Tìm x \(\in\) Z biết:
1) \(\dfrac{1-x}{2017}+\dfrac{2-x}{2016}=\dfrac{3-x}{2015}+\dfrac{4-x}{2014}\)
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NH
5 tháng 3 2023
\(\dfrac{x+4}{2014}+\dfrac{x+3}{2015}=\dfrac{x+2}{2016}+\dfrac{x+1}{2017}\)
\(\dfrac{x+4}{2014}+1+\dfrac{x+3}{2015}+1=\dfrac{x+2}{2016}+1+\dfrac{x+1}{2017}+1\)
\(\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}=\dfrac{x+2018}{2016}+\dfrac{x+2018}{2017}\)
\(\left(x+2018\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)=0\\ x+2018=0\\ x=-2018\)
15 tháng 8 2017
\(\dfrac{1-x}{2017}+\dfrac{2-x}{2016}=\dfrac{3-x}{2015}+\dfrac{4-x}{2014}\)
\(\Leftrightarrow\left(\dfrac{1-x}{2017}+1\right)+\left(\dfrac{2-x}{2016}+1\right)=\left(\dfrac{3-x}{2015}+1\right)+\left(\dfrac{4-x}{2014}+1\right)\)
\(\Leftrightarrow\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}=\dfrac{2018-x}{2015}+\dfrac{2018-x}{2014}\)
\(\Leftrightarrow\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}-\dfrac{2018-x}{2015}-\dfrac{2018-x}{2014}=0\)
\(\Leftrightarrow\left(2018-x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
Mà \(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\ne0\)
\(\Leftrightarrow2018-x=0\Leftrightarrow x=2018\)
Vậy ....
DH
15 tháng 8 2017
\(\dfrac{1-18x}{2017}+\dfrac{2-18x}{2016}=\dfrac{3-18x}{2015}+\dfrac{4-18x}{2014}\)
\(\Rightarrow\left(\dfrac{1-18x}{2017}+1\right)+\left(\dfrac{2-18x}{2016}+1\right)=\left(\dfrac{3-18x}{2015}+1\right)+\left(\dfrac{4-18x}{2014}+1\right)\)
\(\Rightarrow\dfrac{2018-18x}{2017}+\dfrac{2018-18x}{2016}-\dfrac{2018-18x}{2015}-\dfrac{2018-18x}{2014}=0\)
\(\Rightarrow\left(2018-18x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
\(\Rightarrow2018-18x=0\Rightarrow x=\dfrac{1009}{9}\)
Vậy.............
Chúc bạn học tốt!!!
HD
15 tháng 8 2017
\(\dfrac{1-18x}{2017}+\dfrac{2-18x}{2016}=\dfrac{3-18x}{2015}+\dfrac{4-18x}{2014}\)
\(\Rightarrow\left(\dfrac{1-18x}{2017}+1\right)+\left(\dfrac{2-18x}{2016}+1\right)=\left(\dfrac{3-18x}{2015}+1\right)+\left(\dfrac{4-18x}{2014}+1\right)\)
\(\Rightarrow\dfrac{2018-18x}{2017}+\dfrac{2018-18x}{2016}-\dfrac{2018-18x}{2015}-\dfrac{2018-18x}{2014}=0\)
\(\Rightarrow\left(2018-18x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
\(\Rightarrow2018-18x=0\)
\(\Rightarrow18x=2018-0\)
\(\Rightarrow18x=2018\)
\(\Rightarrow x=2018:18\)
\(\Rightarrow x=\dfrac{1009}{9}\)
DH
15 tháng 8 2017
\(\dfrac{1-2x}{2017}+\dfrac{2-2x}{2016}=\dfrac{3-2x}{2015}+\dfrac{4-2x}{2014}\)
\(\Rightarrow\left(\dfrac{1-2x}{2017}+1\right)+\left(\dfrac{2-2x}{2016}+1\right)=\left(\dfrac{3-2x}{2015}+1\right)+\left(\dfrac{4-2x}{2014}+1\right)\)
\(\Rightarrow\dfrac{2018-2x}{2017}+\dfrac{2018-2x}{2016}-\dfrac{2018-2x}{2015}-\dfrac{2018-2x}{2014}=0\)
\(\Rightarrow\left(2018-2x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
Vì \(2017>2016>2015>2014\) nên
\(\dfrac{1}{2017}< \dfrac{1}{2016}< \dfrac{1}{2015}< \dfrac{1}{2014}\)
\(\Rightarrow\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}< 0\)
\(\Rightarrow2018-2x=0\Rightarrow x=1009\)
Vậy...........
Chúc bạn học tốt!!!
HD
15 tháng 8 2017
\(\dfrac{1-2x}{2017}+\dfrac{2-2x}{2016}=\dfrac{3-2x}{2015}+\dfrac{4-2x}{2014}\)
\(\Rightarrow\left(\dfrac{1-2x}{2017}+1\right)+\left(\dfrac{2-2x}{2016}+1\right)=\left(\dfrac{3-2x}{2015}+1\right)+\left(\dfrac{4-2x}{2014}+1\right)\)
\(\Rightarrow\dfrac{2018-2x}{2017}+\dfrac{2018-2x}{2016}-\dfrac{2018-2x}{2015}-\dfrac{2018-2x}{2014}=0\)
\(\Rightarrow\left(20418-2x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
\(Ta\) \(có\)\(:\) \(2017>2016>2015>2014\)
\(\Rightarrow\dfrac{1}{2017}< \dfrac{1}{2016}< \dfrac{1}{2015}< \dfrac{1}{2014}\)
\(\Rightarrow\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}< 0\)
\(\Rightarrow2018-2x=0\)
\(\Rightarrow2x=2018-0\)
\(\Rightarrow2x=2018\)
\(\Rightarrow x=2018:2\)
\(\Rightarrow x=1009\)
16 tháng 8 2021
Đặt \(a=\sqrt{x-2015};b=\sqrt{y-2016};c=\sqrt{z-2017}\left(a,b,c>0\right)\)
Khi đó phương trình trở thành:
\(\dfrac{a-1}{a^2}+\dfrac{b-1}{b^2}+\dfrac{c-1}{c^2}=\dfrac{3}{4}\\ \Leftrightarrow\left(\dfrac{1}{4}-\dfrac{1}{a}+\dfrac{1}{a^2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{b}+\dfrac{1}{b^2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{c}+\dfrac{1}{c^2}\right)=0\\ \Leftrightarrow\left(\dfrac{1}{2}-\dfrac{1}{a}\right)^2+\left(\dfrac{1}{2}-\dfrac{1}{b}\right)^2+\left(\dfrac{1}{2}-\dfrac{1}{c}\right)^2=0\\ \Leftrightarrow a=b=c=2\\ \Leftrightarrow x=2019;y=2020;z=2021\)
Tick plz
NG
23 tháng 1 2021
\(\dfrac{x-1}{2016}+\dfrac{x-2}{2015}-\dfrac{x-3}{2014}=\dfrac{x-4}{2013}\)
\(\Leftrightarrow\dfrac{x-1}{2016}+\dfrac{x-2}{2015}=\dfrac{x-4}{2013}+\dfrac{x-3}{2014}\)
\(\Leftrightarrow\left(\dfrac{x-1}{2016}-1\right)+\left(\dfrac{x-2}{2015}-1\right)=\left(\dfrac{x-4}{2013}-1\right)+\left(\dfrac{x-3}{2014}-1\right)\)
\(\Leftrightarrow\dfrac{x-2017}{2016}+\dfrac{x-2017}{2015}=\dfrac{x-2017}{2013}+\dfrac{x-2017}{2014}\)
\(\Leftrightarrow\dfrac{x-2017}{2016}+\dfrac{x-2017}{2015}-\dfrac{x-2017}{2013}-\dfrac{x-2017}{2014}=0\)
\(\Leftrightarrow x-2017.\left(\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2013}\right)=0\)
\(\text{Mà }\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2103}\ne0\Rightarrow x-2017=0\)
\(\Leftrightarrow x=2017\) \(\text{Vậy }x=2017\)
V
3 tháng 9 2018
\(=>\left(\dfrac{x+4}{2014}+1\right)+\left(\dfrac{x+3}{2015}+1\right)=\left(\dfrac{x+2}{2016}+1\right)+\left(\dfrac{x+1}{2017}+1\right)\)
=> \(\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}=\dfrac{x+2018}{2016}+\dfrac{x+2018}{2017}\)
=> (x+2018).\(\left(\dfrac{1}{2014}+\dfrac{1}{2015}\right)=\left(x+2018\right).\left(\dfrac{1}{2016}+\dfrac{1}{2017}\right)\)
=> (x+2018).\(\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)\) = 0
Mà \(\dfrac{1}{2014}>0;\dfrac{1}{2015}>0;\dfrac{1}{2016}>0;\dfrac{1}{2017}>0\)
=>\(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\ne0\)
=> \(x+2018=0\)
=>x = 0-2018
=> x = 0+(-2018)
=> x = -2018
TB
2
TS
21 tháng 4 2018
\(\dfrac{x-1}{2017}+\dfrac{x-2}{2016}=\dfrac{x-3}{2015}+\dfrac{x-4}{2014}\)
\(\Rightarrow\dfrac{x-1}{2017}+\dfrac{x-2}{2016}-\dfrac{x-3}{2015}-\dfrac{x-4}{2014}=0\)
\(\Rightarrow\dfrac{x-1}{2017}-1+\dfrac{x-2}{2016}-1-\dfrac{x-3}{2015}+1-\dfrac{x-4}{2014}+1=0\)
\(\Rightarrow\left(\dfrac{x-1}{2017}-1\right)+\left(\dfrac{x-2}{2016}-1\right)-\left(\dfrac{x-3}{2015}-1\right)-\left(\dfrac{x-4}{2014}-1\right)=0\)
\(\Rightarrow\dfrac{x-2018}{2017}+\dfrac{x-2018}{2016}-\dfrac{x-2018}{2015}-\dfrac{x-2018}{2014}=0\)
\(\Rightarrow x-2018.\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
Vì \(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\ne0\)
Để \(x-2018.\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
\(\Rightarrow x-2018=0\)
\(x=2018\)
21 tháng 4 2018
Ta có :
\(\dfrac{x-1}{2017}+\dfrac{x-2}{2016}=\dfrac{x-3}{2015}+\dfrac{x-4}{2014}\)
\(\Leftrightarrow\)\(\left(\dfrac{x-1}{2017}-1\right)+\left(\dfrac{x-2}{2016}-1\right)=\left(\dfrac{x-3}{2015}-1\right)+\left(\dfrac{x-4}{2014}-1\right)\) ( trừ 2 vế cho 2 )
\(\Leftrightarrow\)\(\dfrac{x-2018}{2017}+\dfrac{x-2018}{2016}=\dfrac{x-2018}{2015}+\dfrac{x-2018}{2014}\)
\(\Leftrightarrow\)\(\dfrac{x-2018}{2017}+\dfrac{x-2018}{2016}-\dfrac{x-2018}{2015}-\dfrac{x-2018}{2014}=0\)
\(\Leftrightarrow\)\(\left(x-2018\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
Vì \(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\ne0\)
Nên \(x-2018=0\)
\(\Rightarrow\)\(x=2018\)
Vậy \(x=2018\)
Chúc bạn học tốt ~
4 tháng 2 2021
a) Ta có: \(A=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{2014}\right)\left(1-\dfrac{1}{2015}\right)\left(1-\dfrac{1}{2016}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2013}{2014}\cdot\dfrac{2014}{2015}\cdot\dfrac{2015}{2016}\)
\(=\dfrac{1}{2016}\)
b) Ta có: \(\dfrac{x-2}{12}+\dfrac{x-2}{20}+\dfrac{x-2}{30}+\dfrac{x-2}{42}+\dfrac{x-2}{56}+\dfrac{x-2}{72}=\dfrac{16}{9}\)
\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)=\dfrac{16}{9}\)
\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right)=\dfrac{16}{9}\)
\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{3}-\dfrac{1}{9}\right)=\dfrac{16}{9}\)
\(\Leftrightarrow\left(x-2\right)\cdot\dfrac{2}{9}=\dfrac{16}{9}\)
\(\Leftrightarrow x-2=\dfrac{16}{9}:\dfrac{2}{9}=\dfrac{16}{9}\cdot\dfrac{9}{2}=8\)
hay x=10
Vậy: x=10
\(\dfrac{1-x}{2017}+\dfrac{2-x}{2016}=\dfrac{3-x}{2015}+\dfrac{4-x}{2014}\)
\(\left(\dfrac{1-x}{2017}+1\right)+\left(\dfrac{2-x}{2016}+1\right)=\left(\dfrac{3-x}{2015}+1\right)+\left(\dfrac{4-x}{2014}+1\right)\)
\(\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}=\dfrac{2018-x}{2015}+\dfrac{2018-x}{2014}\)
\(\Leftrightarrow\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}-\dfrac{2018-x}{2015}-\dfrac{2018-x}{2014}=0\)
\(\Leftrightarrow\left(2018-x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
Mà \(\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)\ne0\)
\(\Leftrightarrow2018-x=0\Leftrightarrow x=2018\)
Vậy ...