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\(\Leftrightarrow\dfrac{x+1}{2012}+1+\dfrac{x+2}{2011}+1+\dfrac{x+3}{2010}+1=\dfrac{x-1}{2014}+1+\dfrac{x-2}{2015}+1+\dfrac{x-3}{2016}+1\)
=>x+2013=0
hay x=-2013
\(\dfrac{x+1}{2012}+1+\dfrac{x+2}{2011}+1+\dfrac{x+3}{2010}+1=\dfrac{x-1}{2014}+1+\dfrac{x-2}{2015}+1+\dfrac{x-3}{2016}+1\)
\(\Leftrightarrow\left(x+2013\right)\left(\dfrac{1}{2022}+\dfrac{1}{2011}+\dfrac{2}{2010}-\dfrac{1}{2014}-\dfrac{1}{2015}-\dfrac{1}{2016}\ne0\right)=0\Leftrightarrow x=-2013\)
\(\dfrac{x+1}{2012}+\dfrac{x+2}{2011}+\dfrac{x+3}{2010}=\dfrac{x-1}{2014}+\dfrac{x-2}{2015}+\dfrac{x-3}{2016}\)
\(\Leftrightarrow\left(\dfrac{x+1}{2012}+1\right)+\left(\dfrac{x+2}{2011}+1\right)+\left(\dfrac{x+3}{2010}\right)=\left(\dfrac{x-1}{2014}+1\right)+\left(\dfrac{x-2}{2015}+1\right)+\left(\dfrac{x-3}{2016}+1\right)\)
\(\Leftrightarrow\dfrac{x+2013}{2012}+\dfrac{x+2013}{2011}+\dfrac{x+2013}{2010}-\dfrac{x+2013}{2014}-\dfrac{x+2013}{2015}-\dfrac{x+2013}{2016}=0\)
\(\Leftrightarrow\left(x+2013\right)\left(\dfrac{1}{2012}+\dfrac{1}{2011}+\dfrac{1}{2010}-\dfrac{1}{2014}-\dfrac{1}{2015}-\dfrac{1}{2016}\right)=0\)
\(\Leftrightarrow x+2013=0\)
\(\Leftrightarrow x=-2013\)
\(\dfrac{x}{2012}+\dfrac{x+1}{2013}+\dfrac{x+2}{2014}+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}=5\)
\(\Leftrightarrow\dfrac{x}{2012}+\dfrac{x+1}{2013}+\dfrac{x+2}{2014}+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}-5=0\)
\(\Leftrightarrow\dfrac{x}{2012}-1+\dfrac{x+1}{2013}-1+\dfrac{x+2}{2014}-1+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}-1=0\)
\(\Leftrightarrow\dfrac{x-2012}{2012}+\dfrac{x-2012}{2013}+\dfrac{x-2012}{2014}+\dfrac{x-2012}{2015}+\dfrac{x-2012}{2016}=0\)
\(\Leftrightarrow\left(x-12\right).\left(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}\right)=0\)
\(\Leftrightarrow x-12=0\)
\(\Leftrightarrow x=12\)
a) \(5x-3=7\)
\(\Leftrightarrow5x=7+3\)
\(\Leftrightarrow5x=10\)
\(\Leftrightarrow x=\dfrac{10}{5}\)
\(\Leftrightarrow x=2\)
Vậy \(S=\left\{2\right\}\)
b) \(\left(x+3\right)\left(x-4\right)=0\)
\(\Leftrightarrow x+3=0\) hoặc \(x-4=0\)
*) \(x+3=0\)
\(x=0-3\)
\(x=-3\)
*) \(x-4=0\)
\(x=0+4\)
\(x=4\)
Vậy \(S=\left\{-3;4\right\}\)
c) \(\left|x^2+2014\right|=1\)
\(\Leftrightarrow x^2+2014=1\) hoặc \(x^2+2014=-1\)
*) \(x^2+2014=1\)
\(\Leftrightarrow x^2=1-2014\)
\(\Leftrightarrow x^2=-2013\) (vô lý)
*) \(x^2+2014=-1\)
\(\Leftrightarrow x^2=-1-2014\)
\(\Leftrightarrow x^2=-2015\) (vô lý)
Vậy \(S=\varnothing\)
d) \(\dfrac{2}{x+1}-\dfrac{1}{x-3}=\dfrac{3x-11}{x^2-2x-3}\) (1)
ĐKXĐ: \(x\ne-1;x\ne3\)
\(\left(1\right)\Leftrightarrow2\left(x-3\right)-\left(x+1\right)=3x-11\)
\(\Leftrightarrow2x-6-x-1=3x-11\)
\(\Leftrightarrow-2x=-11+7\)
\(\Leftrightarrow-2x=-4\)
\(\Leftrightarrow x=2\) (nhận)
Vậy \(S=\left\{2\right\}\)
b) \(\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}=\dfrac{1}{18}\\< =>\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}=\dfrac{1}{18}\\ < =>\dfrac{1}{x+1}-\dfrac{1}{x+5}=\dfrac{1}{18}\\ quyđồngmẫuvàkhửmẫu\\ x^{2^{ }}+6x-27=0\\ giảipttìmđược:x=3;x=-9\)
a) \(\frac{x-2015}{1}+\frac{x-2014}{2}+\frac{x-2013}{3}+...+\frac{x-1}{2015}+\frac{x}{2016}=0\\ \Leftrightarrow\frac{x-2015}{1}-1+\frac{x-2014}{2}-1+...+\frac{x-1}{2015}-1+\frac{x}{2016}-1=-2016\)
\(\Leftrightarrow\frac{\left(x-2016\right).1}{1}+\frac{\left(x-2016\right).1}{2}+\frac{\left(x-2016\right).1}{3}+...+\frac{\left(x-2016\right).1}{2015}+\frac{\left(x-2016\right).1}{2016}=-2016\)
\(\Leftrightarrow\left(x-2016\right)\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}\right)=-2016\)
tới đây mình chịu. mình nghĩ là phương trình bạn cho là bằng 2016 chứ, như thế giải mới được, còn như này thì mình bó tay
b)
\(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}=\frac{1}{8}\\ \Leftrightarrow\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}=\frac{1}{8}\\ \Leftrightarrow\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+6}=\frac{1}{8}\\ \Leftrightarrow\frac{4}{\left(x+2\right)\left(x+6\right)}=\frac{1}{8}\)
\(\Leftrightarrow\frac{4}{\left(x+2\right)\left(x+6\right)}=\frac{4}{32}\\ \Rightarrow\left(x+2\right)\left(x+6\right)=32\)
\(\Leftrightarrow x^2+8x+12-32=0\\ \Leftrightarrow x^2+8x-20=0\\ \Leftrightarrow\left(x+10\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x+10=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-10\\x=2\end{matrix}\right.\)
vậy phương trình có tập nghiệm là S={-10;2}
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) => \(x-2014=0\) \(\Leftrightarrow x=2014\)
|x-9|=2x+5
Xét 3 TH
TH1: x>9 => x-9=2x+5 =>-9-5=x =>x=-14 (L)
TH2: x<9 => 9-x=2x+5 => 9-5=3x =>x=4/3(t/m)
TH3: x=9 =>0=23(L)
Vậy x= 4/3
Ta có:\(\dfrac{1-2x}{4}-2\le\dfrac{1-5x}{8}+x\\ \)
\(\dfrac{2-4x-16}{8}\le\dfrac{1-5x+8x}{8}\)
\(-4x-14\le1+3x\\ \Leftrightarrow7x+15\ge0\\ \Leftrightarrow x\ge-\dfrac{15}{7}\)
\(\dfrac{1}{x+1}\)-\(\dfrac{5}{x-2}\)=\(\dfrac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow\)\(\dfrac{x-2}{\left(x+1\right)\left(x-2\right)}\)-\(\dfrac{5\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}\)=\(\dfrac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow\)x-2-5(x+1)=15
\(\Leftrightarrow\) x-2-5x-5=15
\(\Leftrightarrow\)x-5x=15+2+5
\(\Leftrightarrow\)-4x=22
\(\Leftrightarrow\)x=-\(\dfrac{11}{2}\)
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) \(\dfrac{x-\dfrac{3x-4}{5}}{15}=\dfrac{5x-\dfrac{3-x}{2}}{5}-x+1\)
\(\Leftrightarrow\dfrac{\dfrac{5x-3x+4}{5}}{15}=\dfrac{\dfrac{10x-3+x}{2}}{5}-x+1\)
\(\Leftrightarrow\dfrac{2x+4}{75}=\dfrac{11x-3}{10}-x+1\)
\(\Leftrightarrow\dfrac{4x+8}{150}=\dfrac{165x-45}{150}-\dfrac{150x-150}{150}\)
\(\Leftrightarrow4x+8=165x-45-150x+150\)
\(\Leftrightarrow4x-165x+150x=-45+150-8\)
\(\Leftrightarrow-11x=97\)
\(\Leftrightarrow x=-\dfrac{97}{11}\)
\(S=\left\{-\dfrac{97}{11}\right\}\)
\(\dfrac{x-1}{2016}+\dfrac{x-2}{2015}+\dfrac{x-3}{2014}+...+\dfrac{x-2016}{1}=2016\)
\(\Leftrightarrow\dfrac{x-1}{2016}-1+\dfrac{x-2}{2015}-1+\dfrac{x-3}{2014}-1+...+\dfrac{x-2016}{1}-1=0\)
\(\Leftrightarrow\dfrac{x-2017}{2016}+\dfrac{x-2017}{2015}+\dfrac{x-2017}{2014}+...+\dfrac{x-2017}{1}=0\)
\(\Leftrightarrow\left(x-2017\right)\left(\dfrac{1}{2016}+\dfrac{1}{2015}+...+1\right)=0\)
\(\Leftrightarrow x-2017=0\) (do \(\dfrac{1}{2016}+\dfrac{1}{2015}+...+1\ne0\))
\(\Rightarrow x=2017\)
Phương trình =2016. Mình quên ghi