\(\Rightarrow\left(\frac{1}{2}\right)^{50}< \left(\frac{1}{16}\right)^{10}\)
So sánh các lũy thừa sau:
a) 3111 và 1714
b) 1617 và 822
c) 10750 và 7375
d) 291 và 535
e) \(\left(\frac{1}{32}\right)^{7^{ }}\) và \(\left(\frac{1}{16}\right)^9\)
a) Ta có : 3111 < 3211 = (25)11 = 255
1714>1614 = (24)14=256
=> 3111 <255<256<1714
=>3111<1714
b)Ta có : 1617 = (24)17 = 268
822 = (23)22 = 266
Vì 268>266 nên 1617 >822
c) Ta có : 10750 <10850= (4.27)50 = 450 .2750 = 2100 . 3150
7375 >7275 = (8.9)75 = 875 . 975 = 2225 . 3150
=> 10750 <2100 .3150 <2225.3150<7375
=> 10750 <7375
d) Ta có : 291 >290 = (25)18 = 3218
535<536 = (52)18 = 2518
Vì 3218 >2518 nên 291 > 535.
e) Ta có : \(\left(\frac{1}{32}\right)^7=\frac{1}{32^7}=\frac{1}{2^{35}}\)
\(\left(\frac{1}{16}\right)^9=\frac{1}{16^9}=\frac{1}{2^{36}}\)
Vì \(\frac{1}{2^{35}}>\frac{1}{2^{36}}\) nên \(\left(\frac{1}{32}\right)^7>\left(\frac{1}{16}\right)^9\)
VIẾT CÁC BIỂU THỨC DƯỚI DẠNG LUỸ THỪA CỦA 1 SỐ HỮU TỈ
\(a,4.\left(\frac{1}{32}\right)^{-2}:\left(2^3.\frac{1}{16}\right)\)
\(b,5^2.3^5.\left(\frac{3}{5}\right)^2\)
2 SO SÁNH
\(a,10^{20}và9^{10}\)
\(b,\left(-5\right)^3và\left(-3\right)^{50}\)
\(c,64^3và16^{12}\)
\(d,\left(\frac{1}{16}\right)^{10}và\left(\frac{1}{2}\right)^{50}\)
Câu 1 :
a) \(4.\left(\frac{1}{32}\right)^{-2}:\left(2^3.\frac{1}{16}\right)\)
\(=2^2.32^2:\left(\frac{1}{8}.16\right)=\left(2.32\right)^2:2=64^2:2\)
\(=2048=2^{11}\)
b) \(5^2.3^5.\left(\frac{3}{5}\right)^2\)
\(=\left(5.\frac{3}{5}\right)^2.3^5=3^2.3^5=3^7\)
VIẾT CÁC BIỂU THỨC DƯỚI DẠNG LUỸ THỪA CỦA 1 SỐ HỮU TỈ
\(a,4\cdot\left(\frac{1}{32}\right)^{-2}:\left(2^3\cdot\frac{1}{16}\right)\\ =4\cdot1024:\left(8\cdot\frac{1}{16}\right)\\ =4\cdot1024:\frac{1}{2}\\ =2\cdot1024\\ =2\cdot2^{10}\\ =2^{11}\)
\(b,5^2\cdot3^5\cdot\left(\frac{3}{5}\right)^2\\ =5^2\cdot\left(\frac{3}{5}\right)^2\cdot3^5\\ =3^2\cdot3^5\\ =3^7\)
2 SO SÁNH
\(a,10^{20}\text{ và }9^{10}\)
Có: \(9^{10}=\left(3^2\right)^{10}=3^{20}\)
\(\Rightarrow10^{20}>3^{20}\\ \text{hay}\text{ }10^{20}>9^{10}\)
\(b,\left(-5\right)^3\text{ và }\left(-3\right)^{50}\)
Có: \(\left(-3\right)^{50}=3^{50}\)
\(\Rightarrow\left(-5\right)^3< 3^{50}\\ \text{hay }\left(-5\right)^3< \left(-3\right)^{50}\)
\(c,64^3\text{ và }16^{12}\)
Có: \(64^3=\left(4^3\right)^3=4^9;16^{12}=\left(4^2\right)^{12}=4^{24}\)
\(\Rightarrow4^9< 4^{24}\\ hay\text{ }64^3< 16^{12}\)
\(d,\left(\frac{1}{16}\right)^{10}\text{ và }\left(\frac{1}{2}\right)^{50}\)
Có: \(\left(\frac{1}{2}\right)^{50}=\left(\frac{1}{2}\right)^{5\cdot10}=\left[\left(\frac{1}{2}\right)^5\right]^{10}=\left(\frac{1}{32}\right)^{10}\)
\(\Rightarrow\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{32}\right)^{10}\\ \text{hay }\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
Bài 1 : So sánh
\(\left(\frac{1}{10}\right)^{15}\) và \(\left(\frac{3}{10}\right)^{20}\)
Bài 2 : So sánh
A = \(\left(\frac{13^{15}+1}{13^{16}+1}\right)\) và B = \(\left(\frac{13^{16}+1}{13^{17}+1}\right)\)
Bài 1:
Ta có:
\(\left(\frac{1}{10}\right)^{15}=\left(\frac{1}{5}\right)^{3.5}=\left(\frac{1}{125}\right)^5\)
\(\left(\frac{3}{10}\right)^{20}=\left(\frac{3}{10}\right)^{4.5}=\left(\frac{81}{10000}\right)^5\)
Lại có:
\(\frac{1}{125}=\frac{80}{10000}< \frac{81}{10000}\Rightarrow\left(\frac{1}{125}\right)^5< \left(\frac{81}{10000}\right)^5\)
\(\Rightarrow\left(\frac{1}{10}\right)^{15}< \left(\frac{3}{10}\right)^{20}\)
Bài 2:
Ta có:
\(A=\frac{13^{15}+1}{13^{16}+1}\Rightarrow13A=\frac{13^{16}+13}{13^{16}+1}=1+\frac{12}{13^{16}+1}\)
\(B=\frac{13^{16}+1}{13^{17}+1}\Rightarrow13B=\frac{13^{17}+13}{13^{17}+1}=1+\frac{12}{13^{17}+1}\)
Mà \(\frac{12}{13^{16}+1}>\frac{12}{13^{17}+1}\)
\(\Rightarrow1+\frac{12}{13^{16}+1}>1+\frac{12}{13^{17}+1}\)
\(\Rightarrow13A>13B\Rightarrow A>B\)
bài 1 tính
a)\(-\frac{1}{4}\) b)\(\left(-2\frac{1}{3}\right)^2\) c)(0,5)3 d)\(\left(-1\frac{1}{3}\right)^4\)
bai 2 tìm x , biết
a)x:\(\left(-\frac{1}{3}\right)^3\)=\(-\frac{1}{3}\) b)\(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\) c)\(\left(\frac{4}{5}\right):x=\left(\frac{4}{5}\right)^7\) d)\(3x+1=27\)
bài 3 so sánh
a)\(10^{20}va9^{10}\) b)\(\left(-5\right)^3^0va\left(-3\right)^{50}\) c)\(64^8va16^{12}\) d)\(\left(\frac{1}{16}\right)^{10}va\left(\frac{1}{2}\right)^{50}\)
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: \(\left(\frac{1}{16}\right)^{10}=\left(\frac{1}{2^4}\right)^{10}=\frac{1}{2^{40}}\)
\(\left(\frac{1}{2}\right)^{50}=\frac{1}{2^{50}}\)
\(\Rightarrow\frac{1}{2^{40}}>\frac{1}{2^{50}}\)
\(\Rightarrow\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
3.a) Ta có: 910=(32)10=320
Mà 1020<320
Nên 1020<910
c)Ta có:648 =(82)8=816
1612=(23)12=836
vì 816<836
Nên 648<162
So sánh
\(\left(\frac{1}{16}\right)^{10}\) và \(\left(\frac{1}{2}\right)^{50}\)
Ai nhanh mình tick
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\(\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
Ta có:
\(16^{10}=\left(2^4\right)^{10}=2^{4\cdot10}=2^{40}< 2^{50}\)
=>\(\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
Vậy ......