So sánh và B biết
a.A=100^100+1/ 100^99+1 và B=100^101+1 /100^100+1
b.A=13^15+1/13^16+1 và B= 13^16+1/13^17+1
c.A= 1999^1999+1/1999^1998+1 và B=1999^2000+1/1999^1999+1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
14 tháng 8 2019
a) +) Có \(A=\frac{13^{15}+1}{13^{16}+1}\)=> 13A = \(\frac{13\left(13^{15}+1\right)}{13^{16}+1}\)
= \(\frac{13^{16}+13}{13^{16}+1}=\frac{13^{16}+1+12}{13^{16}+1}=1+\frac{12}{13^{16}+1}\)(1)
+) Có \(B=\frac{13^{16}+1}{13^{17}+1}\)=> 13B =\(\frac{13\left(13^{16}+1\right)}{13^{17}+1}\)
=\(\frac{13^{17}+13}{13^{17}+1}=1+\frac{12}{13^{17}+1}\)(2)
+) Từ (1) và (2) => \(1+\frac{12}{13^{16}+1}>1+\frac{12}{13^{17}+1}\)
<=> 13A>13B <=> A> B
b) +) Có A=\(\frac{1999^{1999}+1}{1999^{1998}+1}\) => \(\frac{A}{1999}=\frac{1999^{1999}+1}{1999^{1999}+1999}=\frac{1999^{1999}+1999-1998}{1999^{1999}+1999}\)
=\(1-\frac{1998}{1999^{1999}+1999}\) (1)
+) Có B =\(\frac{1999^{2000}+1}{1999^{1999}+1}\)
=> \(\frac{B}{1999}=\frac{1999^{2000}+1}{1999^{2000}+1999}=1-\frac{1998}{1999^{2000}+1999}\)(2)
+) Từ (1) và (2) => \(1-\frac{1998}{1999^{1999}+1999}\)< \(1-\frac{1998}{1999^{2000}+1999}\)
<=> \(\frac{A}{1999}< \frac{B}{1999}\) <=> A< B
16 tháng 10 2022
c: \(\dfrac{A}{10}=\dfrac{100^{100}+1}{100^{100}+10}=1-\dfrac{9}{100^{100}+10}\)
\(\dfrac{B}{10}=\dfrac{100^{69}+1}{100^{69}+10}=1-\dfrac{9}{100^{69}+10}\)
Ta có: 100^100+10>100^69+10
=>-9/(100^100+10)<-9/(100^69+10)
=>A/10<B/10
=>A<B
15 tháng 7 2018
\(\left(1+2+3+...+100\right).\left(1^2+2^2+3^3+...+100^2\right).\left(65.111-13.15.37\right)\)
\(=\left(1+2+3+...+100\right).\left(1^2+2^2+3^3+...+100^2\right).\left(7215-7215\right)\)
\(=\left(1+2+3+...+100\right).\left(1^2+2^2+3^3+...+100^2\right).0\)
\(=0\)
\(1999.1999.1998-1998.1998.1999\)
\(=1999.1998.\left(1999-1998\right)\)
\(=1999.1998.1\)
Tham khảo nhé~
18 tháng 3 2017
Bài 1:
a) Ta có: \(13A=\dfrac{13^{16}+13}{13^{16}+1}=1+\dfrac{12}{13^{16}+1}\)
\(13B=\dfrac{13^{17}+13}{13^{17}+1}=1+\dfrac{12}{13^{17}+1}\)
Vì \(\dfrac{12}{13^{16}+1}>\dfrac{12}{13^{17}+1}\Rightarrow1+\dfrac{12}{13^{16}+1}>1+\dfrac{12}{13^{17}+1}\)
\(\Rightarrow13A>13B\)
\(\Rightarrow A>B\)
Vậy A > B
b) Ta có: \(1999C=\dfrac{1999^{2000}+1999}{1999^{2000}+1}=1+\dfrac{1998}{1999^{2000}+1}\)
\(1999D=\dfrac{1999^{1999}+1999}{1999^{1999}+1}=1+\dfrac{1998}{1999^{1999}+1}\)
\(\dfrac{1998}{1999^{2000}+1}< \dfrac{1998}{1999^{1999}+1}\Rightarrow1+\dfrac{1998}{1999^{2000}+1}< 1+\dfrac{1999}{1999^{1999}+1}\)
\(\Rightarrow1999C< 1999D\)
\(\Rightarrow C< D\)
Vậy C < D
ML
9 tháng 7 2017
a)
\(1-\frac{1998}{1999}=\frac{1}{1999}\)
\(1-\frac{1999}{2000}=\frac{1}{2000}\)
Vì \(\frac{1}{1999}>\frac{1}{2000}\)nên \(\frac{1998}{1999}< \frac{1999}{2000}\)
b) Ta có :
\(\frac{1999}{2001}< 1\)
\(\frac{12}{11}>1\)
Nên \(\frac{1999}{2001}< \frac{12}{11}\)
c)
\(1-\frac{13}{27}=\frac{14}{27}\)
\(1-\frac{27}{41}=\frac{14}{41}\)
Vì \(\frac{14}{27}>\frac{14}{41}\)nên \(\frac{13}{27}< \frac{27}{41}\)
d)
Ta có phân số trung gian là \(\frac{23}{45}\).
Ta có : \(\frac{23}{47}< \frac{23}{45}\) ; \(\frac{24}{45}>\frac{23}{45}\)
Nên \(\frac{23}{47}< \frac{24}{45}\)
a. Có: \(\frac{100^{101}+1}{100^{100}+1}>1\Rightarrow\frac{100^{101}+1}{100^{100}+1}>\frac{100^{101}+\left(1+99\right)}{100^{100}+\left(1+99\right)}\)
\(\Rightarrow B>\frac{100^{101}+100}{100^{100}+100}\\ \Rightarrow B>\frac{100\left(100^{100}+1\right)}{100\left(100^{99}+1\right)}\\ \Rightarrow B>\frac{100^{100}+1}{100^{99}+1}=A\\ \Leftrightarrow A< B\)
Vậy A < B
b. Có: \(\frac{13^{16}+1}{13^{17}+1}< 0\Rightarrow\frac{13^{16}+1}{13^{17}+1}< \frac{13^{16}+\left(1+12\right)}{13^{17}+\left(1+12\right)}\)
\(\Rightarrow B< \frac{13^{16}+13}{13^{17}+13}\\ \Rightarrow B< \frac{13\left(13^{15}+1\right)}{13\left(13^{16}+1\right)}\\ \Rightarrow B< \frac{13^{15}+1}{13^{16}+1}=A\\ \Leftrightarrow A>B\)
Vậy A > B
c. Có: \(\frac{1999^{2000}+1}{1999^{1999}+1}>1\Rightarrow\frac{1999^{2000}+1}{1999^{1999}+1}>\frac{1999^{2000}+\left(1+1998\right)}{1999^{1999}+\left(1+1998\right)}\)
\(\Rightarrow B>\frac{1999^{2000}+1999}{1999^{1999}+1999}\\ \Rightarrow B>\frac{1999\left(1999^{1999}+1\right)}{1999\left(1999^{1998}+1\right)}\\ \Rightarrow B>\frac{1999^{1999}+1}{1999^{1998}+1}=A\\ \Leftrightarrow A< B\)
Vậy A < B