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\(\dfrac{x-3}{2011}+\dfrac{x-2}{2012}=\dfrac{x-2012}{2}+\dfrac{x-2011}{3}\)
\(\Leftrightarrow\dfrac{x-3}{2011}+\dfrac{x-2}{2012}-2=\dfrac{x-2012}{2}+\dfrac{x-2011}{3}-2\)
\(\Leftrightarrow\left(\dfrac{x-3}{2011}-1\right)+\left(\dfrac{x-2}{2012}-1\right)=\left(\dfrac{x-2012}{2}-1\right)+\left(\dfrac{x-2011}{3}-1\right)\)
\(\Leftrightarrow\dfrac{x-2014}{2011}+\dfrac{x-2014}{2012}-\dfrac{x-2014}{2}-\dfrac{x-2014}{3}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)
\(\Leftrightarrow x-2014=0\)
\(\Leftrightarrow x=0\)
\(\dfrac{x-3}{2011}+\dfrac{x-2}{2012}=\dfrac{x-2012}{2}+\dfrac{x-2011}{3}\)
<=>\(\dfrac{x-3}{2011}-1+\dfrac{x-2}{2012}-1=\dfrac{x-2012}{2}-1+\dfrac{x-2011}{3}-1\)
<=>\(\dfrac{x-2014}{2011}+\dfrac{x-2014}{2012}=\dfrac{x-2014}{2}+\dfrac{x-2014}{3}\)
<=>\(\dfrac{x-2014}{2011}+\dfrac{x-2014}{2012}-\dfrac{x-2014}{2}-\dfrac{x-2014}{3}=0\)
<=>\(\left(x-2014\right)\left(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)
vì 1/2011+1/2012-1/2-1/3 khác 0
=>x-2014=0<=>x=2014
vậy....................
Giải bpt sau:
\(\dfrac{x+3}{2011}\)+\(\dfrac{x+2}{2012}\)+\(\dfrac{x+1}{2013}\)≥\(\dfrac{3x}{2014}\)
\(\dfrac{x+3}{2011}+\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\ge\dfrac{3x}{2014}\)
\(\dfrac{x+3}{2011}+1+\dfrac{x+2}{2012}+1+\dfrac{x+1}{2013}+1\ge\dfrac{3x}{2014}+3\)
\(\dfrac{x+2014}{2011}+\dfrac{x+2014}{2012}+\dfrac{x+2014}{2013}\ge3\left(\dfrac{x+2014}{2014}\right)\)
\(\left(x+2014\right)\left(\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}-\dfrac{3}{2014}\right)\ge0\)
Mà \(\left(\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}-\dfrac{3}{2014}\right)>0\) (bạn có thể chứng minh nếu thích )
Nên \(x+2014\ge0\)
\(\Leftrightarrow x\ge-2014\)
Vậy
Vì you ghi sai phương trình nên tui sửa lại đề nghen!!!
Ta có \(\dfrac{x+1}{2013}+\dfrac{x+2}{2012}=\dfrac{x+3}{2011}+\dfrac{x+4}{2010}\)
\(\Leftrightarrow\dfrac{x+1}{2013}+1+\dfrac{x+2}{2012}+1=\dfrac{x+3}{2011}+1+\dfrac{x+4}{2010}+1\)
\(\Leftrightarrow\dfrac{x+2014}{2013}+\dfrac{x+2014}{2012}=\dfrac{x+2014}{2011}+\dfrac{x+2014}{2010}\)
\(\Leftrightarrow\dfrac{x+2014}{2013}+\dfrac{x+2014}{2012}-\dfrac{x+2014}{2011}-\dfrac{x+2014}{2010}=0\)
\(\Leftrightarrow\left(x+2014\right)\left(\dfrac{1}{2013}+\dfrac{1}{2012}-\dfrac{1}{2011}-\dfrac{1}{2010}\right)=0\)
\(\Leftrightarrow x+2014=0\)
\(\Leftrightarrow x=-2014\)
Vậy \(x=-2014\)
\(\dfrac{x-1}{2012}+\dfrac{x-2}{2011}+\dfrac{x-3}{2010}+...+\dfrac{x-2012}{1}=2012\)
<=>\(\dfrac{x-1}{2012}-1+\dfrac{x-2}{2011}-1+\dfrac{x-3}{2010}-1+...+\dfrac{x-2012}{1}-1=0\)
<=>\(\dfrac{x-2013}{2012}+\dfrac{x-2013}{2011}+\dfrac{x-2013}{2010}+...+\dfrac{x-2013}{1}=0\)
<=>\(\left(x-2013\right)\left(\dfrac{1}{2012}+\dfrac{1}{2011}+...+1\right)=0\)
do 1/2012+1/2011....+1 khác 0 =>x-2013=0<=>x=2013
vậy..........................
\(\dfrac{x-1}{2012}+\dfrac{x-2}{2011}+\dfrac{x-3}{2010}+...+\dfrac{x-2012}{1}=2012\)
\(\left(\dfrac{x-1}{2012}+\dfrac{x-2}{2011}+\dfrac{x-3}{2010}+...+\dfrac{x-2012}{1}\right)-2012=0\)
\(\Rightarrow\dfrac{x-2013}{2012}+\dfrac{x-2013}{2011}+\dfrac{x-2013}{2010}+...+\dfrac{x-2013}{1}=0\)
\(\Rightarrow x-2013\left(\dfrac{1}{2012}+\dfrac{1}{2011}+\dfrac{1}{2010}+...+\dfrac{1}{1}\right)=0\)
Vì \(x-2013\left(\dfrac{1}{2012}+\dfrac{1}{2011}+\dfrac{1}{2010}+...+\dfrac{1}{1}\right)=0\)nên x - 2013 hoặc \(\dfrac{1}{2012}+\dfrac{1}{2011}+\dfrac{1}{2010}+...+\dfrac{1}{1}\) = 0. Nhưng \(\dfrac{1}{2012}+\dfrac{1}{2011}+\dfrac{1}{2010}+...+\dfrac{1}{1}\ne0\) nên x - 2013 = 0. Vì vậy x = 2013.
Vậy...
\(\dfrac{x-1}{2012}+\dfrac{x-2}{2011}+\dfrac{x-3}{2010}+...+\dfrac{x-2012}{1}=2012\)
\(\Leftrightarrow\dfrac{x-1}{2012}-1+\dfrac{x-2}{2011}-1+...+\dfrac{x-2012}{1}-1=0\)
\(\Leftrightarrow\dfrac{x-2013}{2012}+\dfrac{x-2013}{2011}+...+\dfrac{x-2013}{1}=0\)
\(\Leftrightarrow\left(x-2013\right)\left(\dfrac{1}{2012}+\dfrac{1}{2011}+...+\dfrac{1}{1}\right)=0\)
Dễ thấy: \(\dfrac{1}{2012}+\dfrac{1}{2011}+...+\dfrac{1}{1}>0\)
\(\Rightarrow x-2013=0\Rightarrow x=2013\)
Sao lại trừ 1 vậy bạn ??? mình không hiểu cho lắm mong bạn giúp đỡ
\(\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}+...+\dfrac{x-2012}{2}=2012\)
\(\Rightarrow\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}+...+\dfrac{x-2012}{2}-2012=0\)
\(\Rightarrow\dfrac{x-1}{2013}-1+\dfrac{x-2}{2012}-1+\dfrac{x-3}{2011}-1+...+\dfrac{x-2012}{2}-1=0\)
\(\Rightarrow\dfrac{x-2014}{2013}+\dfrac{x-2014}{2012}+\dfrac{x-2014}{2011}+...+\dfrac{x-2014}{2}=0\)
\(\Rightarrow\left(x-2014\right)\left(\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}+...+\dfrac{1}{2}\right)=0\)
Mà \(\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}+...+\dfrac{1}{2}\ne0\)
\(\Rightarrow x-2014=0\)
\(\Rightarrow x=2014\)
\(a.\dfrac{3x-2}{5}+\dfrac{x-1}{9}=\dfrac{14x-3}{15}-\dfrac{2x+1}{9}\\ \Leftrightarrow\dfrac{27x-18}{45}+\dfrac{5x-5}{45}=\dfrac{42x-9}{45}-\dfrac{10x+5}{45}\\ \Rightarrow27x-18+5x-5=42x-9-10x-5\\ \Leftrightarrow32x-23=32x-14\\ \Leftrightarrow0x=9\\ \Rightarrow Phươngtrìnhvônghiệm\\ \Rightarrow S=\phi\)
\(b.\dfrac{x+3}{2}-\dfrac{2-x}{3}-1=\dfrac{x+5}{6}\\ \Leftrightarrow\dfrac{3x-9}{6}-\dfrac{4-2x}{6}-\dfrac{6}{6}=\dfrac{x+5}{6}\\ \Rightarrow3x-9-4+2x-6=x+5\\ \Leftrightarrow5x-19=x+5\\ \Leftrightarrow4x=24\\ \Rightarrow x=6\\ \Rightarrow S=\left\{6\right\}\)
\(c.\dfrac{x+5}{2010}+\dfrac{x+4}{2011}+\dfrac{x+3}{2012}+\dfrac{x+2}{2013}=-4\\ \Leftrightarrow\dfrac{x+5}{2010}+1+\dfrac{x+4}{2011}+1+\dfrac{x+3}{2012}+1+\dfrac{x+2}{2013}+1=-4+4\\ \Rightarrow\dfrac{2015+x}{2010}+\dfrac{2015+x}{2011}+\dfrac{2015+x}{2012}+\dfrac{2015+x}{2013}=0\\ \Leftrightarrow\left(2015+x\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\right)=0\)
Do \(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}>0\)
nên \(2015+x=0\Rightarrow x=-2015\)
Câu d tương tự...thêm rồi chuyển vế sang :v
Lời giải:
Ta có:
\(\frac{x-1}{2012}+\frac{x-2}{2011}+\frac{x-3}{2010}+...+\frac{x-2012}{1}=2012\)
\(\Leftrightarrow \left(\frac{x-1}{2012}-1\right)+\left(\frac{x-2}{2011}-1\right)+\left(\frac{x-3}{2010}-1\right)+...+\left(\frac{x-2012}{1}-1\right)=0\)
\(\Leftrightarrow \frac{x-2013}{2012}+\frac{x-2013}{2011}+...+\frac{x-2013}{1}=0\)
\(\Leftrightarrow (x-2013)\left(\frac{1}{2012}+\frac{1}{2011}+...+1\right)=0\)
Dễ thấy \(\frac{1}{2012}+\frac{1}{2011}+...+1\neq 0\Rightarrow x-2013=0\)
\(\Leftrightarrow x=2013\)
Vậy PT có nghiệm \(x=2013\)
1, \(\dfrac{x-3}{2011}+\dfrac{x-2}{2012}=\dfrac{x-2012}{2}+\dfrac{x-2011}{3}\\ \\ < =>\dfrac{x-3}{2011}-1+\dfrac{x-2}{2012}-1=\dfrac{x-2012}{2}-1+\dfrac{x-2011}{3}-1\\ \\ < =>\dfrac{x-2014}{2011}+\dfrac{x-2014}{2012}-\dfrac{x-2014}{2}-\dfrac{x-2014}{3}=0\\ \\ < =>\left(x-2014\right).\left(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\\ \\ < =>x-2014=0< =>x=2014\)
2, \(x^2+1=x\\ \\ < =>x^2-x+1=0\\ \\ < =>x^2-x+\dfrac{1}{4}+\dfrac{3}{4}=0\\ \\ < =>\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\)
có vế trái luôn dương, vế phải = 0 => vô nghiệm