So sanh hai so :(1/2)^30va(1/3)^20
So sanh cac so sau bang hai cach :
a, 2^50 va 16^20
b,2^30 va 3^20
c, 2^20 va 2 ^16.15
d, 33^44 va 44^33 ( 1 cach )
e,72^45 - 72^44 va 72^44 - 72^43
so sanh 2 phan so
a=10^19+1/10^20+1;b=10^20+1+10^21+1
S = 1+3+3^2+3^3+3^4+ ....+3^20
SO SANH S VOI 1/2 . 3 ^ 31
Ta có: \(S=1+3+3^2+...+3^{20}\)
\(\Rightarrow3S=3+3^2+3^3+...+3^{21}\)
\(\Rightarrow3S-S=\left(3+3^2+3^3+...+3^{21}\right)-\left(1+3+3^2+...+3^{20}\right)\)
\(\Rightarrow2S=3^{21}-1\)
\(\Rightarrow S=\left(3^{21}-1\right).\frac{1}{2}\)
\(\Rightarrow S=3^{21}.\frac{1}{2}-\frac{1}{2}\)
Vì \(3^{21}.\frac{1}{2}-\frac{1}{2}< 3^{21}.\frac{1}{2}\) nên \(A< \frac{1}{2}.3^{21}\)
Vậy \(A< \frac{1}{2}.3^{21}\)
so sanh A=20^10+1/20^10-1 Va B=20^10-1/20^10-3
Ta thấy B=20^10-1/20^10-3 là phân số lớn hơn 1.
Theo tính chất nếu a/b>1 thì a/b > a+n/b+n ( n khác 0 )
Ta có : 20^10-1/20^10-3 > 20^10-1+2/20^10-3+2
<=> B > 20^10+1/20^10-3 = A
<=> B > A
Vậy B > A
cho a=(1-1/2)*(1-1/3)*(1-1/4)*...(1-1/19)*(1-1/20)
b=(1-1/4)*(1-1/9)*(1-1/16)*...*(1-1/18)*(1-1/180)
so sanh a va 1/21
so sanh b va 11/21
so sanh a=20^10/20^10-1 voi b=20^10-1/20^10-3
so sanh
A=20^10+1/20^10-1 VÀ B=20^10-1/20^10-3
Ta có công thức sau: \(\frac{a}{b}\) > \(\frac{a+m}{b+m}\) ( m khác 0;\(\frac{a}{b}\)>1)
Vì 20^10-1>20^10-3 => B>1
Áp dụng vào bài giải ta có:
A=\(\frac{\left(20^{10}-1\right)+2}{\left(20^{10}-3\right)+2}\) < \(\frac{20^{10}-1}{20^{10}-3}\)= B
Vậy A < B
Cho a=(1-1/2)*(1-1/3)*(1-1/4)*...*(1-1/19)*(1-1/20). So sanh a voi 1/21
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)\)
\(A=\left(\frac{2}{2}-\frac{1}{2}\right)\left(\frac{3}{3}-\frac{1}{3}\right)...\left(\frac{19}{19}-\frac{1}{19}\right)\left(\frac{20}{20}-\frac{1}{20}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{18}{19}.\frac{19}{20}\)
\(A=\frac{1.2.3...18.19}{2.3.4...19.20}\)
\(A=\frac{1}{20}\Leftrightarrow A>\frac{1}{21}\)
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{20}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}......\frac{19}{20}=\frac{1}{20}>\frac{1}{21}\)
\(\text{Vậy: A lớn hơn 1/21}\)
So sanh K=1/2+1/22 +1/23 +......+1/220 vơi H=1
\(\Rightarrow\)2K=\(1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{19}}\)\(\Rightarrow2K-k=k=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{19}}-k\)
\(\Rightarrow k=1-\frac{1}{2^{20}}< 1\)
\(\Rightarrow k< H\)
Vậy......