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\(\dfrac{3-x}{2016}-1=\dfrac{2-x}{2017}+\dfrac{1-x}{2018}\)
\(\Leftrightarrow\dfrac{3-x}{2016}+1=\dfrac{2-x}{2017}+1+\dfrac{1-x}{2018}+1\)
\(\Leftrightarrow\dfrac{2019-x}{2016}=\dfrac{2019-x}{2017}+\dfrac{2019-x}{2018}\)
\(\left(2019-x\right)\left(\dfrac{1}{2016}-\dfrac{1}{2017}-\dfrac{1}{2018}\right)=0\)
2019 -x =0 ; x =2019
\(\dfrac{x+4}{2015}+\dfrac{x+3}{2016}=\dfrac{x+2}{2017}+\dfrac{x+1}{2018}\)
\(\Leftrightarrow\left(\dfrac{x+4}{2015}+1\right)+\left(\dfrac{x+3}{2016}+1\right)=\left(\dfrac{x+2}{2017}+1\right)+\left(\dfrac{x+1}{2018}+1\right)\)
\(\Leftrightarrow\dfrac{x+2019}{2015}+\dfrac{x+2019}{2016}=\dfrac{x+2019}{2017}+\dfrac{x+2019}{2018}\)
\(\Leftrightarrow\dfrac{x+2019}{2015}+\dfrac{x+2019}{2016}-\dfrac{x+2019}{2017}-\dfrac{x+2019}{2018}=0\)
\(\Leftrightarrow\left(x+2019\right)\left(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}-\dfrac{1}{2018}\right)=0\)
Mà \(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}-\dfrac{1}{2018}\ne0\)
\(\Leftrightarrow x+2019=0\)
\(\Leftrightarrow x=-2019\)
Vậy...
\(\dfrac{1}{2019}:2017.x=-\dfrac{1}{2017}\)
\(\dfrac{1}{2019.2017}x=-\dfrac{1}{2017}\)
x=\(-\dfrac{1}{2017}:\dfrac{1}{2019.2017}\)=-2019
Vậy x=-2019
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 = \(\dfrac{-1}{2019}\)
\(\dfrac{2017}{1}+\dfrac{2016}{2}+...+\dfrac{2}{2016}+\dfrac{1}{2017}\)
\(=\left(\dfrac{2016}{2}+1\right)+\left(\dfrac{2015}{3}+1\right)+...+\left(\dfrac{2}{2016}+1\right)+\left(\dfrac{1}{2017}+1\right)+1\)
\(=\dfrac{2018}{2}+\dfrac{2018}{3}+...+\dfrac{2018}{2017}+\dfrac{2018}{2018}\)
\(=2018\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2018}\right)\)
Theo đề, ta có: \(x=\dfrac{2018\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2018}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2018}}=2018\)
\(A=\left|2x-\dfrac{1}{3}\right|+1007\)
\(\left|2x-\dfrac{1}{3}\right|\ge0\)
\(\Rightarrow\left|2x-\dfrac{1}{3}\right|+1007\ge1007\)
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
a: \(\Leftrightarrow\dfrac{7}{2}x-\dfrac{3}{4}=\dfrac{1}{2}x+\dfrac{5}{2}\)
\(\Leftrightarrow3x=\dfrac{5}{2}+\dfrac{3}{4}=\dfrac{10}{4}+\dfrac{3}{4}=\dfrac{13}{4}\)
=>x=13/12
b: \(\Leftrightarrow x\cdot\left(\dfrac{2}{3}-\dfrac{1}{2}\right)=-\dfrac{1}{3}+\dfrac{2}{5}\)
\(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{-5+6}{15}=\dfrac{1}{15}\)
\(\Leftrightarrow x=\dfrac{1}{15}:\dfrac{1}{6}=\dfrac{2}{5}\)
c: \(\Leftrightarrow x\cdot\dfrac{1}{3}+x\cdot\dfrac{2}{5}+\dfrac{2}{5}=0\)
\(\Leftrightarrow x\cdot\dfrac{11}{15}=-\dfrac{2}{5}\)
\(\Leftrightarrow x=-\dfrac{2}{5}:\dfrac{11}{15}=\dfrac{-2}{5}\cdot\dfrac{15}{11}=\dfrac{-30}{55}=\dfrac{-6}{11}\)
d: \(\Leftrightarrow-\dfrac{1}{3}x+\dfrac{1}{2}+\dfrac{2}{3}-x-\dfrac{1}{2}=5\)
\(\Leftrightarrow-\dfrac{4}{3}x+\dfrac{2}{3}=5\)
\(\Leftrightarrow-\dfrac{4}{3}x=5-\dfrac{2}{3}=\dfrac{13}{3}\)
\(\Leftrightarrow x=\dfrac{13}{3}:\dfrac{-4}{3}=\dfrac{-13}{4}\)
e: \(\Leftrightarrow\left(\dfrac{x+2015}{5}+1\right)+\left(\dfrac{x+2016}{4}+1\right)=\left(\dfrac{x+2017}{3}+1\right)+\left(\dfrac{x+2018}{2}+1\right)\)
=>x+2020=0
hay x=-2020
a: \(\dfrac{3x+2}{5x+7}=\dfrac{3x-1}{5x+1}\)
\(\Leftrightarrow\left(3x+2\right)\left(5x+1\right)=\left(3x-1\right)\left(5x+7\right)\)
\(\Leftrightarrow15x^2+3x+10x+2=15x^2+21x-5x-7\)
=>16x-7=13x+2
=>3x=9
hay x=3
b: \(\dfrac{x+1}{2016}+\dfrac{x}{2017}=\dfrac{x+2}{2015}+\dfrac{x+3}{2014}\)
\(\Leftrightarrow\left(\dfrac{x+1}{2016}+1\right)+\left(\dfrac{x}{2017}+1\right)=\left(\dfrac{x+2}{2015}+1\right)+\left(\dfrac{x+3}{2014}+1\right)\)
=>x+2017=0
hay x=-2017
e: \(\left(2x-3\right)^2=144\)
=>2x-3=12 hoặc 2x-3=-12
=>2x=15 hoặc 2x=-9
=>x=15/2 hoặc x=-9/2
\(\Leftrightarrow\left(\dfrac{x+1}{2019}+1\right)+\left(\dfrac{x+2}{2018}+1\right)=\left(\dfrac{x+3}{2017}+1\right)+\left(\dfrac{x+4}{2016}+1\right)\)
\(\Leftrightarrow\dfrac{x+2020}{2019}+\dfrac{x+2020}{2018}-\dfrac{x+2020}{2017}-\dfrac{x+2020}{2016}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{2017}-\dfrac{1}{2016}\right)=0\)
\(\Leftrightarrow x=-2020\)(do \(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{2017}-\dfrac{1}{2016}\ne0\))
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!!!
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
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
\(\dfrac{1}{2017}+\dfrac{1}{2018}-\dfrac{1}{2019}-\dfrac{1}{2020}>0\)
\(\Rightarrow x+1=0\)
\(\Rightarrow x=-1\)
x=-1