Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{199.200}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{199}-\frac{1}{200}=1-\frac{1}{200}< 1\)
Ta có:
\(A=5+5^2+5^3+5^4+...+5^{200}\)
\(5A=5.\left(5+5^2+5^3+...+5^{200}\right)\)
\(5A=5^2+5^3+5^4+...+5^{201}\)
\(5A-A=\left(5^2+5^3+5^4+...+5^{200}+5^{201}\right)-\left(5+5^2+5^3+5^4+...+5^{200}\right)\)
\(4A=5^2+5^3+5^4+...+5^{200}+5^{201}-5-5^2-5^3-5^4-...-5^{200}\)
\(4A=\left(5^2-5^2\right)+\left(5^3-5^3\right)+\left(5^4-5^4\right)+...+\left(5^{200}-5^{200}\right)+5^{201}-5\)
\(4A=0+0+0+...+0+5^{201}-5\)
\(4A=5^{201}-5\)
\(A=\frac{5^{201}-5}{4}\)
Vì \(5^{201}-5< 5^{201}\)
\(\Rightarrow\frac{5^{201}-5}{4}< \frac{5^{201}}{4}< 5^{201}\)
hay \(A< 5^{201}\)
Vậy \(A< 5^{201}\)
A = 5 + 52 + 53 + 54 + ... + 5200
5A = 52 + 53 + 54 + 55 + ... + 5201
5A - A = (52 + 53 + 54 + 55 + ... + 5201) - (5 + 52 + 53 + 54 + ... + 5200)
4A = 5201 - 5 < 5201
=> A < 5201
Bài 1:
19920 < 20020 = (23 . 52)20 = 260 . 540
200315 > 200015 = (24 . 53)15 = 260 . 545
Do: 260 . 540 < 260 . 545 => 19920 < 260 . 540 < 260 . 545 < 200315 => 19920 < 200315
Bài 2:
a/ (3 . x - 9) . 312 = 315 => 3 . x - 9 = 315 : 312
=> 3 . x - 9 = 27 => 3 . x = 27 + 9
=> 3 . x = 36 => x = 12
b/ (7 . x + 6) . 55 = 58 => 7 . x + 6 = 58 : 55
=> 7 . x + 6 = 125 => 7 . x = 125 - 6
=> 7 . x = 119 => x = 17
c/ (x - 5)4 = (x - 5)6
<=> x - 5 = 1 hoặc x - 5 = -1 hoặc x - 5 = 0
=> x = 6 hoặc x = 4 hoặc x = 5