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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: \(10A=10.\left(\frac{10^{2014}+1}{10^{2015}+1}\right)=\frac{10^{2015}+10}{10^{2015}+1}=\frac{10^{2015}+1+9}{10^{2015}+1}=1+\frac{9}{10^{2015}+1}\)
\(10B=10.\left(\frac{10^{2015}+1}{10^{2016}+1}\right)=\frac{10^{2016}+10}{10^{2016}+1}=\frac{10^{2016}+1+9}{10^{2016}+1}=1+\frac{9}{10^{2016}+1}\)
Vì 1 = 1; 9 = 9 ta so sánh mẫu:
Ta có: 102015 < 102016 => 102015+1 < 102016+1
=> \(1+\frac{9}{10^{2015}+1}>1+\frac{9}{10^{2016}+1}\)
=> 10A > 10B
=> A > B.
Ta có:
\(10A=\frac{10^{2015}+20200}{10^{2015}+2020}=1+\frac{18180}{10^{2015}+2020}\)
\(10B=\frac{10^{2016}+20200}{10^{2016}+2020}=1+\frac{18180}{10^{2016}+2020}\)
Vì \(10^{2016}+2020>2^{2015}+2020\)
=> \(\frac{18180}{10^{2016}+2020}< \frac{18180}{10^{2015}+2020}\)
=> \(1+\frac{18180}{10^{2016}+2020}< 1+\frac{18180}{10^{2015}+2020}\)
=> 10B < 10A
=> B<A
Ta có :
\(A=\frac{10^{2016}+1}{10^{2015}+1}=\frac{\left(10^{2016}+1\right).10}{\left(10^{2015}+1\right).10}=\frac{10^{2017}+10}{10^{2016}+10}=\frac{10^{2017}+10}{10^{2016}+10}\)
Vì \(10^{2017}=10^{2017}\)và \(10>1\)nên \(10^{2017}+10>10^{2017}+1\)( 1 )
Vì \(10^{2016}=10^{2016}\)và \(10>1\)nên \(10^{2016}+10>10^{2016}+1\)( 2 )
Từ ( 1 ) và ( 2 ) , suy ra : \(\frac{10^{2017}+10}{10^{2016}+10}>\frac{10^{2017}+1}{10^{2016}+1}\)
Vậy \(A>B\)
\(B=\frac{10^{2016}+1}{10^{2017}+1}=\frac{10^{2016}+1+9}{10^{2017}+1+9}=\frac{10^{2016}+10}{10^{2017}+10}=\frac{10.\left(10^{2015}+1\right)}{10.\left(10^{2016}+1\right)}=\frac{10^{2015}+1}{10^{2016}+1}\)
lm tương tự vs B ta có
\(A=\frac{10^{2015}+1}{10^{2014}+1}\)
suy ra A>B
mình nhầm câu b:
Áp dụng....
A=10^11-1/10^12-1<10^11-1+11/10^12-1+11=10^11+10/10^12+10=10.(10^10+1)/10.(10^11+1)
=10^10+1/10^11+1=B
Vậy A<B(câu này mới đúng còn câu b mình làm chung với câu a là sai)
a) Với a<b=>a+n/b+n >a/b
Với a>b=>a+n/b+n<a/b
Với a=b=>a+n/b+n=a/b
b) Áp dụng t/c a/b<1=>a/b<a+m/b+m(a,b,m thuộc z,b khác 0)ta có:
A=(10^11)-1/(10^12)-1=(10^11)-1+11/(10^12)-1+11=(10^11)+10/(10^12)+10=10.[(10^10)+1]/10.[(10^11)+1]
=(10^10)+1/(10^11)+1=B
Vậy A=B
A=\(\frac{10^{2015}+1}{10^{2016}+1}\)=>10A=\(\frac{10.\left(10^{2015}+1\right)}{10^{2016}+1}\)= \(\frac{10^{2016}+10}{10^{2016}+1}\)=\(\frac{\left(10^{2016}+1\right)+9}{10^{2016}+1}\)=\(\frac{10^{2016}+1}{10^{2016}+1}+\frac{9}{10^{2016}+1}\)=1+\(\frac{9}{10^{2016}+1}\)
B=\(\frac{10^{2016}+1}{10^{2017}+1}\)=>10B=\(\frac{10.\left(10^{2016}+1\right)}{10^{2017+1}}=\frac{10^{2017}+10}{10^{2017}+1}\)= \(\frac{\left(10^{2017}+1\right)+9}{10^{2017}+1}\)=\(\frac{10^{2017}+1}{10^{2017}+1}+\frac{9}{10^{2017}+1}\)= 1+\(\frac{9}{10^{2017}+1}\)
Vì \(10^{2016}+1< 10^{17}+1\)=>\(\frac{9}{10^{2016}+1}\)>\(\frac{9}{10^{2017}+1}\)nên \(1+\frac{9}{10^{2016}+1}>1+\frac{9}{10^{2017}+1}\)=>10A>10B
Vậy A>B
Cảm ơn bạn nhìu nhé.