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![](https://rs.olm.vn/images/avt/0.png?1311)
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Ta có :
A = \(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\)+ ... + \(\frac{1}{99}-\frac{1}{100}\)
= \(\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...\frac{1}{99}\right)\)- \(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}...+\frac{1}{100}\right)\)
= \(\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...\frac{1}{99}\right)\)+ \(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}...+\frac{1}{100}\right)\)
- \(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)x 2
= \(\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)- \(\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
= \(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)= B
Vậy , A = B
~ Chúc bạn học giỏi ! ~
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{a\times b}+a+b=a\times b$
$\frac{a+b}{a\times b}+(a+b)+\frac{1}{a\times b}+1=a\times b+1$
$(a+b)\times (\frac{1}{a\times b}+1)+(\frac{1}{a\times b}+1)=a\times b+1$
$(\frac{1}{a\times b}+1)\times (a+b+1)=a\times b+1$
$\frac{(a\times b+1)\times (a+b+1)}{a\times b}=a\times b+1$
$(a\times b+1)\times (\frac{a+b+1}{a\times b}-1)=0$
$\Rightarrow a\times b+1=0$ hoặc $\frac{a+b+1}{a\times b}=1$
Hiển nhiên $a\times b+1>0$ với $a,b$ là số tự nhiên.
$\Rightarrow \frac{a+b+1}{a\times b}=1$
$\Rightarrow a+b+1=a\times b$
$a\times b-a-b=1$
$a\times (b-1)-(b-1)=2$
$(b-1)\times (a-1)=2=1\times 2=2\times 1$
TH1:
$a-1=2, b-1=1\Rightarrow a=3; b=2$
TH2:
$a-1=1, b-1=2\Rightarrow a=2; b=3$
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 1 / 1008 + 1 / 2013 - 1 / 2016 x 2017
A = 1 / 1008 + 1 / 2013 - 1 / 2016 x 1 / 2017
B = 1 / 2014 + 1 / 2016 + 1 / 2017 + 1 / 2014 x 2016
B = 1 / 2014 + 1 / 2016 + 1 / 2017 + 1 / 2014 x 1 / 2016
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(A=\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+...+\frac{1}{2011\cdot2011}\)
có :
\(\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)
\(\frac{1}{4\cdot4}< \frac{1}{3\cdot4}\)
...
\(\frac{1}{2011\cdot2011}< \frac{1}{2010\cdot2011}\)
nên :
\(A< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2010\cdot2011}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}\)
\(\Rightarrow A< 1-\frac{1}{2011}\)
\(\Rightarrow A< \frac{2010}{2011}< 1\)
b, \(A=\frac{2010}{2011}=1-\frac{1}{2011}\)
\(\frac{3}{4}=1-\frac{1}{4}\)
\(\frac{1}{4}>\frac{1}{2011}\)
nên :
\(A>\frac{3}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(Giải\)
\(\Rightarrow A=\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}\)\(+\frac{1}{4}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2014}\)
\(A=0+0+0+...+0+0\)
\(\Rightarrow A=0\)
\(a.\)\(A< 1\)
b. \(A< \frac{3}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=1011-1/1012-1<1
=>A=1011-1/1012-1<1011+1.10/1012+1.10
=1011+10/1012+10
=10(1010+1)/10(1011+1)
=1010+1/1011+1=B
=>A<B
ví dụ
a là 1
b là 2
ta có
1/1 - 1/2 và 1/1x2
= 1/2 và 1/2
khi đó ta thấy 1/2 = 1/2
và 1/1 - 1/2 = 1/1x2
giúp mik với ạ