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A=-2015/2015x2016
A=-1/2016
B=-2014/2014x2015
B=-1/2015
vi 2016>2015,-1/2016>-1/2015
vay A>B
b) Ta có: \(A=\dfrac{10^{2009}+1}{10^{2010}+1}\)
\(\Leftrightarrow10A=\dfrac{10^{2010}+10}{10^{2010}+1}=1+\dfrac{9}{10^{2010}+1}\)
Ta có: \(B=\dfrac{10^{2010}+1}{10^{2011}+1}\)
\(\Leftrightarrow10B=\dfrac{10^{2011}+10}{10^{2011}+1}=1+\dfrac{9}{10^{2011}+1}\)
Ta có: \(10^{2010}+1< 10^{2011}+1\)
\(\Leftrightarrow\dfrac{9}{10^{2010}+1}>\dfrac{9}{10^{2011}+1}\)
\(\Leftrightarrow\dfrac{9}{10^{2010}+1}+1>\dfrac{9}{10^{2011}+1}+1\)
\(\Leftrightarrow10A>10B\)
hay A>B
a,\(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\)
\(=>5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)
\(=>5A-A=1-\frac{1}{5^{100}}=>A=\frac{1-\frac{1}{5^{100}}}{4}\)
b, Ta có \(1-\frac{1}{5^{100}}< 1=>\frac{1-\frac{1}{5^{100}}}{4}< \frac{1}{4}\)hay \(A< \frac{1}{4}\)
Ta có công thức :
\(\frac{a}{b}< \frac{a+c}{b+c}\)\(\left(\frac{a}{b}< 1;a,b,c\inℕ^∗\right)\)
Áp dụng vào ta có :
\(B=\frac{10^{2014}+1}{10^{2015}+1}< \frac{10^{2014}+1+9}{10^{2015}+1+9}=\frac{10^{2014}+10}{10^{2015}+10}=\frac{10\left(10^{2013}+1\right)}{10\left(10^{2014}+1\right)}=\frac{10^{2013}+1}{10^{2014}+1}=A\)
\(\Rightarrow\)\(B< A\) hay \(A>B\)
Vậy \(A>B\)
Chúc bạn học tốt ~
áp dụng tính chất
nếu a/b>1thì a/b<(a+n)/(b+n)
=)))))))))))))))))
ta có :
\(10A=\frac{10^{2014}+10}{10^{2014}+1}=\frac{\left(10^{2014}+1\right)+9}{10^{2014}+1}=1+\frac{9}{10^{2014}+1}\)
\(10B=\frac{10^{2015}+10}{10^{2015}+1}=\frac{\left(10^{2015}+1\right)+9}{10^{2015}+1}=1+\frac{9}{10^{2015}+1}\)
ta thấy \(10^{2014}+1< 10^{2015}+1\Rightarrow\frac{9}{10^{2014}+1}>\frac{9}{10^{2015}+1}\Rightarrow10A>10B\Rightarrow A>B\)
\(10A=\dfrac{10^{2015}+2016+9\cdot2016}{10^{2015}+2016}=1+\dfrac{18144}{10^{2015}+2016}\)
\(10B=\dfrac{10^{2016}+9+18144}{10^{2016}+2016}=1+\dfrac{18144}{10^{2016}+2016}\)
mà \(\dfrac{18144}{10^{2015}+2016}>\dfrac{18144}{10^{2016}+2016}\)
nên A>B
có :
\(B=\frac{10^{2015}+1}{10^{2014}+1}>1\)
\(\Rightarrow\frac{10^{2015}+1}{10^{2014}+1}>\frac{10^{2015}+1+9}{10^{2014}+1+9}\) \(=\frac{10^{2015}+10}{10^{2014}+10}=\frac{10.\left(10^{2014}+1\right)}{10.\left(10^{2013}+1\right)}\)
\(=\frac{10^{2014}+1}{10^{2013}+1}=A\)
\(\Rightarrow B>A\)
Vậy B > A
k cho mk nhé
Lời giải:
a) Xét hiệu \(\frac{a+n}{b+n}-\frac{a}{b}=\frac{(a+n).b-a(b+n)}{b(b+n)}=\frac{n(b-a)}{b(b+n)}\)
Nếu $b>a$ thì $\frac{a+n}{b+n}-\frac{a}{b}>0\Rightarrow \frac{a+n}{b+n}>\frac{a}{b}$
Nếu $b<a$ thì $\frac{a+n}{b+n}-\frac{a}{b}<0\Rightarrow \frac{a+n}{b+n}<\frac{a}{b}$
Nếu $b=a$ thì $\frac{a+n}{b+n}-\frac{a}{b}=0\Rightarrow \frac{a+n}{b+n}=\frac{a}{b}$
b) Rõ ràng $10^{11}-1< 10^{12}-1$.
Đặt $10^{11}-1=a; 10^{12}-1=b; 11=n$ thì: $a< b$; $A=\frac{a}{b}$ và $B=\frac{10^{11}+10}{10^{12}+10}=\frac{a+n}{b+n}$
Áp dụng kết quả phần a:
$b>a\Rightarrow \frac{a+n}{b+n}>\frac{a}{b}$ hay $B>A$
2)Ta có: \(2^{332}< 2^{333}=\left(2^3\right)^{111}=8^{111}\)
\(3^{223}>3^{222}=\left(3^2\right)^{111}=9^{111}\)
Vì \(8^{111}< 9^{111}\) mà \(2^{332}< 8^{111},3^{223}>9^{111}\) nên suy ra \(2^{332}< 3^{223}\)
Vậy \(2^{332}< 3^{223}\)
1) \(A=\dfrac{10^{2013}+1}{10^{2014}+1}\Rightarrow10A=\dfrac{10^{2014}+10}{10^{2014}+1}=\dfrac{10^{2014}+1}{10^{2014}+1}+\dfrac{9}{10^{2014}+1}=1+\dfrac{9}{10^{2014}+1}\)
\(B=\dfrac{10^{2014}+1}{10^{2015}+1}\Rightarrow10B=\dfrac{10^{2015}+10}{10^{2015}+1}=\dfrac{10^{2015}+1}{10^{2015}+1}+\dfrac{9}{10^{2015}+1}=1+\dfrac{9}{10^{2015}+1}\)Vì: \(10^{2014}+1< 10^{2015}+1\Rightarrow\dfrac{9}{10^{2014}+1}>\dfrac{9}{10^{2015}+1}\Rightarrow1+\dfrac{9}{10^{2014}+1}>1+\dfrac{9}{10^{2015}+1}\)
Nên suy ra \(10A>10B\Rightarrow A>B\)