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Nga Rau má
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Khánh Linh
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Nguyễn Lê Phước Thịnh
1 tháng 6 2022 lúc 22:46

\(\dfrac{x+2017}{x+2018}=\dfrac{2022}{2023}\)

\(\Leftrightarrow2023x+4080391=2022x+4080396\)

=>x=5

Khánh Linh
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Nguyễn Thanh Hằng
8 tháng 8 2017 lúc 11:52

Ta có :

\(\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}=\left(1-\dfrac{1}{2018}\right)+\left(1-\dfrac{1}{2019}\right)+\left(1-\dfrac{1}{2020}\right)\)

\(=\left(1+1+1\right)-\left(\dfrac{1}{2018}+\dfrac{1}{2019}+\dfrac{1}{2020}\right)\)

\(=3-\left(\dfrac{1}{2018}+\dfrac{1}{2019}+\dfrac{1}{2020}\right)< 3\)

\(\Leftrightarrow\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}< 3\)

TV Cuber
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Nguyễn Ngọc Huy Toàn
10 tháng 5 2022 lúc 13:02

\(T=\dfrac{-2\left|x-2018\right|-2021}{2020+\left|x-2018\right|}\)

Để T lớn nhất thì \(2020+\left|x-2018\right|\) nhỏ nhất

Mà \(2020+\left|x-2018\right|\ge2020;\forall x\) 

--> \(Min=2020\) khi \(x=2018\)

Khi đó \(T=\dfrac{-2\left|2018-2018\right|-2021}{2020+\left|0\right|}=\dfrac{-2.0-2021}{2020}=-\dfrac{2021}{2020}\) 

--> \(Max_T=-\dfrac{2021}{2020}\) khi \(x=2018\)

P/s: hongg bt đúng hem nha:v

Hoàng Đình Bảo
10 tháng 5 2022 lúc 14:04

$T=\frac{-2|x-2018|-2021}{2020+|x-2018|}=\frac{-2(|x-2018|+2020)+2019}{2020+|x-2018|}=-2+\frac{2019}{2020+|x-2018|}$

Lại có $|x-2018| \ge 0$ nên 

$T=-2+\frac{2019}{2020+|x-2018|} \le -2+\frac{2019}{2020}=-\frac{2021}{2020}$

Vậy $GTLN=-\frac{2021}{2020}$

Dấu $"="$ xảy ra khi và chỉ khi: $|x-2018|=0\Leftrightarrow x=2018$

 

nito
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Ng Ngọc
13 tháng 8 2023 lúc 15:01

\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}=\dfrac{x+1}{6}\)

\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}-\dfrac{x+1}{6}=0\)

\(\left(x+1\right)\left(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\right)=0\)

\(\)vì \(\dfrac{1}{3}>\dfrac{1}{6};\dfrac{1}{4}>\dfrac{1}{6};\dfrac{1}{5}>\dfrac{1}{6}=>\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}>0\)

\(=>x+1=0\)

\(=>x=-1\)

b,

\(\dfrac{x+1}{2020}+\dfrac{x+2}{2019}=\dfrac{x+3}{2018}+\dfrac{x+4}{2017}\)

\(\left(\dfrac{x+1}{2020}+1\right)+\left(\dfrac{x+2}{2019}+1\right)=\left(\dfrac{x+3}{2018}+1\right)+\left(\dfrac{x+4}{2017}+1\right)\)

\(\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}=\dfrac{x+2021}{2018}+\dfrac{x+2021}{2017}\)

\(=>\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}-\dfrac{x+2021}{2018}-\dfrac{x+2021}{2017}=0\)

\(=>\left(x+2021\right)\left(\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}\right)=0\)

Vì \(\dfrac{1}{2020}< \dfrac{1}{2018};\dfrac{1}{2019}< \dfrac{1}{2017}=>\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}< 0\)

\(=>x+2021=0\)

\(=>x=-2021\)

 

c,

\(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)

\(\left(\dfrac{x+2}{327}+1\right)+\left(\dfrac{x+3}{326}+1\right)+\left(\dfrac{x+4}{325}+1\right)+\left(\dfrac{x+5}{324}+1\right)+\left(\dfrac{x+349}{5}-4\right)=0\)

\(\dfrac{x+329}{327}+\dfrac{x+329}{326}+\dfrac{x+329}{325}+\dfrac{x+329}{324}+\dfrac{x+329}{5}=0\)

\(=>\left(x+329\right)\left(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}\right)=0\)

Vì \(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}>0\)

\(=>x+329=0\)

\(=>x=-329\)

Banh Bao Chien
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Trần Minh An
14 tháng 12 2017 lúc 15:54

Ta có: \(\dfrac{x+1}{2017}+\dfrac{x+1}{2018}=\dfrac{x+1}{2019}+\dfrac{x+1}{2020}\)

\(\Rightarrow\left(\dfrac{x+1}{2017}+\dfrac{x+1}{2018}\right)-\left(\dfrac{x+1}{2019}+\dfrac{x+1}{2020}\right)=0\)

\(\Rightarrow\dfrac{x+1}{2017}+\dfrac{x+1}{2018}-\dfrac{x+1}{2019}-\dfrac{x+1}{2020}=0\)

\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{2017}+\dfrac{1}{2018}-\dfrac{1}{2019}-\dfrac{1}{2020}\right)=0\)

\(\dfrac{1}{2017}>\dfrac{1}{2018}>\dfrac{1}{2019}>\dfrac{1}{2020}>0\) nên

\(\dfrac{1}{2017}+\dfrac{1}{2018}-\dfrac{1}{2019}-\dfrac{1}{2020}>0\)

\(\Rightarrow x+1=0\)

\(\Rightarrow x=-1\)

nguyễn thảo
14 tháng 12 2017 lúc 21:24

x=-1

x&#x2212;42021+x&#x2212;32020=x&#x2212;22019+x&#x2212;12018" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline-block; float:none; font-size:15.82px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-break:break-word; word-spacing:normal" class="MathJax_CHTML mjx-chtml">

x&#x2212;42021+x&#x2212;32020&#x2212;x&#x2212;22019&#x2212;x&#x2212;12018=0" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline-table; float:none; font-size:15.82px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-break:break-word; word-spacing:normal" class="MathJax_CHTML mjx-chtml">

x&#x2212;42021)+(1+x&#x2212;32020)&#x2212;(1+x&#x2212;22019)&#x2212;(1+x&#x2212;12018)=0" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline-block; float:none; font-size:15.82px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-break:break-word; word-spacing:normal" class="MathJax_CHTML mjx-chtml">⇔ x+20172021+x+20172020&#x2212;x+20172019&#x2212;x+20172018=0" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline-block; float:none; font-size:15.82px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-break:break-word; word-spacing:normal" class="MathJax_CHTML mjx-chtml">

12021+12020&#x2212;12019&#x2212;12018)=0" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline-block; float:none; font-size:15.82px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-break:break-word; word-spacing:normal" class="MathJax_CHTML mjx-chtml">

⇔ x + 2017 = 0

⇔ x = -2017

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Nguyễn Phương Uyên
17 tháng 3 2020 lúc 7:13

\(\frac{x-1}{2020}+\frac{x-2}{2021}=\frac{x+1}{2018}+\frac{x+2}{2017}\)

\(\Leftrightarrow\frac{x-1}{2020}+1+\frac{x-2}{2021}-1=\frac{x+1}{2018}+1+\frac{x+2}{2017}+1\)

\(\Leftrightarrow\frac{x+2019}{2020}+\frac{x+2019}{2021}=\frac{x+2019}{2018}+\frac{x+2019}{2017}\)

\(\Leftrightarrow\left(x+2019\right)\left(\frac{1}{2020}+\frac{1}{2021}-\frac{1}{2018}-\frac{1}{2017}\right)=0\)

mà \(\frac{1}{2020}+\frac{1}{2021}-\frac{1}{2018}-\frac{1}{2017}\ne0\)

\(\Leftrightarrow x+2019=0\)

\(\Leftrightarrow x=-2019\)

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Trần Quang Dũng
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Minh Hiếu
19 tháng 4 2022 lúc 20:55

a) \(2\left(\dfrac{2}{3.5}+\dfrac{4}{5.9}+...+\dfrac{16}{n\left(n+16\right)}\right)=\dfrac{16}{25}\)

\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{n}-\dfrac{1}{n+16}=\dfrac{8}{25}\)

\(\dfrac{1}{3}-\dfrac{1}{n+16}=\dfrac{8}{25}\)

\(\dfrac{n+13}{3\left(n+16\right)}=\dfrac{8}{25}\)

\(24n+384=25n+325\)

\(25n-24n=384-325\)

\(n=59\)

Minh Hiếu
19 tháng 4 2022 lúc 20:57

b) Sai đề nha

\(\left\{{}\begin{matrix}\dfrac{2018}{2019}< 1\\\dfrac{2019}{2020}< 1\\\dfrac{2020}{2021}< 1\\\dfrac{2021}{2022}< 1\end{matrix}\right.\)

\(\Rightarrow\dfrac{2018}{2019}+\dfrac{2019}{2020}+\dfrac{2020}{2021}+\dfrac{2021}{2022}< 4\)

Đào Khánh	Linh
19 tháng 4 2022 lúc 21:01

chị ơi hình như chị nhầm rồi P/s cuối phải là 1/n.(n+6)thì phải

Minh Ngọc
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Hồ Kim Ngọc
16 tháng 4 2023 lúc 10:02

(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = (\(\dfrac{2021}{2}+1\))+(\(\dfrac{2020}{3}+1\))+....+(\(\dfrac{1}{2022}+1\))

(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = \(\dfrac{2023}{2}\)+\(\dfrac{2023}{3}\)+....+ \(\dfrac{2023}{2022}\)

(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = 2023.( \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\))

vậy x= 2023