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Ta có:
\(\dfrac{x}{2014}+\dfrac{x+1}{2015}+\dfrac{x+2}{2016}+\dfrac{x+3}{2017}+\dfrac{x+4}{2018}=5\)
\(\Leftrightarrow\left(\dfrac{x}{2014}-1\right)+\left(\dfrac{x+1}{2015}-1\right)+\left(\dfrac{x+2}{2016}-1\right)+\left(\dfrac{x+3}{2017}-1\right)+\left(\dfrac{x+4}{2018}-1\right)=0\)\(\Leftrightarrow\dfrac{x-2014}{2014}+\dfrac{x-2014}{2015}+\dfrac{x-2014}{2016}+\dfrac{x-2014}{2017}+\dfrac{x-2014}{2018}=0\)\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}\right)=0\) (1)
Mà \(\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}>0\) (2)
Từ (1) và (2) => \(x-2014=0\) \(\Leftrightarrow x=2014\)
\(\frac{x+1}{2019}+\frac{x+2}{2018}=\frac{x+2017}{3}+\frac{x+2016}{4}\)
\(\Leftrightarrow\frac{x+1}{2019}+1+\frac{x+2}{2018}+1=\frac{x+2017}{3}+1+\frac{x+2016}{4}+1\)
\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}-\frac{x+2020}{3}-\frac{x+2020}{4}=0\)
\(\Leftrightarrow\left(x+2020\right).\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{3}-\frac{1}{4}\right)=0\)
Mà \(\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{3}-\frac{1}{4}\right)\ne0\)
\(\Rightarrow x+2020=0\Leftrightarrow x=-2020\)
Vậy...
\(a,\dfrac{1+x}{2017}+\dfrac{2+x}{2016}+\dfrac{3+x}{2015}=-3\)
\(\Leftrightarrow\dfrac{1+x}{2017}+1+\dfrac{2+x}{2016}+1+\dfrac{3+x}{2015}+1=-3+3\)
\(\Leftrightarrow\dfrac{1+x+2017}{2017}+\dfrac{2+x+2016}{2016}+\dfrac{3+x+2015}{2015}=0\)
\(\Leftrightarrow\dfrac{x+2018}{2017}+\dfrac{x+2018}{2016}+\dfrac{x+2018}{2015}=0\)
\(\Leftrightarrow\left(x+2018\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}+\dfrac{1}{2015}\right)=0\)
\(\Leftrightarrow x+2018=0\)
\(\Leftrightarrow x=-2018\)
b,\(\dfrac{x-\dfrac{3x-4}{5}}{15}=\dfrac{5x-\dfrac{3-x}{2}}{5}-x+1\)
\(\Leftrightarrow\dfrac{\dfrac{5x-3x+4}{5}}{15}=\dfrac{\dfrac{10x-3+x}{2}}{5}-x+1\)
\(\Leftrightarrow\dfrac{\dfrac{2x+4}{5}}{15}=\dfrac{\dfrac{11x-3}{2}}{5}-\dfrac{5x-5}{5}\)
\(\Leftrightarrow\dfrac{2x+4}{75}=\dfrac{11x-3}{10}-\dfrac{10x-10}{10}\)
\(\Leftrightarrow\dfrac{2x+4}{75}=\dfrac{11x-3-10x+10}{10}\)
\(\Leftrightarrow\dfrac{2x+4}{75}=\dfrac{x+7}{10}\)
\(\Leftrightarrow10\left(2x+4\right)=75\left(x+7\right)\)
\(\Leftrightarrow20x+40=75x+525\)
\(\Leftrightarrow20x-75x=525-40\)
\(\Leftrightarrow-55x=485\)
\(\Leftrightarrow x=-\dfrac{97}{11}\)
a) \(\dfrac{1+x}{2017}+\dfrac{2+x}{2016}+\dfrac{3+x}{2015}=-3\)
\(\Leftrightarrow\dfrac{1+x}{2017}+1+\dfrac{2+x}{2016}+1+\dfrac{3+x}{2015}+1=0\)
\(\Leftrightarrow\dfrac{x+2018}{2017}+\dfrac{x+2018}{2016}+\dfrac{x+2018}{2015}=0\)
\(\Leftrightarrow\left(x+2018\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}+\dfrac{1}{2015}\right)=0\)
\(\Rightarrow x+2018=0\)
\(\Leftrightarrow x=-2018\)
b) \(\dfrac{x-\dfrac{3x-4}{5}}{15}=\dfrac{5x-\dfrac{3-x}{2}}{5}-x+1\)
\(\Leftrightarrow\dfrac{\dfrac{5x-3x+4}{5}}{15}=\dfrac{\dfrac{10x-3+x}{2}}{5}-x+1\)
\(\Leftrightarrow\dfrac{2x+4}{75}=\dfrac{11x-3}{10}-x+1\)
\(\Leftrightarrow\dfrac{4x+8}{150}=\dfrac{165x-45}{150}-\dfrac{150x-150}{150}\)
\(\Leftrightarrow4x+8=165x-45-150x+150\)
\(\Leftrightarrow4x-165x+150x=-45+150-8\)
\(\Leftrightarrow-11x=97\)
\(\Leftrightarrow x=-\dfrac{97}{11}\)
\(S=\left\{-\dfrac{97}{11}\right\}\)
b) \(x,y\ge1\Rightarrow xy\ge1\)
BĐT đã cho tương đương với:
\(\left(\dfrac{1}{1+x^2}-\dfrac{1}{1+xy}\right)+\left(\dfrac{1}{1+y^2}-\dfrac{1}{1+xy}\right)\ge0\)
\(\Leftrightarrow\dfrac{xy-x^2}{\left(1+x^2\right)\left(1+xy\right)}+\dfrac{xy-y^2}{\left(1+y^2\right)\left(1+xy\right)}\ge0\)
\(\Leftrightarrow+\dfrac{x\left(y-x\right)}{\left(1+x^2\right)\left(1+xy\right)}+\dfrac{y\left(x-y\right)}{\left(1+y^2\right)\left(1+xy\right)}\ge0\)
\(\Leftrightarrow\dfrac{\left(y-x\right)^2\left(xy-1\right)}{\left(1+x^2\right)\left(1+y^2\right)\left(1+xy\right)}\ge0\)
BĐT cuối luôn đúng nên ta có đpcm
Đẳng thức xảy ra khi x=y hoặc xy=1
\(\dfrac{x+1}{2019}+\dfrac{x+2}{2018}=\dfrac{x+2017}{3}+\dfrac{x+2016}{4}\)
\(\Leftrightarrow\left(\dfrac{x+1}{2019}+1\right)+\left(\dfrac{x+2}{2018}+1\right)=\left(\dfrac{x+2017}{3}+1\right)+\left(\dfrac{x+2016}{4}+1\right)\)
\(\Leftrightarrow\dfrac{x+2020}{2019}+\dfrac{x+2020}{2018}-\dfrac{x+2020}{3}-\dfrac{x+2020}{4}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{3}-\dfrac{1}{4}\right)=0\)
\(\Leftrightarrow x+2020=0\) ( do \(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{3}-\dfrac{1}{4}\ne0\))
\(\Leftrightarrow x=-2020\)
Vậy phương trình có tập nghiệm S = \(\left\{-2020\right\}\)
Sửa đề: \(\dfrac{x-4}{2019}+\dfrac{x-3}{2018}=\dfrac{x-2}{2017}+\dfrac{x-1}{2016}\)
\(\Leftrightarrow\dfrac{x-4}{2019}+1+\dfrac{x-3}{2018}+1=\dfrac{x-2}{2017}+1+\dfrac{x-1}{2016}+1\)
\(\Leftrightarrow\dfrac{x+2015}{2019}+\dfrac{x+2015}{2018}=\dfrac{x+2015}{2017}+\dfrac{x+2015}{2016}\)
\(\Leftrightarrow\left(x+2015\right)\left(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{2017}-\dfrac{1}{2016}\right)=0\)
\(\Leftrightarrow x=-2015\) vì \(\left(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{2017}-\dfrac{1}{2016}\right)\ne0\)
\(\dfrac{2-x}{2015}-1=\dfrac{1-x}{2016}-\dfrac{x}{2017}\\ \Leftrightarrow\dfrac{2-x-2015}{2015}=\left(\dfrac{1-x}{2016}-1\right)+\left(1-\dfrac{x}{2017}\right)\\ \Leftrightarrow\dfrac{2017-x}{2015}-\dfrac{2017-x}{2016}-\dfrac{2017-x}{2017}=0\\ \Leftrightarrow\left(2017-x\right)\left(\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)=0\\ \Leftrightarrow2017-x=0\left(Vì\text{ }\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\ne0\right)\\ \Leftrightarrow x=2017\)
Vậy tập nghiệm của phương trình là \(S=\left\{2017\right\}\)
a)\(\dfrac{x-2}{3}-\dfrac{x-3}{4}=1\Leftrightarrow\dfrac{4x-8-3x+9}{12}=1\) ⇔x+1=12⇔x=11 Vậy phương trình đã cho có tập nghiệm S=\(\left\{11\right\}\) b)\(\dfrac{x-1}{2015}+\dfrac{x-2}{2014}+\dfrac{x-5}{2011}+\dfrac{x+1}{2017}=4\) \(\Leftrightarrow\left(\dfrac{x-1}{2015}-1\right)+\left(\dfrac{x-2}{2014}-1\right)+\left(\dfrac{x-5}{2011}-1\right)+\left(\dfrac{x+1}{2017}-1\right)=4-4\) \(\Leftrightarrow\dfrac{x-1-2015}{2015}+\dfrac{x-2-2014}{2014}+\dfrac{x-5-2011}{2011}+\dfrac{x+1-2017}{2017}=0\) \(\Leftrightarrow\dfrac{x-2016}{2015}+\dfrac{x-2016}{2014}+\dfrac{x-2016}{2011}+\dfrac{x-2016}{2017}=0\)
\(\Leftrightarrow\left(x-2016\right)\left(\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2011}+\dfrac{1}{2017}\right)=0\)
\(\Leftrightarrow x-2016=0\) (vì \(\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2011}+\dfrac{1}{2017}\ne0\) )
⇔x=2016
Vậy phương trình đã cho có tập nghiệm S=\(\left\{2016\right\}\)
c)3(x-1)-5(x+4)+6(2-x)=7 ⇔3x-3-5x-20+12-6x=7⇔3x-5x-6x=7-12+20+3⇔-8x=18⇔\(x=\dfrac{-9}{4}\)
Vậy phương trình đã cho có tập nghiệm S=\(\left\{\dfrac{-9}{4}\right\}\)
a) ĐKXĐ: \(x\ne\pm2\)
Ta có: \(\dfrac{x}{x+2}=\dfrac{x^2+4}{x^2-4}\)
\(\Leftrightarrow\dfrac{x}{x+2}=\dfrac{x^2+4}{\left(x+2\right)\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{x^2+4}{\left(x+2\right)\left(x-2\right)}\)
\(\Rightarrow x\left(x-2\right)=x^2+4\)
\(\Leftrightarrow x^2-2x=x^2+4\)
\(\Leftrightarrow-2x=4\Leftrightarrow x=-2\)(KTMĐK)
Vậy phương trình vô nghiệm
b) ĐKXĐ: \(x\ne3;x\ne-1\)
Ta có: \(\dfrac{x}{2x-6}+\dfrac{x}{2x+2}+\dfrac{2x}{\left(x+1\right)\left(3-x\right)}=0\)
\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x+1\right)}-\dfrac{2x}{\left(x+1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\dfrac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}-\dfrac{2.2x}{2\left(x+1\right)\left(x-3\right)}=0\)
\(\Rightarrow x\left(x+1\right)+x\left(x-3\right)-2.2x=0\)
\(\Leftrightarrow x^2+x+x^2-3x-4x=0\)
\(\Leftrightarrow2x^2-6x=0\)
\(\Leftrightarrow2x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TMĐK\right)\\x=3\left(KTMĐK\right)\end{matrix}\right.\)
Vậy phương trình có nghiệm là \(x=0\)
\(\dfrac{x+1}{2015}+\dfrac{x+2}{2016}=\dfrac{x+3}{2017}+\dfrac{x+4}{2018}\)
<=>\(\dfrac{x+1}{2015}-1+\dfrac{x+2}{2016}-1=\dfrac{x+3}{2017}-1+\dfrac{x+4}{2018}-1\)
<=>\(\dfrac{x-2014}{2015}+\dfrac{x-2014}{2016}=\dfrac{x-2014}{2017}+\dfrac{x-2014}{2018}\)
<=>\(\dfrac{x-2014}{2015}+\dfrac{x-2014}{2016}-\dfrac{x-2014}{2017}-\dfrac{x-2014}{2018}=0\)
<=>\(\left(x-2014\right)\left(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}-\dfrac{1}{2018}\right)=0\)
vì 1/2015+1/2016-1/2017-1/2018 khác 0
=>x-2014=0<=>x=2014
vậy.....................
chúc bạn học totts ^^
\(\dfrac{x+1}{2015}+\dfrac{x+2}{2016}=\dfrac{x+3}{2017}+\dfrac{x+4}{2018}\)
\(\Leftrightarrow\dfrac{x+1}{2015}-1+\dfrac{x+2}{2016}-1=\dfrac{x+3}{x017}-1+\dfrac{x+4}{2018}-1\)
\(\Leftrightarrow\dfrac{x+1-2015}{2015}+\dfrac{x+2-2016}{2016}=\dfrac{x+3-2017}{2017}+\dfrac{x+4-2018}{2018}\)\(\Leftrightarrow\dfrac{x-2014}{2015}+\dfrac{x-2014}{2016}=\dfrac{x-2014}{2017}+\dfrac{x-2014}{2018}\)
\(\Leftrightarrow\dfrac{x-2014}{2015}+\dfrac{x-2014}{2016}-\dfrac{x-2014}{2017}-\dfrac{x-2014}{2018}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}-\dfrac{1}{2018}\right)=0\)
Vì: \(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}-\dfrac{1}{2018}\ne0\)
\(\Rightarrow x-2014=0\)
\(\Rightarrow x=2014\)
Vậy........