a)Chứng minh \(\dfrac{1}{2}< \dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+..........+\dfrac{1}{100}< 1\)
b)So sánh A=\(\dfrac{10^{19}+1}{10^{20}+1}\) B=\(\dfrac{10^{20}+1}{10^{21}+1}\)
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7 tháng 5 2021 lúc 13:27
So sánh:
a/ \(A=\dfrac{17^{18}+1}{17^{19}+1};B=\dfrac{17^{17}+1}{17^{18}+1}\)
b/ \(A=\dfrac{10^8-2}{10^8+2};B=\dfrac{10^8}{10^8+4}\)
c/ \(A=\dfrac{20^{10}+1}{20^{10}-1};B=\dfrac{20^{10}-1}{20^{10}-3}\)
GIÚP MÌNH VỚI
Giải:
a) A=1718+1/1719+1
17A=1719+17/1719+1
17A=1719+1+16/1719+1
17A=1+16/1719+1
Tương tự:
B=1717+1/1718+1
17B=1718+17/1718+1
17B=1718+1+16/1718+1
17B=1+16/1718+1
Vì 16/1719+1<16/1718+1 nên 17A<17B
⇒A<B
b) A=108-2/108+2
A=108+2-4/108+2
A=1+-4/108+2
Tương tự:
B=108/108+4
B=108+4-4/108+1
B=1+-4/108+1
Vì -4/108+2>-4/108+1 nên A>B
c)A=2010+1/2010-1
A=2010-1+2/2010-1
A=1+2/2010-1
Tương tự:
B=2010-1/2010-3
B=2010-3+2/2010-3
B=1+2/2010-3
Vì 2/2010-3>2/2010-1 nên B>A
⇒A<B
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so sánh
1) A = \(\dfrac{10^{11}-1}{10^{12}-1}\) và B =\(\dfrac{10^{10}+1}{10^{11}+1}\)
2) A = \(\dfrac{2018^9+1}{2018^{10}-1}\) và B = \(\dfrac{2018^{19}+1}{2018^{20}+1}\)
3) A = \(\dfrac{2018^{19}+1}{2018^{20}+1}\) và B = \(\dfrac{2018^{20}+1}{2018^{21}+1}\)
So sánh A và B : \(A=\dfrac{20^{10}+1}{20^{10}-1}\) và \(B=\dfrac{20^{10}-1}{20^{10}-3}\)
Giải:
Ta có:
A=2010+1/2010-1
A=2010-1+2/2010-1
A=1+2/2010-1
Tương tự:
B=2010-1/2010-3
B=2010-3+2/2010-3
B=1+2/2010-3
Vì 2/2010-1<2/2010-3 nên A<B
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so sánh:
A= \(\dfrac{20^{10}+1}{20^{10}-1}\)và B=\(\dfrac{20^{10}-1}{20^{10}-3}\)
Lời giải:
$A=\frac{20^{10}-1+2}{20^{10}-1}=1+\frac{2}{20^{10}-1}$
$B=\frac{20^{10}-3+2}{20^{10}-3}=1+\frac{2}{20^{10}-3}$
Vì $20^{10}-1> 20^{10}-3$
$\Rightarrow \frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}$
$\Rightarrow 1+\frac{2}{20^{10}-1}< 1+\frac{2}{20^{10}-3}$
$\Rightarrow A< B$
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So sánh: A=\(\dfrac{20^{10}+1}{20^{10}-1}\) và B= \(\dfrac{20^{10}-1}{20^{10}-3}\)
Cảm ơn ạ!
\(A=\dfrac{20^{10}+1}{20^{10}-1}=\dfrac{20^{10}-1}{20^{10}-1}+\dfrac{2}{20^{10}-1}=1+\dfrac{2}{20^{10}-1}\)
\(B=\dfrac{20^{10}-1}{20^{10}-3}=\dfrac{20^{10}-3}{20^{10}-3}+\dfrac{2}{20^{10}-3}=1+\dfrac{2}{20^{10}-3}\)
\(\dfrac{2}{20^{10}-1}>\dfrac{2}{20^{10}-3}\Leftrightarrow A>B\)
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a) dfrac{2}{1^2}.dfrac{6}{2^2}.dfrac{12}{3^2}.dfrac{20}{4^2}.dfrac{30}{5^2}.....dfrac{110}{10^2}.x-20
b) left(1+dfrac{1}{2}+dfrac{1}{3}+...+dfrac{1}{2013}right).x+2013dfrac{2014}{1}+dfrac{2015}{2}+...+dfrac{4025}{2012}+dfrac{4026}{2013}
c) left(dfrac{1}{1.2}+dfrac{1}{3.4}+...+dfrac{1}{99.100}right).xdfrac{2012}{51}+dfrac{2012}{52}+...+dfrac{2012}{99}+dfrac{2012}{100}
Đọc tiếp
a) \(\dfrac{2}{1^2}.\dfrac{6}{2^2}.\dfrac{12}{3^2}.\dfrac{20}{4^2}.\dfrac{30}{5^2}.....\dfrac{110}{10^2}.x=-20\)
b) \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2013}\right).x+2013=\dfrac{2014}{1}+\dfrac{2015}{2}+...+\dfrac{4025}{2012}+\dfrac{4026}{2013}\)
c) \(\left(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right).x=\dfrac{2012}{51}+\dfrac{2012}{52}+...+\dfrac{2012}{99}+\dfrac{2012}{100}\)
Cho hai phân số \(A=\dfrac{10^{19}+1}{10^{20}+1}\);\(B=\dfrac{10^{20}+1}{10^{21}+1}\). So sánh A và B
1.Tính nhanh:
A dfrac{dfrac{2}{3}-dfrac{1}{4}+dfrac{5}{11}}{dfrac{5}{12}+1-dfrac{7}{11}}
2. Cho: B dfrac{1}{4}+dfrac{1}{5}+dfrac{1}{6}+...+dfrac{1}{19} .Hãy chứng tỏ rằng B 1.
3. Rút gọn:
a) C left(1-dfrac{1}{2}right).left(1-dfrac{1}{3}right).left(1-dfrac{1}{4}right)....left(1-dfrac{1}{20}right)
b) D 1+dfrac{1}{2}+dfrac{1}{2^2}+dfrac{1}{2^3}+...+dfrac{1}{2^{2012}}
4. So sánh: Edfrac{20^{10}+1}{20^{10}-1} và F dfrac{20^{10}-1}{20^{10}-3}
5. Tính giá trị của biểu thức:
M dfrac{dfrac{3}{...
Đọc tiếp
1.Tính nhanh:
A= \(\dfrac{\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}}{\dfrac{5}{12}+1-\dfrac{7}{11}}\)
2. Cho: B =\(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{19}\) .Hãy chứng tỏ rằng B > 1.
3. Rút gọn:
a) C= \(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)....\left(1-\dfrac{1}{20}\right)\)
b) D= \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2012}}\)
4. So sánh: E=\(\dfrac{20^{10}+1}{20^{10}-1}\) và F =\(\dfrac{20^{10}-1}{20^{10}-3}\)
5. Tính giá trị của biểu thức:
M= \(\dfrac{\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{11}}\)
Bài 1: Cho A=\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
a) Chứng minh: A=\(\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+...+\dfrac{1}{100}\)
b) Chứng minh: A<\(\dfrac{5}{6}\)
a) $A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}$
$=>A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}$
$=>A=(1+\dfrac{1}{3}+...+\dfrac{1}{99})-(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100})$
$=>A=(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{99}+\dfrac{1}{100})-(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}.2)$
$=>A=(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100})-(1+\dfrac{1}{2}+...+\dfrac{1}{50})$
$=>A=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}$
b) Ta có : $A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}$
$=>A=(1-\dfrac{1}{2}+\dfrac{1}{3})-(\dfrac{1}{4}-\dfrac{1}{5})-...-(\dfrac{1}{98}-\dfrac{1}{99})-\dfrac{1}{100}$
$=>A<1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}$
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