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TM
14 tháng 12 2017
Ta có: \(\dfrac{x+1}{2017}+\dfrac{x+1}{2018}=\dfrac{x+1}{2019}+\dfrac{x+1}{2020}\)
\(\Rightarrow\left(\dfrac{x+1}{2017}+\dfrac{x+1}{2018}\right)-\left(\dfrac{x+1}{2019}+\dfrac{x+1}{2020}\right)=0\)
\(\Rightarrow\dfrac{x+1}{2017}+\dfrac{x+1}{2018}-\dfrac{x+1}{2019}-\dfrac{x+1}{2020}=0\)
\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{2017}+\dfrac{1}{2018}-\dfrac{1}{2019}-\dfrac{1}{2020}\right)=0\)
Vì \(\dfrac{1}{2017}>\dfrac{1}{2018}>\dfrac{1}{2019}>\dfrac{1}{2020}>0\) nên
\(\dfrac{1}{2017}+\dfrac{1}{2018}-\dfrac{1}{2019}-\dfrac{1}{2020}>0\)
\(\Rightarrow x+1=0\)
\(\Rightarrow x=-1\)
PV
2
MN
19 tháng 11 2018
\(\dfrac{1}{2019}:2017.x=-\dfrac{1}{2017}\)
\(\dfrac{1}{2019.2017}x=-\dfrac{1}{2017}\)
x=\(-\dfrac{1}{2017}:\dfrac{1}{2019.2017}\)=-2019
Vậy x=-2019
HN
19 tháng 11 2018
2017 . x = \(\dfrac{1}{2019}:\left(\dfrac{-1}{2017}\right)\)
2017 . x = \(\dfrac{1}{2019}.\left(-2017\right)\)
2017 . x = \(-\dfrac{2017}{2019}\)
x = \(-\dfrac{2017}{2019}:2017\)
x = \(-\dfrac{2017}{2019}.\dfrac{1}{2017}\)
x = \(\dfrac{-1}{2019}\)
KH
1 tháng 2 2019
\(2019-\left|x-2019\right|=x\)
\(\Leftrightarrow\left|x-2019\right|=2019-x\)
\(\Leftrightarrow\left[{}\begin{matrix}2019-x=x-2019\\2019-x=2019-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-2x=-4038\\0x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2019\\x=0\end{matrix}\right.\)
Vậy \(x=2019;x=0\)
1 tháng 2 2019
\(a)\)\(2019-\left|x-2019\right|=x\)
\(\Leftrightarrow-\left|x-2019\right|-x=-2019\)
TH1: \(x-2019\ge0\Rightarrow x\ge2019\)
\(-\left(x-2019\right)-x=-2019\\ \Leftrightarrow-x+2019-x=-2019\\ \Leftrightarrow-x-x=-2019-2019\\ \Leftrightarrow-2x=-4038\\ \Leftrightarrow x=2019\left(TM\right)\)
TH2: \(x-2019< 0\Rightarrow x< 2019\)
\(-\left[-\left(x-2019\right)\right]-x=-2019\\ \Leftrightarrow x-2019-x=-2019\\ \Leftrightarrow x-x=-2019+2019\\ \Leftrightarrow0x=0\left(VSN\right)\)
Vậy ......
VV
1
13 tháng 12 2022
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x-1}{2018}=\dfrac{3-y}{2019}=\dfrac{x-1+3-y}{2018+2019}=1\)
=>x-1=2018 và 3-y=2019
=>x=2019; y=-2016
DH
24 tháng 6 2017
Bài 1:
a, \(\dfrac{x+5}{x}=\dfrac{4}{3}\)
\(\Rightarrow3x+15=4x\\ \Rightarrow4x-3x=15\\ \Rightarrow x=15\)
b, \(\dfrac{x-20}{x-10}=\dfrac{x+40}{x+70}\)
\(\Rightarrow\left(x-20\right).\left(x+70\right)=\left(x+40\right)\left(x-10\right)\)
\(\Rightarrow x^2+70x-20x-1400=x^2-10x+40x-400\)
\(\Rightarrow x^2-x^2+70x-20x+10x-40x=-400+1400\)
\(\Rightarrow20x=1000\Rightarrow x=50\)
c, \(4^x=\dfrac{1.2.3.....31}{4.6.8.....64}\)
\(\Rightarrow4^x=\dfrac{1}{2.2.2.2.....2.2.64}\) (có 30 số 2)
\(\Rightarrow4^x=\dfrac{1}{2^{30}.4^3}\Rightarrow4^x=\dfrac{1}{4^{15}.4^3}\)
\(\Rightarrow4^x=\dfrac{1}{4^{18}}\)
\(\Rightarrow4^x=4^{-18}\)
Vì \(4\ne-1;4\ne0;4\ne1\) nên \(x=-18\)
Chúc bạn học tốt!!!
24 tháng 6 2017
a , \(\dfrac{x+5}{x}=\dfrac{4}{3}\Leftrightarrow3\left(x+5\right)=4x\)
<=> 3x+15=4x
<=> x= 15
b , \(\dfrac{x-20}{x-10}=\dfrac{x+40}{x+70}\)
<=> \(\dfrac{x-10}{x-10}-\dfrac{10}{x-10}=\dfrac{x+70}{x+70}-\dfrac{30}{x+70}\)
<=> \(1-\dfrac{10}{x-10}=1-\dfrac{30}{x+70}\)
<=> \(\dfrac{10}{x-10}=\dfrac{30}{x+70}\Leftrightarrow\dfrac{1}{x-10}=\dfrac{3}{x+70}\)
<=> (x+70)=3(x-10)
<=> x+70 = 3x-30
<=> 100=2x
<=> x= 50
TV
15 tháng 6 2018
Ta có:
\(\left\{{}\begin{matrix}x^2+xy+\dfrac{y^2}{3}=2019\\z^2+\dfrac{y^2}{3}=1011\\x^2+xz+z^2=1008\end{matrix}\right.\Leftrightarrow x^2+xy+\dfrac{y^2}{3}=z^2+\dfrac{y^2}{3}+x^2+xz+z^2\)
\(\Rightarrow xy=2z^2+xz\Leftrightarrow xy+xz=2z^2+2xz\)
\(\Rightarrow x\left(y+z\right)=2z\left(x+z\right)\Leftrightarrow\dfrac{2z}{x}=\dfrac{y+z}{x+z}\left(đpcm\right)\)
8 tháng 11 2018
1) Vì \(\left|x-2018\right|\) \(\ge\) \(\forall\) x \(\in\) Z
=> \(\left|x-2018\right|+2019\) \(\ge\) 2019
Vậy để biểu thức đạt GTNN \(\Leftrightarrow\)\(\left|x-2018\right|\) = 0
=> x - 2018 = 0
=> x = 0 + 2018
=> x = 2018
Thay x vào biểu thức, ta có:
\(\left|2018-2018\right|\) + 2019
= 0 + 2019
= 2019
18 tháng 11 2022
R=|2x-4|+|2x+5|+1
=|4-2x|+|2x+5|+1
=>R>=|4-2x+2x+5|+1=10
Dấu = xảy ra khi (2x-4)(2x+5)<=0
=>-5/2<=x<=2
c: Q=|x+1/3|+|2/3-x|>=|x+1/3+2/3-x|=1
Dấu = xảy ra khi (x+1/3)(x-2/3)<=0
=>-1/3<=x<=2/3
4 tháng 8 2017
\(A=\left|2x-\dfrac{1}{3}\right|+1007\)
\(\left|2x-\dfrac{1}{3}\right|\ge0\)
\(\Rightarrow\left|2x-\dfrac{1}{3}\right|+1007\ge1007\)
Dấu "=" xảy ra khi:
\(\left|2x-\dfrac{1}{3}\right|=0\Rightarrow2x=\dfrac{1}{3}\Rightarrow x=\dfrac{1}{6}\)
\(\Rightarrow MIN_A=1007\) khi \(x=\dfrac{1}{6}\)
B tương tự
\(C=\left|2018-x\right|+\left|2017-x\right|\)
\(C=\left|2018-x\right|+\left|x-2017\right|\)
Áp dụng BĐT:
\(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)
\(\Rightarrow C\ge\left|2018-x+x-2017\right|\)
\(C\ge1\)
Dấu "=" xảy ra khi:
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}2018-x\ge0\Rightarrow x\le2018\\x-2017\ge0\Rightarrow x\ge2017\end{matrix}\right.\\\left\{{}\begin{matrix}2018-x< 0\Rightarrow x>2018\\x-2017< 0\Rightarrow x< 2017\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow2017\le x\le2018\)
D tương tự
\(\dfrac{x+1}{x-1}=\dfrac{x-2019}{x+2019}\)
\(\Leftrightarrow1+\dfrac{2}{x-1}=1-\dfrac{4038}{x+2019}\)
\(\Leftrightarrow\dfrac{2}{1-x}=\dfrac{4038}{x+2019}\)
\(\Leftrightarrow2x+4038=4038-4038x\)
\(\Leftrightarrow2x=-4038x\)
\(\Leftrightarrow x=0\)
Vậy x = 0