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a) Ta có A = \(\frac{2^{2018}+1}{2^{2019}+1}\)
=> 2A = \(\frac{2^{2019}+2}{2^{2019}+1}=1+\frac{1}{2^{2019}+1}\)
Lại có B = \(\frac{2^{2017}+1}{2^{2018}+1}\)
=> 2B = \(\frac{2^{2018}+2}{2^{2018}+1}=\frac{2^{2018}+1+1}{2^{2018}+1}=1+\frac{1}{2^{2018}+1}\)
Vì \(\frac{1}{2^{2018}+1}>\frac{1}{2^{2019}+1}\Rightarrow1+\frac{1}{2^{2018}+1}>1+\frac{1}{2^{2019}+1}\Rightarrow2B>2A\Rightarrow B>A\)
TL :
Ko biết thì đừng làm
Nhớ làm hết , chi tiết mới đc 1 SP
HT
a) \(\frac{2^{10}+1}{2^{10}-1}\)và \(\frac{2^{10}-1}{2^{10}-3}\)
Ta có chính chất phân số trung gian là \(\frac{2^{10}+1}{2^{10}-3}\)
\(\frac{2^{10}+1}{2^{10}-1}>\frac{2^{10}+1}{2^{10}-3}\) ; \(\frac{2^{10}-1}{2^{10}-3}< \frac{2^{10}+1}{2^{10}-3}\)
Vì \(\frac{2^{10}+1}{2^{10}-1}>\frac{2^{10}+1}{2^{10}-3}>\frac{2^{10}-1}{2^{10}-3}\)
Nên \(\frac{2^{10}+1}{2^{10}-1}>\frac{2^{10}-1}{2^{10}-3}\)
b) \(A=\frac{2011}{2012}+\frac{2012}{2013}\)và \(B=\frac{2011+2012}{2012+2013}\)
Ta có : \(A=\frac{2011}{2012}+\frac{2012}{2013}>\frac{2011}{2013}+\frac{2012}{2013}=\frac{2011+2012}{2013}>\frac{2011+2012}{2012+2013}=B\)
Vậy A > B
Có gì sai cho sorry
a,
\(\frac{2^{10}+1}{2^{10}-1}=1+\frac{2}{2^{10}-1}< 1+\frac{2}{2^{10}-3}=\frac{2^{10}-1}{2^{10}-3}\)
b,
\(\frac{2011}{2012}+\frac{2012}{2013}>\frac{2011}{2012+2013}+\frac{2012}{2012+2013}=\frac{2011+2012}{2012+2013}\)
Đề hình như sai rùi bn, ở A mẫu phải là 108 - 1 chứ
Áp dụng a/b < 1 => a/b < a+m/b+m (a;b;m thuộc N*)
Ta có:
\(B=\frac{10^8}{10^8-3}< \frac{10^8+2}{10^8-3+2}=\frac{10^8+2}{10^8-1}=A\)
=> B < A
a) Ta có :
\(A=\frac{10^{2010}+1}{10^{2011}+1}\)
\(\Rightarrow10A=\frac{10^{2011}+10}{10^{2011}+1}=\frac{\left(10^{2011}+1\right)+9}{10^{2011}+1}=1+\frac{9}{10^{2011}+1}\)
\(B=\frac{10^{2011}+1}{10^{2012}+1}\)
\(\Rightarrow10B=\frac{10^{2012}+10}{10^{2012}+1}=\frac{\left(10^{2012}+1\right)+9}{10^{2012}+1}=1+\frac{9}{10^{2012}+1}\)
Vì \(\frac{9}{10^{2011}+1}>\frac{9}{10^{2012}+1}\)nên \(10A>10B\)
\(\Rightarrow A>B\)
Vậy : \(A>B\)
b) Ta có :
\(\left(\frac{-1}{2}\right)^{11}=\frac{-1^{11}}{2^{11}}=\frac{-1}{2^{11}}\)
\(\left(\frac{-1}{2}\right)^{13}=\frac{-1^{13}}{2^{13}}=\frac{-1}{2^{13}}\)
Vì \(\frac{-1}{2^{11}}>\frac{-1}{2^{13}}\)nên \(\left(\frac{-1}{2}\right)^{11}>\left(\frac{-1}{2}\right)^{13}\)
Vậy : \(\left(\frac{-1}{2}\right)^{11}>\left(\frac{-1}{2}\right)^{13}\)
\(B=\frac{10^{2011}+1}{10^{2012}+1}< \frac{10^{2011}+1+9}{10^{2012}+1+9}\)
\(B=\frac{10^{2011}+1}{10^{2012}+1}< \frac{10^{2011}+10}{10^{2012}+10}\)
\(B=\frac{10^{2011}+1}{10^{2012}+1}< \frac{10\cdot\left(10^{2010}+1\right)}{10\cdot\left(10^{2011}+1\right)}=\frac{10^{2010}+1}{10^{2011}+1}=A\)
Vậy : B < A
(a+b+c)2=a2+b2+c2+2ab+2cb+2ac>a2+b2+c2
(a+b+c)2/3>(a2+b2+c2)/3
ta có \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
tương đương \(2a^2+2b^2+2c^2\ge2ab+2bc+2ac\)
cộng 2 vế với a2+b2+c2, ta có\(3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)
chia cả 2 vế cho 9,ta có bđt cần chứng minh