so sánh
\(\frac{a-1}{a}\)và\(\frac{b-1}{b}\)(a,b >0)
So sánh A và B bằng cách so sánh với 1:
\(A=\frac{2010}{2011}+\frac{2011}{2012}\)và \(B=\frac{2010+2011}{2011+2012}\)
ta có :
\(B=\frac{2010+2011}{2011+2012}=\frac{2010}{2011+2012}+\frac{2011}{2011+2012}\)
ta có : \(\frac{2010}{2011}>\frac{2010}{2011+2012}\)
\(\frac{2011}{2012}>\frac{2011}{2011+2012}\)
=> \(\frac{2010}{2011}+\frac{2011}{2012}>\frac{2010+2011}{2011+2012}\)
hay A>B
Bài 1: Cho S = 1 + 3 + 32+ 33 + 34 +...+ 32017. Hãy tính: A = 2S + 2018
Bài 2: So sánh \(\frac{2017}{555^{333}}\)và \(\frac{2017}{333^{555}}\)
Bài 3: Cho \(\frac{a}{2b}=\frac{2b-2015c}{2c}=\frac{2016c-b}{2a}\)( a ; b ; c \(\ne\)0 ) và a + b + c \(\ne\)0.
Chứng minh: a = b
Bài 4: So sánh \(\frac{1}{444^{555}}\)và \(\frac{1}{555^{444}}\)
Cho A =\(\frac{100^{2007}+1}{100^{2008}+1}\)và cho B=\(\frac{100^{2006}+1}{100^{2007}+1}\).Hãy so sánh A và B
\(A=\frac{100^{2007}+1}{100^{2008}+1}\Rightarrow100.A=\frac{100^{2008}+100}{100^{2008}+1}=\frac{100^{2008}+1+99}{100^{2008}+1}=1+\frac{99}{100^{2008}+1}\)
\(B=\frac{100^{2006}+1}{100^{2007}+1}\Rightarrow100.B=\frac{100^{2007}+100}{100^{2007}+1}=\frac{100^{2007}+1+99}{100^{2007}+1}=1+\frac{99}{100^{2007}+1}\)
Vì \(\frac{99}{100^{2007}+1}>\frac{99}{100^{2008}+1};1=1\Rightarrow1+\frac{99}{100^{2007}+1}>1+\frac{99}{100^{2008}+1}\)hay \(A>B\)
Vậy \(A>B\)
so sánh A và B
A=\(\frac{1}{2}\)và B =\(\frac{\sqrt{x}-2}{3\sqrt{x}}\)
So sánh A và B biết:
\(A=\frac{10^{2011}+1}{10^{2012}+1};B=\frac{10^{2012}+1}{10^{2013}+1}\)
So sánh 2 phân số sau $\frac{10^{2011}+10}{10^{2012}+10}v\text{à}\frac{10^{2012}-10}{10^{2013}-10}$102011+10102012+10 và102012−10102013−10
kick dzô chữ xanh là được!! OK
Ta có :
10. A = \(\frac{10.\left(10^{2011}+1\right)}{10^{2012}+1}\)
= \(\frac{10^{2012}+10}{10^{2012}+1}\)
= \(\frac{10^{2012}+1+9}{10^{2012}+1}\)
= \(\frac{10^{2012}+1}{10^{2012}+1}-\frac{9}{10^{2012}+1}\)
= 1 - \(\frac{9}{10^{2012}+1}\)
10 . B = \(\frac{10.\left(10^{2012}+1\right)}{10^{2013}+1}\)
= \(\frac{10^{2013}+10}{10^{2013}+1}\)
= \(\frac{10^{2013}+1+9}{10^{2013}+1}\)
= 1 - \(\frac{9}{10^{2013}+1}\)
Vì \(\frac{9}{10^{2012}+1}\) >\(\frac{9}{10^{2013}+1}\) nên 10.A > 10.B
=> A >B
Vậy ...........
So sánh A và B biết:
A=\(\frac{10^{17}+1}{10^{18}+1}\), B=\(\frac{10^{18}+1}{10^{19}+1}\)
Ta có: \(A=\frac{10^{18}+1}{10^{19}+1}>\frac{10.\left(10^{17}+1\right)}{10.\left(10^{18}+1\right)}=\frac{10^{17}+1}{10^{18}+1}\)
Vậy A < B
A=\(\frac{2016^{2016}+1}{2016^{2017}+1}\)
B=\(\frac{2016^{2015}+1}{2016^{2016}+1}\)
So sánh A và B
Vì 20162016 + 1 < 20162017 + 1
\(\Rightarrow\frac{2016^{2016}+1}{2016^{2017}+1}< \frac{2016^{2016}+1+2015}{2016^{2016}+1+2015}=\frac{2016^{2016}+2016}{2016^{2017}+2016}=\frac{2016\left(2016^{2015}+1\right)}{2016\left(2016^{2016}+1\right)}\)
\(=\frac{2016^{2015}+1}{2016^{2016}+1}=B\)
\(\Rightarrow\)A < B
Ta có :
\(A=\frac{2016^{2016}+1}{2016^{2017}+1}< \frac{2016^{2016}+2015+1}{2016^{2017}+2015+1}=\frac{2016^{2016}+2016}{2016^{2017}+2016}=\frac{2016.\left(2016^{2015}+1\right)}{2016.\left(2016^{2016}+1\right)}\)
\(=\frac{2016^{2015}+1}{2016^{2016}+1}=B\)
\(\Rightarrow A< B\)
\(A=\frac{2016^{2016}+1}{2016^{2017}+1}< \frac{2016^{2016}+2015+1}{2016^{2017}+2015+1}\)
\(A=\frac{2016^{2016}+2016}{2016^{2017}+2016}\)
\(=\frac{2016.\left(2016^{2015}+1\right)}{2016.\left(2016^{2016}+1\right)}\)
Mả \(\frac{2016^{2015}+1}{2016^{2016}+1}=B\)
\(\Leftrightarrow A< B\)
So sánh A và B biết:
A=\(\frac{10^{17}+1}{10^{18}+1}\), B=\(\frac{10^{18}+1}{10^{19}+1}\)
Vì \(\frac{10^{18}+1}{10^{19}+1}< 1\Rightarrow B=\frac{10^{18}+1}{10^{19}+1}< \frac{10^{18}+1+9}{10^{19}+1+9}\)
\(\Rightarrow B< \frac{10^{18}+10}{10^{19}+10}\)
\(\Rightarrow B< \frac{10\left(10^{17}+1\right)}{10\left(10^{18}+1\right)}\)
\(\Rightarrow B< \frac{10^{17}+1}{10^{18}+1}\)
\(\Rightarrow B< A\)
Vậy A > B.
So sánh A và B:
a) A = \(\frac{10^{19}+1}{10^{20}+1}\); B = \(\frac{10^{20}+1}{10^{21}+1}\)
b) A = \(\frac{9^{99}+1}{9^{100}+1}\); B = \(\frac{10^{98}-1}{10^{99}-1}\)