So Sánh:
a. 5/721 và 5/834
b. 4/37 và 5/36
c. 1994/1995 và 1999/2000
d. 489/487 và 487/485
e. 123*125+119/124*125-177 và 1
f. 1/1*2+1/2*3+1/3*4+.......+1/199*200 và 1
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1: 243^5=(3^5)^5=3^25
3*27^8=3*(3^3)^8=3^25
=>243^5=3*27^8
6: 125^5=(5^3)^5=5^15
25^7=(5^2)^7=5^14
=>125^5>25^7(15>14)
5: 78^12-78^11=78^11(78-1)=78^11*77
78^11-78^10=78^10*77
mà 11>10
nên 78^12-78^11>78^11-78^10
1: 243^5=(3^5)^5=3^25
3*27^8=3*3^24=3^25=243^5
3: 3^300=27^100
2^200=4^100
mà 27>4
nên 3^300>2^200
4: 15^2=3^2*5^2
81^3*125^3=3^12*5^9
=>15^2<81^3*125^3
6: 125^5=5^15
25^7=5^14
mà 15>14
nên 125^5>25^7
A = \(\dfrac{3^{123}+1}{3^{125}+1}\) Vì 3123 + 1 < 2125 + 1 Nên A = \(\dfrac{3^{123}+1}{3^{125}+1}\)< \(\dfrac{3^{123}+1+2}{3^{125}+1+2}\)
A < \(\dfrac{3^{123}+3}{3^{125}+3}\) = \(\dfrac{3.\left(3^{122}+1\right)}{3.\left(3^{124}+1\right)}\) = \(\dfrac{3^{122}+1}{3^{124}+1}\) = B
Vậy A < B
a: \(2\dfrac{2}{10}=2,2\)
\(4\dfrac{31}{100}=4,31\)
9/10=0,9
51/10=5,1
75/100=0,75
125/100=1,25
1075/1000=1,075
\(14\dfrac{17}{100}=14,17\)
b: 1/2=0,5
3/4=0,75
4/5=0,8
5/8=0,625
\(2\dfrac{1}{2}=2,5\)
\(3\dfrac{1}{4}=3,25\)
\(1\dfrac{3}{5}=1,6\)
\(4\dfrac{3}{8}=4,375\)
Bài 1:
a: Sửa đề: 1/3^200
1/2^300=(1/8)^100
1/3^200=(1/9)^100
mà 1/8>1/9
nên 1/2^300>1/3^200
b: 1/5^199>1/5^200=1/25^100
1/3^300=1/27^100
mà 25^100<27^100
nên 1/5^199>1/3^300
a) Vì \(\dfrac{1}{24}< \dfrac{1}{83}\)
⇒ \(\dfrac{1}{24^9}>\dfrac{1}{83^{13}}\)
a) \(\left(\dfrac{1}{24}\right)^9>\left(\dfrac{1}{27}\right)^9=\dfrac{1}{3^{27}}\)
\(\left(\dfrac{1}{83}\right)^{13}< \left(\dfrac{1}{81}\right)^{13}=\dfrac{1}{3^{52}}\)
Mà \(\dfrac{1}{3^{27}}>\dfrac{1}{3^{52}}\)
\(\Rightarrow\left(\dfrac{1}{24}\right)^9>\left(\dfrac{1}{83}\right)^{13}\)
b) \(3^{300}=\left(3^3\right)^{100}=27^{100}\)
\(5^{199}< 5^{200}=\left(5^2\right)^{100}=25^{100}\)
Mà \(25^{100}< 27^{100}\)
\(\Rightarrow5^{199}< 3^{300}\)
\(\Rightarrow\dfrac{1}{5^{199}}>\dfrac{1}{3^{300}}\)
a) Vì \(721< 834\Rightarrow\frac{5}{721}>\frac{5}{834}\)
b) Ta có \(\frac{4}{37}< \frac{5}{37}< \frac{5}{36}\Rightarrow\frac{4}{37}< \frac{5}{36}\)
c) Ta có \(\frac{1994}{1995}=1-\frac{1}{1995}\)
\(\frac{1999}{2000}=1-\frac{1}{2000}\)
Vì \(\frac{1}{1995}>\frac{1}{2000}\Rightarrow1-\frac{1}{1995}< 1-\frac{1}{2000}\Rightarrow\frac{1994}{1995}< \frac{1999}{2000}\)
d) Ta có :\(\frac{489}{487}=1+\frac{2}{487}\)
\(\frac{487}{485}=1+\frac{2}{485}\)
Vì \(\frac{2}{485}>\frac{2}{487}\Rightarrow1+\frac{2}{485}>1+\frac{2}{487}\Rightarrow\frac{489}{487}>\frac{487}{485}\)
e) Ta có : \(\frac{123.125+119}{124.125-177}=\frac{123.125+119}{\left(123+1\right).125-177}=\frac{123.125+119}{123.125+125-177}=\frac{123.125+119}{123.125-52}\)
\(=\frac{123.125-52+171}{123.125-52}=1+\frac{171}{123.125-52}>1\)
f) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{199.200}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{199}-\frac{1}{200}=1-\frac{1}{200}< 1\)