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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
\(\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}+...+\dfrac{x-2012}{2}=2012\)
\(\Rightarrow\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}+...+\dfrac{x-2012}{2}-2012=0\)
\(\Rightarrow\dfrac{x-1}{2013}-1+\dfrac{x-2}{2012}-1+\dfrac{x-3}{2011}-1+...+\dfrac{x-2012}{2}-1=0\)
\(\Rightarrow\dfrac{x-2014}{2013}+\dfrac{x-2014}{2012}+\dfrac{x-2014}{2011}+...+\dfrac{x-2014}{2}=0\)
\(\Rightarrow\left(x-2014\right)\left(\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}+...+\dfrac{1}{2}\right)=0\)
Mà \(\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}+...+\dfrac{1}{2}\ne0\)
\(\Rightarrow x-2014=0\)
\(\Rightarrow x=2014\)
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
\(\dfrac{x-3}{2011}+\dfrac{x-2}{2012}=\dfrac{x-2012}{2}+\dfrac{x-2011}{3}\)
\(\Leftrightarrow\dfrac{x-3}{2011}-1+\dfrac{x-2}{2012}-1=\dfrac{x-2012}{2}-1+\dfrac{x-2011}{3}-1\)
\(\Leftrightarrow\dfrac{x-2014}{2011}+\dfrac{x-2014}{2012}=\dfrac{x-2014}{2}+\dfrac{x-2014}{3}\)
\(\Leftrightarrow\dfrac{x-2014}{2011}+\dfrac{x-2014}{2012}-\dfrac{x-2014}{2}-\dfrac{x-2014}{3}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)
\(\Leftrightarrow x-2014=0\)
\(\Leftrightarrow x=2014\)
Vậy x = 2014.
Đề sai rồi bạn!
Mình sửa lại đề nha!
\(\dfrac{x-3}{2011}+\dfrac{x-2}{2012}=\dfrac{x-2012}{2}+\dfrac{x-2011}{3}\)
\(\Leftrightarrow\left(\dfrac{x-3}{2011}-1\right)+\left(\dfrac{x-2}{2012}-1\right)=\left(\dfrac{x-2012}{2}-1\right)+\left(\dfrac{x-2011}{3}-1\right)\)
\(\Leftrightarrow\dfrac{x-2014}{2011}+\dfrac{x-2014}{2012}=\dfrac{x-2014}{2}+\dfrac{x-2014}{3}\)
\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)
\(\Leftrightarrow x-2014=0\)
\(\Leftrightarrow x=2014\)
Vậy S={2014}
\(\frac{x-3}{2011}+\frac{x-2}{2012}=\frac{x-2012}{2}+\frac{x-2011}{3}\)
\(\Rightarrow\left(\frac{x-3}{2011}-1\right)+\left(\frac{x-2}{2012}-1\right)=\left(\frac{x-2012}{2}-1\right)+\left(\frac{x-2011}{3}-1\right)\)
\(\Rightarrow\frac{x-2014}{2011}+\frac{x-2014}{2012}=\frac{x-2014}{2}+\frac{x-2014}{3}\)
\(\Rightarrow\frac{x-2014}{2011}+\frac{x-2014}{2012}-\frac{x-2014}{2}-\frac{x-2014}{3}=0\)
\(\Rightarrow\left(x-2014\right)\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2}-\frac{1}{3}\right)=0\)
Mà \(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2}-\frac{1}{3}\ne0\)
\(\Rightarrow x-2014=0\)
\(\Rightarrow x=2014\)
Vậy x = 2014
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 \(x=-2014\)
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 \(x=2013\)
\(a.\dfrac{3x-2}{5}+\dfrac{x-1}{9}=\dfrac{14x-3}{15}-\dfrac{2x+1}{9}\\ \Leftrightarrow\dfrac{27x-18}{45}+\dfrac{5x-5}{45}=\dfrac{42x-9}{45}-\dfrac{10x+5}{45}\\ \Rightarrow27x-18+5x-5=42x-9-10x-5\\ \Leftrightarrow32x-23=32x-14\\ \Leftrightarrow0x=9\\ \Rightarrow Phươngtrìnhvônghiệm\\ \Rightarrow S=\phi\)
\(b.\dfrac{x+3}{2}-\dfrac{2-x}{3}-1=\dfrac{x+5}{6}\\ \Leftrightarrow\dfrac{3x-9}{6}-\dfrac{4-2x}{6}-\dfrac{6}{6}=\dfrac{x+5}{6}\\ \Rightarrow3x-9-4+2x-6=x+5\\ \Leftrightarrow5x-19=x+5\\ \Leftrightarrow4x=24\\ \Rightarrow x=6\\ \Rightarrow S=\left\{6\right\}\)
\(c.\dfrac{x+5}{2010}+\dfrac{x+4}{2011}+\dfrac{x+3}{2012}+\dfrac{x+2}{2013}=-4\\ \Leftrightarrow\dfrac{x+5}{2010}+1+\dfrac{x+4}{2011}+1+\dfrac{x+3}{2012}+1+\dfrac{x+2}{2013}+1=-4+4\\ \Rightarrow\dfrac{2015+x}{2010}+\dfrac{2015+x}{2011}+\dfrac{2015+x}{2012}+\dfrac{2015+x}{2013}=0\\ \Leftrightarrow\left(2015+x\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\right)=0\)
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
\(1.\) Giả sử : \(a\ge b\ge c\Rightarrow a+b\ge a+c\ge b+c\)
Ta có : \(\dfrac{c}{a+b}\le\dfrac{c}{b+c};\dfrac{b}{a+c}\le\dfrac{b}{b+c};\dfrac{a}{b+c}=\dfrac{a}{b+c}\)
\(\Rightarrow\dfrac{c}{a+b}+\dfrac{b}{a+c}+\dfrac{a}{b+c}\le\dfrac{b+c}{b+c}+\dfrac{a}{b+c}=1+\dfrac{a}{b+c}< 1+1=2\left(đpcm\right)\)
\(2.\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\)
\(\Leftrightarrow\dfrac{yz+xz+xy}{xyz}=\dfrac{1}{x+y+z}\)
\(\Leftrightarrow\left(x+y+z\right)\left(xy+yz+xz\right)=xyz\)
\(\Leftrightarrow x^2y+x^2z+xy^2+y^2z+xyz+xyz+yz^2+xz^2=0\)
\(\Leftrightarrow xy\left(x+y+z\right)+yz\left(x+y+z\right)+xz\left(x+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)y\left(x+z\right)+xz\left(x+z\right)=0\)
\(\Leftrightarrow\left(x+z\right)\left(xy+y^2+yz+xz\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(x+z\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-y\\y=-z\\x=-z\end{matrix}\right.\)
+) Với : \(x=-y\) , ta có :
Đpcm \(\Leftrightarrow-\dfrac{1}{y^{2011}}+\dfrac{1}{y^{2011}}+\dfrac{1}{z^{2011}}=\dfrac{1}{-y^{2011}+y^{2011}+z^{2011}}\)
\(\Leftrightarrow\dfrac{1}{z^{2011}}=\dfrac{1}{z^{2011}}\left(luôn-đúng\right)\)
Tương tự với 2 TH còn lại .
\(\RightarrowĐCPM\)
\(\dfrac{x-3}{2011}+\dfrac{x-2}{2012}=\dfrac{x-2012}{2}+\dfrac{x-2011}{3}\)
\(\Leftrightarrow\dfrac{x-3}{2011}+\dfrac{x-2}{2012}-2=\dfrac{x-2012}{2}+\dfrac{x-2011}{3}-2\)
\(\Leftrightarrow\left(\dfrac{x-3}{2011}-1\right)+\left(\dfrac{x-2}{2012}-1\right)=\left(\dfrac{x-2012}{2}-1\right)+\left(\dfrac{x-2011}{3}-1\right)\)
\(\Leftrightarrow\dfrac{x-2014}{2011}+\dfrac{x-2014}{2012}-\dfrac{x-2014}{2}-\dfrac{x-2014}{3}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)
\(\Leftrightarrow x-2014=0\)
\(\Leftrightarrow x=0\)
\(\dfrac{x-3}{2011}+\dfrac{x-2}{2012}=\dfrac{x-2012}{2}+\dfrac{x-2011}{3}\)
<=>\(\dfrac{x-3}{2011}-1+\dfrac{x-2}{2012}-1=\dfrac{x-2012}{2}-1+\dfrac{x-2011}{3}-1\)
<=>\(\dfrac{x-2014}{2011}+\dfrac{x-2014}{2012}=\dfrac{x-2014}{2}+\dfrac{x-2014}{3}\)
<=>\(\dfrac{x-2014}{2011}+\dfrac{x-2014}{2012}-\dfrac{x-2014}{2}-\dfrac{x-2014}{3}=0\)
<=>\(\left(x-2014\right)\left(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)
vì 1/2011+1/2012-1/2-1/3 khác 0
=>x-2014=0<=>x=2014
vậy....................