So sánh A và B
\(A=\frac{100^{2015}+1}{100^{2014}+1}\)
\(B=\frac{100^{2016}+1}{100^{2015}+1}\)
Gấp nha!
A=\(\frac{100^{2015}+1}{100^{2016}+1}\)
B=\(\frac{100^{2016}+1}{100^{2017}+1}\)
So sánh A và B
a, Cho A=\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+...+\frac{1}{99}+\frac{1}{100}\) . So Sánh A với 1
b, B=\(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\). So sánh B với \(\frac{1}{2}\)
c, cho M=\(\frac{2013}{2014}+\frac{2014}{2015}\)và N=\(\frac{2013+2014}{2014+2015}\). So sánh M và N
Câu a, p/s cuối cùng là \(\frac{1}{100}\)nha mí bn
a) Ta có :
\(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{100}\)
\(>\frac{1}{10}+\frac{1}{100}.90=\frac{1}{10}+\frac{90}{100}=1\)
vậy A > 1
b) \(B=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\)
\(>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{1}{20}.10=\frac{1}{2}\)
Vậy B > \(\frac{1}{2}\)
So sánh :
A = \(\frac{100^{2015}+1}{100^{2016}+1}\) và B = \(\frac{100^{2016}+1}{100^{2017}+1}\)
Ta có:
A=100^2015+1/100^2016+1 suy ra 100A=100^2016+100/100^2016+1=100^2016+1+99/100^2016+1=1/99/100^2016+1
Lại có
B=100^2016+1/100^2017+1 suy ra 100B=100^2017+100/100^2017+1=100^2017+1+99/100^2017+1=1/99/100^2017+1
Vì1/99/100^2016+1>1/99/100^2017+1 suy ra A>B
so sánh
\(\frac{100^{2015^{ }}+1}{100^{2005}+1}\) và\(\frac{100^{2016}+1}{100^{2006}+1}\)
\(\frac{100^{2015}+1}{100^{2015}+1}=1\)
\(\frac{100^{2016}+1}{100^{2016}+1}=1\)
Vì 1 = 1 nên \(\frac{100^{2015}+1}{100^{2015}+1}=\frac{100^{2016}+1}{100^{2016}+1}\)
à mình nhìn nhầm đề
Mình giải nha
Đặt \(A=\frac{100^{2015}+1}{100^{2005}+1}\Rightarrow\frac{A}{100^{10}}=\frac{100^{2015}+1}{100^{2015}+100^{10}}=\frac{100^{2015}+100^{10}-999}{100^{2015}+100^{10}}=1-\frac{999}{100^{2015}+100^{10}}\)
Đặt \(B=\frac{100^{2016}+1}{100^{2006}+1}\Rightarrow\frac{B}{100^{10}}=\frac{100^{2016}+100^{10}-999}{100^{2016}+100^{10}}=1-\frac{999}{100^{2016}+100^{10}}\)
\(1-\frac{999}{100^{2015}+100^{10}}< 1-\frac{999}{100^{2016}+100^{10}}\Rightarrow A< B\)
Rõ ràng\(\frac{100^{2016}+1}{100^{2006}+1}\)<1 nên theo tính chất khi \(\frac{a}{b}\)< 1 => \(\frac{a}{b}\)<\(\frac{a+m}{b+m}\) => \(\frac{100^{2016}+1}{100^{2006}+1}\)<\(\frac{100^{2016}+1+99}{100^{2006}+1+99}\)
<\(\frac{100^{2016}+100}{100^{2006}+100}\)
=>\(\frac{100^{2016}+1}{100^{2006}+1}\)< \(\frac{100^{2016}+100}{100^{2006}+100}\) = \(\frac{100\left(100^{2015}+1\right)}{100\left(100^{2005}+1\right)}\)= \(\frac{\left(100^{2015}+1\right)}{\left(100^{2005}+1\right)}\)
Vậy\(\frac{100^{2016}+1}{100^{2006}+1}\) < \(\frac{\left(100^{2015}+1\right)}{\left(100^{2005}+1\right)}\)
Cho A=100^2016+1/100^2015-1 và B=100^2015+1/100^2014-1
Trong 2 số A và B, số nào lớn hơn?
Giúp mình với nhé!
\(A=\frac{100^{2016}+1}{100^{2015}-1}\)
\(\frac{1}{100}.A=\frac{100^{2016}+1}{100\left(100^{2015}-1\right)}\)
\(=\frac{100^{2016}+1}{100^{2016}-100}\)
\(=\frac{\left(100^{2016}-100\right)+101}{100^{2016}-100}\)
\(=\frac{100^{2016}-100}{100^{2016}-100}\)\(+\frac{101}{100^{2016}-100}\)
\(=1+\frac{101}{100^{2016}-100}\)
\(B=\frac{100^{2015}+1}{100^{2014}-1}\)
\(\frac{1}{100}.B=\frac{100^{2015}+1}{100\left(100^{2014}-1\right)}\)
\(=\frac{100^{2015}+1}{100^{2015}-100}\)
\(=\frac{\left(100^{2015}-100\right)+101}{100^{2015}-100}\)
\(=\frac{100^{2015}-100}{100^{2015}-100}\)\(+\frac{101}{100^{2015}-100}\)
\(=1+\frac{101}{100^{2015}-100}\)
\(\hept{\begin{cases}Vì101>0\\100^{2016}-100>100^{2015}-100>0\end{cases}}\)
\(\Rightarrow\frac{101}{100^{2016}-100}< \frac{101}{100^{2015}-100}\)
\(\Rightarrow1+\frac{101}{100^{2016}-100}< 1+\frac{101}{100^{2015}-100}\)
\(\Rightarrow\frac{1}{100}.A< \frac{1}{100}.B\)
\(\Rightarrow A< B\left(vì\frac{1}{100}>0\right)\)
Vậy A<B
So sánh : \(A=\frac{2015^{2016}+1}{2015^{2015}+1}\) và \(B=\frac{2014^{2015}+1}{2014^{2014}+1}\)
A = \(\frac{2015^{2016}+1}{2015^{2015}+1}=\frac{2015^{2015}+1}{2015^{2015}+1}+\frac{2015}{2015^{2015}+1}=1+\frac{2015}{2015^{2015}+1}\)
B = \(\frac{2014^{2015}+1}{2014^{2014}+1}=\frac{2014^{2014}+1}{2014^{2014}+1}+\frac{2014}{2014^{2014}+1}=1+\frac{2014}{2014^{2014}+1}\)
Rồi bạn tự so sánh nha
A=5^2016+4/5^2015+4 B =\(\frac{5^{2014}+4}{5^{2013}+4}\)
A=\(\frac{3^{98}-1}{3^{100}-1}\)b=\(\frac{3^{98}-1}{3^{100}-1}\)
@@@) Ta có: \(A=\frac{5^{2016}+4}{5^{2015}+4}\Rightarrow\frac{1}{5}A=\frac{5^{2016}+4}{5^{2016}+20}=1+\frac{-16}{5^{2016}+20}\)
\(B=\frac{5^{2014}+4}{5^{2013}+4}\Rightarrow\frac{1}{5}B=\frac{5^{2014}+4}{5^{2014}+20}=1+\frac{-16}{5^{2014}+20}\)
Ta thấy: \(1+\frac{-16}{5^{2016}+20}>1+\frac{-16}{5^{2014}+20}\) =>\(\frac{1}{5}A>\frac{1}{5}B\Rightarrow A>B\)
Bài thứ 2 sai để nhé hai cái đó = nhau mà
Bài 1.So sánh A và B biết: A=\(\frac{10^{17}+1}{10^{18}+1}\) B=\(\frac{10^{18}+1}{10^{19}+1}\)
Bài 2.So sánh S=\(\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2013}\)với 4
Bài 3.Cho A=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)Chứng minh rằng A<\(\frac{3}{4}\)
Bài 4.
a)Tính nhanh tổng sau:A=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\)
b)Tìm x biết:\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x.\left(x+2\right)}=\frac{1008}{2017}\)
mn giúp mk nha mk đang cần gấp
ai nhanh mk sẽ tick cho
tk mn
Bài 1 :
Ta có :
\(A=\frac{10^{17}+1}{10^{18}+1}=\frac{\left(10^{17}+1\right).10}{\left(10^{18}+1\right).10}=\frac{10^{18}+10}{10^{19}+10}\)
Mà : \(\frac{10^{18}+10}{10^{19}+10}>\frac{10^{18}+1}{10^{19}+1}\)
Mà \(A=\frac{10^{18}+10}{10^{19}+10}\)nên \(A>B\)
Vậy \(A>B\)
Bài 2 :
Ta có :
\(S=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2013}\)
\(\Rightarrow S=\frac{2014-1}{2014}+\frac{2015-1}{2015}+\frac{2016-1}{2016}+\frac{2013+3}{2013}\)
\(\Rightarrow S=1-\frac{1}{2014}+1-\frac{1}{2015}+1-\frac{1}{2016}+1+\frac{3}{2013}\)
\(\Rightarrow S=4+\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)\)
Vì \(\frac{1}{2013}>\frac{1}{2014}>\frac{1}{2015}>\frac{1}{2016}\)nên \(\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)>0\)
Nên : \(M>4\)
Vậy \(M>4\)
Bài 3 :
Ta có :
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.......+\frac{1}{100^2}\)
Suy ra : \(A< \frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+....+\frac{1}{99.101}\)
\(\Rightarrow A< \frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{2.4}+......+\frac{2}{99.101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-......-\frac{1}{101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left[\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{99}\right)-\left(\frac{1}{3}+\frac{1}{4}+......+\frac{1}{101}\right)\right]\)
\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{100}-\frac{1}{101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}\right)\)
\(\Rightarrow A< \frac{3}{4}\)
Vậy \(A< \frac{3}{4}\)
Bài 4 :
\(a)A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2015.2017}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{1}{2015.2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{2016}{2017}\)
\(\Rightarrow A=\frac{1008}{2017}\)
Vậy \(A=\frac{1008}{2017}\)
\(b)\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+......+\frac{1}{x\left(x+2\right)}=\frac{1008}{2017}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{x.\left(x+2\right)}=\frac{2016}{2017}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{x+2}=\frac{2016}{2017}\)
\(1-\frac{1}{x+2}=\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{2017}\)
\(\Rightarrow x+2=2017\)
\(\Rightarrow x=2017-2=2015\)
Vậy \(x=2015\)
Cho tổng A gồm 2016 số hạng A=\(\frac{1}{19^1}+\frac{2}{19^2}_{ }+\frac{3}{19^3}+..................+\frac{n}{19^n}+.....+\frac{2016}{19^{2016}}\)
Hãy so sánh A^2016 và A^2015
Ai giải được cho 100 tick
Không cần giải cũng biết đáp án:
Nếu A là số dương thì A^2016>A^2015
Nếu A là số âm thì A^2016 là số dương , A^2015 là số âm nên chắc chắn A^2016>A^2015
k nha