\(2012\cdot|x-2011|+\left(x-2011\right)^2=2013\cdot|2011-x|\)
Tìm x
Những câu hỏi liên quan
Tìm x biết:
\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)\cdot x=\frac{2012}{1}+\frac{2011}{2}+\frac{2010}{3}+...+\frac{2}{2011}+\frac{1}{2012}\)
Tìm x:
2012×|x-2011|+ (x-2011)^2=2013×|2011-x|
2012 . | x - 2011| + (x-2011)2 = 2013 . | 2011 - x|
|x-2011|.|x-2011| + 2012 . | x - 2011| - 2013 . | 2011- x| =0
|x - 2011|.| x - 2011| + 2012 .| x - 2011| - 2013 | x - 2011| = 0
| x- 2011| .| x -2011| - | x - 2011| = 0
| x - 2011|. { | x - 2011| - 1} = 0
\(\left[{}\begin{matrix}\left|x-2011\right|=0\\\left|x-2011\right|-1=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2011\\x=2012\\x=2010\end{matrix}\right.\)
Kết luận x \(\in\) { 2010; 2011; 2012}
Đúng 2
Bình luận (0)
Tính
dfrac{1}{x-y}cdotsqrt{x^4left(x-yright)^2} (xy)
sqrt{27}cdotsqrt{48cdotleft(2-aright)^2} (a2)
left(sqrt{2012}+sqrt{2011}right)cdotleft(sqrt{2012}+sqrt{2011}right)
sqrt{dfrac{64x^2}{49left(y+1right)^2}} (x0;y-1)
sqrt{dfrac{121x^2}{144left(y+2right)}}left(x0;y -2right)
sqrt{dfrac{676x^3}{169xy^2}}left(x0;y 1right)
Đọc tiếp
Tính
\(\dfrac{1}{x-y}\cdot\sqrt{x^4\left(x-y\right)^2}\) (x>y)
\(\sqrt{27}\cdot\sqrt{48\cdot\left(2-a\right)^2}\) (a>2)
\(\left(\sqrt{2012}+\sqrt{2011}\right)\cdot\left(\sqrt{2012}+\sqrt{2011}\right)\)
\(\sqrt{\dfrac{64x^2}{49\left(y+1\right)^2}}\) (x<0;y>-1)
\(\sqrt{\dfrac{121x^2}{144\left(y+2\right)}}\left(x>0;y< -2\right)\)
\(\sqrt{\dfrac{676x^3}{169xy^2}}\left(x>0;y< 1\right)\)
a: \(=\dfrac{1}{x-y}\cdot x^2\cdot\left(x-y\right)=x^2\)
b: \(=\sqrt{27\cdot48}\cdot\left|a-2\right|=36\left(a-2\right)\)
c: \(=\left(\sqrt{2012}+\sqrt{2011}\right)^2\)
d: \(=\dfrac{8}{7}\cdot\dfrac{-x}{y+1}\)
e: \(=\dfrac{11}{12}\cdot\dfrac{x}{-y-2}=\dfrac{-11x}{12\left(y+2\right)}\)
Đúng 0
Bình luận (0)
Giải phương trình 0,05(\(\left(\frac{2x-2}{2011}+\frac{2x}{2012}+\frac{2x+2}{2013}\right)=3,3-\left(\frac{x-1}{2011}+\frac{x}{2012}+\frac{x+1}{2013}\right)\)
bài 2 Tìm GTNN của biểu thức A=\(\text{x^2-5x+y^2+xy-4y+2012}\)
Tìm x
2012.| x-2011| +\(\left(x-2011\right)^2\)=2013.|2011-x|
\(\frac{1}{4028}< \left(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot.......\cdot\frac{2011}{2012}\cdot\frac{2013}{2014}\right)^2< \frac{1}{2015}\)
Tìm x
2012.| x-2011| +\(\left(x-2011\right)^2\)=2013.|2011-x|
=>\(-\left|x-2011\right|+\left(x-2011\right)^2=0\)
\(\Leftrightarrow\left|x-2011\right|\left(\left|x-2011\right|-1\right)=0\)
\(\Leftrightarrow x\in\left\{2011;2012;2010\right\}\)
Đúng 0
Bình luận (0)
tìm x
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{2}{x\cdot\left(x+1\right)}=\frac{2011}{2013}\)
Cái này lớp 6 :
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+......+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)
\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\left(x+1\right)}\right)=\frac{2011}{2013}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
<=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4026}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{2}{4026}=\frac{1}{2013}\)
\(\Leftrightarrow x+1=2013\)
=> x = 2012
Đúng 1
Bình luận (0)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)
\(\Rightarrow2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2011}{2013}\)
\(\Rightarrow2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2011}{2013}\)
\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(\Rightarrow1-\frac{2}{x+1}=\frac{2011}{2013}\)
\(\Rightarrow\frac{2}{x+1}=1-\frac{2011}{2013}\)
\(\Rightarrow\frac{2}{x+1}=\frac{2}{2013}\)
\(\Rightarrow x+1=2013\)
\(\Rightarrow x=2013-1\)
\(\Rightarrow x=2012\)
Vậy \(x=2012\)
~ Ủng hộ nhé
Đúng 1
Bình luận (0)
\((x-1)/2013+(x-2)/2012+(x-3)/2011+(x-4)/2010+(x-5)/2009+(x-6)/2008\)
\(\frac{1}{x^2+9\cdot x+20}+\frac{1}{x^2+11\cdot x+30}+\frac{1}{x^2+13\cdot x+42}=\frac{1}{18}\)
\(\frac{x-1}{2013}+\frac{x-2}{2012}+\frac{x-3}{2011}=\frac{x-4}{2010}+\frac{x-5}{2009}+\frac{x-6}{2008}\) ( có lẽ đề như này )
\(\Leftrightarrow\frac{x-1}{2013}-1+\frac{x-2}{2012}-1+\frac{x-3}{2011}-1=\frac{x-4}{2010}-1+\frac{x-5}{2009}-1+\frac{x-6}{2008}-1\)
\(\Leftrightarrow\frac{x-2014}{2013}+\frac{x-2014}{2012}+\frac{x-2014}{2011}-\frac{x-2014}{2010}-\frac{x-2014}{2009}-\frac{x-2014}{2008}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
\(\Leftrightarrow x-2014=0\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\right)\)
\(\Leftrightarrow x=2014\)
...
Ta có : \(x^2+9x+20=x^2+4x+5x+20=\left(x+4\right)\left(x+5\right)\)
\(x^2+11x+30=x^2+5x+6x+30=\left(x+5\right)\left(x+6\right)\)
\(x^2+13x+42=x^2+6x+7x+42=\left(x+6\right)\left(x+7\right)\)
\(\Rightarrow Pt\Leftrightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\) (*)\(ĐKXĐ:x\ne-4;x\ne-5;x\ne-6;x\ne-7\)
(*) \(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow3.18=x^2+4x+7x+28\)
\(\Leftrightarrow x^2-2x+13x-26=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+13=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\left(tm\right)\\x=-13\left(tm\right)\end{cases}}}\)