So sánh S và \(\frac{2019}{1010}\)
Biết S = \(\frac{2}{1.2}\) \(+\frac{2}{2.3}\) \(+\frac{2}{3.4}\) \(+...+\frac{2}{2020.2021}\)
K
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Những câu hỏi liên quan
15 tháng 1 2019
\(a,ĐKXĐ:x-1\ge0\Leftrightarrow x\ge1\)
Đặt \(\hept{\begin{cases}\sqrt[3]{2-x}=a\\\sqrt{x-1}=b\left(b\ge0\right)\end{cases}\Rightarrow}a^3+b^2=2-x+x-1=1\)
Lại có: \(a=1-b\)
Thay vào được
\(\left(1-b\right)^3+b^2=1\)
\(\Leftrightarrow1-3b+3b^2-b^3+b^2-1=0\)
\(\Leftrightarrow-b^3+4b^2-3b=0\)
\(\Leftrightarrow b^3-4b^2+3b=0\)
\(\Leftrightarrow b\left(b^2-4b+3\right)=0\)
\(\Leftrightarrow b\left(b-1\right)\left(b-3\right)=0\)
\(\Leftrightarrow b=0\left(h\right)b=1\left(h\right)b=3\)(T/m ĐK b>0)
*Với b = 0
\(\Leftrightarrow\sqrt{x-1}=0\)
\(\Leftrightarrow x=1\left(TmĐKXĐ\right)\)
*Với b = 1
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\left(TmĐKXĐ\right)\)
*Với b = 3
\(\Leftrightarrow\sqrt{x-1}=3\)
\(\Leftrightarrow x-1=9\)
\(\Leftrightarrow x=10\)
Vậy \(S\in\left\{1;2;10\right\}\)
15 tháng 1 2019
em chỉ bt bài 2 nha!
\(A=\left(1-\frac{2}{2\cdot3}\right)\left(1-\frac{2}{3\cdot4}\right)...\left(1-\frac{2}{2020\cdot2021}\right)\)
\(\frac{2}{3}\cdot\frac{5}{6}\cdot\frac{9}{10}\cdot...\cdot\frac{2020\cdot2021-2}{2020\cdot2021}\left(1\right)\)
Mặt khác:\(2020\cdot2021-2=2020\left(2022-1\right)+2020-2022\)
\(=2020\cdot2022-2022\)
\(=2022\left(2020-1\right)=2019\cdot2022\left(2\right)\)
Từ (1),(2) ta có:
\(A=\frac{4\cdot1}{2\cdot3}\cdot\frac{5\cdot2}{3\cdot4}\cdot...\cdot\frac{2022\cdot2019}{2020\cdot2021}\)
\(=\frac{\left(4\cdot5\cdot6\cdot...\cdot2022\right)\left(1\cdot2\cdot3\cdot...\cdot2019\right)}{\left(2\cdot3\cdot4\cdot...\cdot2020\right)\left(3\cdot4\cdot5\cdot...\cdot2021\right)}\)
\(=\frac{2021\cdot2022}{2\cdot3}\cdot\frac{1\cdot2}{2020\cdot2021}=\frac{2022}{3\cdot2020}=\frac{2022}{6060}\)
B
8 tháng 2 2020
B=11.2+13.4+15.6+....+12019.2020
⇒2B=21.2+23.4+25.6+....+22019.2020
<1+12.3+13.4+14.5+15.6+....+12018.2019+12019.2020
2B<1+3−22.3+4−33.4+5−44.5+....+2019−20182018.2019+2020−20192019.2020
2B<1+12−13+13−14+...+12019−12020
2B<1+12−12020<1+12
B<34
---------------------
Đặt 22018=a;32019=b;52020=c(a,b,c>0)
A=aa+b+bb+c+cc+a>aa+b+c+ba+b+c+ca+b+c=1
⇒A>1>34>B
PA
30 tháng 10 2016
\(S=\frac{2}{1\times2}+\frac{2}{2\times3}+\frac{2}{3\times4}+...+\frac{2}{98\times99}+\frac{2}{99\times100}\)
\(S=2\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{98\times99}+\frac{1}{99\times100}\right)\)
\(S=2\times\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(S=2\times\left(1-\frac{1}{100}\right)\)
\(S=2\times\frac{99}{100}\)
\(S=\frac{99}{50}\)
30 tháng 10 2016
\(S=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{98.99}+\frac{2}{99.100}\)
\(S=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(S=2.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}+\frac{1}{100}\right)\)
\(S=2.\left(\frac{1}{1}-\frac{1}{100}\right)\\ S=2.\left(\frac{100}{100}+\frac{-1}{100}\right)\\ S=2.\frac{99}{100}\\ S=\frac{99}{50}\)
14 tháng 5 2019
B= 1/1.2+1/2.3+...+1/2019.2020
B=1/1-1/2+1/2-1/3+...+1/2019-1/2020
B=1-1/2020=2020/2020-1/2020=2019/2020
14 tháng 5 2019
\(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1.\)
Với : \(a=2^{2018};.b=3^{2019};,c=5^{2020}.\)
Và : \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2019.2020}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\Leftrightarrow\)
\(B=1-\frac{1}{2020}< 1< A\)
S
12 tháng 5 2017
A = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
=\(1-\frac{1}{50}\)
Vì \(1-\frac{1}{50}< 1\)nên A < 1
B = \(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
=\(\frac{1}{2}-\frac{1}{100}\)
Vì \(\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)nên B < \(\frac{1}{2}\)
Q
12 tháng 5 2017
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}\)
\(\Rightarrow A< 1\)
\(B=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(B=\frac{1}{2}-\frac{1}{100}\)
\(\Rightarrow B< \frac{1}{2}\)
AH
Akai Haruma
Giáo viên
14 tháng 5 2019
Lời giải:
\(B=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{2019.2020}\)
\(\Rightarrow 2B=\frac{2}{1.2}+\frac{2}{3.4}+\frac{2}{5.6}+....+\frac{2}{2019.2020}\)
\(< 1+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{2018.2019}+\frac{1}{2019.2020}\)
\(2B< 1+\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+....+\frac{2019-2018}{2018.2019}+\frac{2020-2019}{2019.2020}\)
\(2B< 1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\( 2B< 1+\frac{1}{2}-\frac{1}{2020}< 1+\frac{1}{2}\)
\(B< \frac{3}{4}\)
---------------------
Đặt \(2^{2018}=a; 3^{2019}=b; 5^{2020}=c(a,b,c>0)\)
\(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}> \frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1\)
\(\Rightarrow A>1> \frac{3}{4}> B\)
2 tháng 6 2018
Bài 1 :
\(A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}< 1\left(1\right)\)
\(B=\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\right)\)\(>\frac{1}{10}+\frac{1}{100}.90=1\left(2\right)\)
Từ (1) và ( 2) ta có \(A< 1\) \(B>1\)NÊN \(A< B\)
Bài 2:
\(S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
\(=\frac{\left(a+b+c\right)-\left(b+c\right)}{b+c}+\)\(\frac{\left(a+b+c\right)-\left(c+a\right)}{c+a}\)\(+\frac{\left(a+b+c\right)-\left(a+b\right)}{a+b}\)
\(=\frac{7-\left(b+c\right)}{b+c}+\frac{7-\left(c+a\right)}{c+a}+\frac{7-\left(a+b\right)}{a+b}\)
\(=7.\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-3\)
\(=7.\frac{7}{10}-3\)\(=\frac{49}{10}-3=\frac{19}{10}\)
\(S=\frac{19}{10}>\frac{19}{11}=1\frac{8}{11}\)
Chúc bạn học tốt ( -_- )
2 tháng 6 2018
Bài 1:
ta có: \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}< 1\)
\(\Rightarrow A< 1\)(1)
ta có: \(\frac{1}{11}>\frac{1}{100};\frac{1}{12}>\frac{1}{100};...;\frac{1}{99}>\frac{1}{100}\)
\(\Rightarrow\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}+\frac{1}{100}\) ( có 90 số 1/100)
\(=\frac{90}{100}=\frac{9}{10}\)
\(\Rightarrow B=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}>\frac{1}{10}+\frac{9}{10}=1\)
\(\Rightarrow B>1\)(2)
Từ (1);(2) => A<B
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S = 2 x ( 1 / 1x2 + 1 / 2x3 + .. + 1 / 2020x 2021 ) = 2 x ( 1/1 - 1/2 + 1/2 - 1/3 + ... + 1/2020 - 1/2021 )
= 2 x ( 1 - 1/2021) = 2 x 2020/2021
= 4040/2021 = (4042 - 2) /2021 = 2 - 2/2021
2019/1010 = ( 2020 - 1 )/1010 = 2 - 1/1010 = 2 - 2/2020
Ta thấy 2/2021 < 2
Nên 2 - 2/2021 > 2 - 2/2020
Vậy S > 2019/1010
Cái này đơn giản mà owner first ưi, ib FB teo giải cho nhé :)