so sanh 2^1000 va 5^400
so sanh 2^1000 vs 5^400
ghi cach giai cho 5 tick
21000 = 210.100 = 1024100
5400 = 54.100 = 625100
1024100 > 625100 nên 21000 > 5400
So sanh A va B biet A =1×3×5×7ו••×9997×9999 va B=2×4×6ו••×9998×1000
so sanh (1/16)^200 va 1/2^1000
Ta có :
\(\left(\frac{1}{16}\right)^{200}=\frac{1}{16^{200}}=\frac{1}{\left(2^4\right)^{200}}=\frac{1}{2^{800}}\)
Vì \(\frac{1}{2^{800}}>\frac{1}{2^{1000}}\) nên \(\left(\frac{1}{16}\right)^{200}>\frac{1}{2^{1000}}\)
Bấm máy thì cả 2 đều = 0
=> (1/16)^200 = (1/2)^1000
Ta có:
\(\left(\frac{1}{16}\right)^{200}=\frac{1}{16^{200}}=\frac{1}{\left(2^4\right)^{200}}=\frac{1}{2^{800}}>\frac{1}{2^{1000}}.\)
so sanh 1/35 va 1000/-35
so sánh : \(\dfrac{1}{35}\) và \(\dfrac{1000}{-35}\)
có : \(\dfrac{1000}{-35}\) = \(\dfrac{-1000}{35}\)
\(\Rightarrow\) \(1\) \(>\) (\(-1000\) )
\(\Rightarrow\) \(\dfrac{1}{35}\) \(< \) \(\dfrac{-1000}{35}\)
vậy : \(\dfrac{1}{35}\) < \(\dfrac{1000}{-35}\) hay \(\dfrac{1000}{-35}\) > \(\dfrac{1}{35}\)
a. so sanh a=1.3.5......999 va b=501\2 .501\2...........1000\2
b. 5 mu2x-3 -2.5mu3=5 mu 2.3
ho tro dum mk trong 1tuan toi nha hom nay 16\3\2019
BAI1 SO SANH 2^600 va 3^400
Nguyễn Văn Tân làm đúng nhưng cách làm như bạn thì sai ùi !
2^600 = ( 2^6 )^100 = 64^100
3^400 = ( 3^4)^100 = 81^100
Vì 64 < 81 nên 64^100 < 81^100
Nên : 2^600 < 3^400
2600 = (26)100 =32100
3400 =(34)100 =81100
Mà 32<81
=> 2600<3400
so sanh A va B; A=1/2^2+1/2^3+.....+1/2^1000; B=1
so sanh A va B; A=1/2+1/2^2+1/2^3+.....+1/2^1000; B=1
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{1000}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{999}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{999}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{1000}}\right)\)
\(A=1-\frac{1}{2^{1000}}< 1=B\)
`Answer:`
Đặt \(C=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{999.1000}\)
Ta thấy:
\(\frac{1}{1.2}>\frac{1}{2^2}\)
\(\frac{1}{2.3}>\frac{1}{2^3}\)
\(\frac{1}{3.4}>\frac{1}{2^4}\)
...
\(\frac{1}{999.1000}>\frac{1}{2^{1000}}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{1000}}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{999.1000}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+\frac{1}{5}-\frac{1}{5}+...+\frac{1}{999}-\frac{1}{1000}\)
\(\Rightarrow A< 1-\frac{1}{1000}\)
Mà \(\frac{1}{1000}>0\)
\(\Rightarrow1-\frac{1}{1000}< 1\)
\(\Rightarrow C< B\)
\(\Rightarrow A< C< B\)
\(\Rightarrow A< B\)
so sanh a
a) 9 mu 20 va 27 mu 13
b) 3 mu 1000 va 2 mu 1500
a, \(9^{20}=\left(3^2\right)^{20}=3^{40}\)
\(27^{13}=\left(3^3\right)^{13}=3^{39}\)
mà \(3^{40}>3^{39}\Leftrightarrow9^{20}=27^{13}\)
vậy \(9^{20}=27^{13}\)
920 = 340 ; 2713 = 339
Vì 40 > 39 nên 340 > 339 và 920 > 2713
b ) 31000 = 30100
21500 = 30100
Vì 100 =100 nên .....