Tìm \(x\) \(\in\) Z sao cho
2011+2010+2009+.....+ \(x\) = 2011
K
Khách
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TN
10 tháng 6 2017
\(pt\Leftrightarrow\frac{1-\sqrt{x-2009}}{x-2009}+\frac{1-\sqrt{y-2010}}{y-2010}+\frac{1-\sqrt{z-2011}}{z-2011}=-\frac{3}{4}\)
\(\Leftrightarrow\left(\frac{1}{x-2009}-\frac{\sqrt{x-2009}}{x-2009}+\frac{1}{4}\right)+\left(\frac{1}{y-2010}-\frac{\sqrt{y-2010}}{y-2010}+\frac{1}{4}\right)+\left(\frac{1}{z-2011}-\frac{\sqrt{z-2011}}{z-2011}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{x-2009}-\frac{1}{\sqrt{x-2009}}+\frac{1}{4}\right)+\left(\frac{1}{y-2010}-\frac{1}{\sqrt{y-2010}}+\frac{1}{4}\right)+\left(\frac{1}{z-2011}-\frac{1}{\sqrt{z-2011}}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{\sqrt{x-2009}}-\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{y-2010}}-\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{z-2011}}-\frac{1}{2}\right)^2=0\)
Xảy ra khi \(\hept{\begin{cases}\frac{1}{\sqrt{x-2009}}=\frac{1}{2}\\\frac{1}{\sqrt{y-2010}}=\frac{1}{2}\\\frac{1}{\sqrt{z-2011}}=\frac{1}{2}\end{cases}}\Rightarrow\hept{\begin{cases}\sqrt{x-2009}=2\\\sqrt{y-2010}=2\\\sqrt{z-2011}=2\end{cases}}\Rightarrow\hept{\begin{cases}x=2013\\y=2014\\z=2015\end{cases}}\)
MA
1
HP
28 tháng 9 2021
\(\dfrac{x-1}{2011}+\dfrac{x-2}{2010}-\dfrac{x-3}{2009}=\dfrac{x-4}{2008}\)
<=> \(\left(\dfrac{x-1}{2011}-1\right)+\left(\dfrac{x-2}{2010}-1\right)-\left(\dfrac{x-3}{2009}-1\right)=\left(\dfrac{x-4}{2008}-1\right)\)
<=> \(\dfrac{x-2012}{2011}+\dfrac{x-2012}{2010}-\dfrac{x-2012}{2009}-\dfrac{x-2012}{2008}=0\)
<=> \(\left(x-2012\right)\left(\dfrac{1}{2011}+\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2008}\right)=0\)
<=> x - 2012 = 0
<=> x = 2012
BT
20 tháng 2 2017
Ta có: /x-2009/2009\(\ge\)0; (y-2010)2010=[(y-2010)1005]2 \(\ge\)0 và 2011/z-2011/\(\ge\)0
Tổng 3 số dương 0 khi và chỉ khi 3 số đó đều=0, khi đó dấu bằng xảy ra.
=> \(\hept{\begin{cases}Ix-2009I^{2009}=0\\\left(y-2010\right)^{2010}=0\\2011Iz-2011I=0\end{cases}}\)
=> x=2009; y=2010; z=2011
NM
Giaỉ phương trình:
\( \sqrt{x-2009}-1/{x-2009}+ \sqrt{y-2010}-1/y-2010+ \sqrt{z-2011}-1/z-2011 =3/4\)
2
NP
8 tháng 4 2015
−1x−2009+y−2010−−−−−−−√−1y−2010+z−2011−−−−−−−√−1z−2011=34
Ta có
x−2009−−−−−−−√−1x−2009+y−2010−−−−−−−√−1y−2010+z−2011−−−−−−−√−1z−2011=34⇔(1x−2009−−−−−−−√−12)2+(1y−2010−−−−−−−√−12)2+(1z−2011−−−−−−−√−12)2=0
⇒x=2013,y=2014,z=2015
G
14 tháng 5 2019
Đặt: \(\hept{\begin{cases}\sqrt{x-2009}=a\\\sqrt{y-2010}=b\\\sqrt{z-2011}=c\end{cases}}\)
Ta có: \(\frac{1}{a}-\frac{1}{a^2}+\frac{1}{b}-\frac{1}{b^2}+\frac{1}{c}-\frac{1}{c^2}-\frac{3}{4}=0\)
\(\Leftrightarrow\frac{1}{a^2}-\frac{1}{a}+\frac{1}{b^2}-\frac{1}{b}+\frac{1}{c^2}-\frac{1}{c}+\frac{3}{4}=0\)
\(\Leftrightarrow\left(\frac{1}{a^2}-\frac{1}{a}+\frac{1}{4}\right)+\left(\frac{1}{b^2}-\frac{1}{b}+\frac{1}{4}\right)+\left(\frac{1}{c^2}-\frac{1}{c}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{a}-\frac{1}{2}\right)^2+\left(\frac{1}{b}-\frac{1}{2}\right)^2+\left(\frac{1}{c}-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow a=b=c=\frac{1}{2}\)
Thay vào tìm x;y;z
24 tháng 9 2019
Đặt: \(\hept{\begin{cases}\sqrt{x-2009}=a\\\sqrt{y-2010}=b\\\sqrt{z-2011}=c\end{cases}}\)
Ta có: \frac{1}{a}-\frac{1}{a^2}+\frac{1}{b}-\frac{1}{b^2}+\frac{1}{c}-\frac{1}{c^2}-\frac{3}{4}=0a1−a21+b1−b21+c1−c21−43=0
\Leftrightarrow\frac{1}{a^2}-\frac{1}{a}+\frac{1}{b^2}-\frac{1}{b}+\frac{1}{c^2}-\frac{1}{c}+\frac{3}{4}=0⇔a21−a1+b21−b1+c21−c1+43=0
\Leftrightarrow\left(\frac{1}{a^2}-\frac{1}{a}+\frac{1}{4}\right)+\left(\frac{1}{b^2}-\frac{1}{b}+\frac{1}{4}\right)+\left(\frac{1}{c^2}-\frac{1}{c}+\frac{1}{4}\right)=0⇔(a21−a1+41)+(b21−b1+41)+(c21−c1+41)=0
\Leftrightarrow\left(\frac{1}{a}-\frac{1}{2}\right)^2+\left(\frac{1}{b}-\frac{1}{2}\right)^2+\left(\frac{1}{c}-\frac{1}{2}\right)^2=0⇔(a1−21)2+(b1−21)2+(c1−21)2=0
\Leftrightarrow a=b=c=\frac{1}{2}⇔a=b=c=21
Thay vào tìm x;y;z
22 tháng 2 2017
Ta có:
\(a_2^2=a_1.a_3;a_3^2=a_2.a_4;...;a^2_{2010}=a_{2009}.a_{2011}\)
\(\Rightarrow\frac{a_1}{a_2}=\frac{a_2}{a_3};\frac{a_2}{a_3}=\frac{a_3}{a_4};...;\frac{a_{2009}}{a_{2010}}=\frac{a_{2010}}{a_{2011}}\)
\(\Rightarrow\frac{a_1}{a_2}=\frac{a_2}{a_3}=\frac{a_3}{a_4}=...=\frac{a_{2010}}{a_{2011}}\)
\(\Rightarrow\frac{a_1^{2010}}{a_2^{2010}}=\frac{a_2^{2010}}{a_3^{2010}}=...=\frac{a_{2010}^{2010}}{a_{2011}^{2010}}=\frac{a_1^{2010}+a_2^{2010}+...+a_{2010}^{2010}}{a_2^{2010}+a_3^{2010}+...+a_{2011}^{2010}}\) (1)
Ta lại có:
\(\frac{a_1^{2010}}{a_2^{2010}}=\frac{a_1}{a_2}.\frac{a_1}{a_2}...\frac{a_1}{a_2}=\frac{a_1}{a_2}.\frac{a_2}{a_3}...\frac{a_{2009}}{a_{2010}}.\frac{a_{2010}}{a_{2011}}=\frac{a_1}{a_{2011}}\) (2)
Từ (1) và (2) ta suy ra
\(\frac{a_1^{2010}+a_2^{2010}+...+a_{2010}^{2010}}{a_2^{2010}+a_3^{2010}+...+a_{2011}^{2010}}=\frac{a_1}{a_{2011}}\)
22 tháng 2 2017
Ta có :
\(a_2^2=a_1.a_3\Rightarrow\frac{a_1}{a_2}=\frac{a_2}{a_3}\)
\(a^2_3=a_2.a_4\Rightarrow\frac{a_2}{a_3}=\frac{a_3}{a_4}\)
\(............\)
\(a^2_{2010}=a_{2009}.a_{2011}\Rightarrow\frac{a_{2009}}{a_{2010}}=\frac{a_{2010}}{a_{2011}}\)
\(\Rightarrow\frac{a_1}{a_2}=\frac{a_2}{a_3}=........=\frac{a_{2009}}{a_{2010}}=\frac{a_{2010}}{a_{2011}}\)
Đặt \(\frac{a_1}{a_2}=\frac{a_2}{a_3}=.......=\frac{a_{2010}}{a_{2011}}=k\)
\(\Rightarrow a_1=a_2.k\)
\(\Rightarrow a_1=a_3.k^2\)
\(\Rightarrow a_1=a_4.k^3\)
\(...............\)
\(\Rightarrow a_1=a_{2011}.k^{2010}\)
\(\Rightarrow\frac{a_1}{a_{2011}}=k^{2010}\) (1)
Ta có : \(k^{2010}=\left(\frac{a_1}{a_2}\right)^{2010}=\left(\frac{a_2}{a_3}\right)^{2010}=...=\left(\frac{a_{2010}}{a_{2011}}\right)^{2010}=\frac{a_1^{2010}}{a_2^{2010}}=\frac{a_2^{2010}}{a_3^{2010}}=....=\frac{a_{2010}^{2010}}{a_{2011}^{2010}}\)
\(=\frac{a_1^{2010}+a_2^{2010}+a_3^{2010}+....+a^{2010}_{2010}}{a_2^{2010}+a_3^{2010}+a_4^{2010}+....+a_{2011}^{2010}}\) ( theo TC DTSBN ) (2)
Từ (1) ; (2) \(\Rightarrow\frac{a_1^{2010}+a_2^{2010}+....+a_{2010}^{2010}}{a_2^{2010}+a_3^{2010}+....+a_{2011}^{2010}}=\frac{a_1}{a_{2011}}\) (đpcm)
\(2011+2010+2009+...+x=2011\)
\(2011+\left(2010+2009+...+x\right)=2011\)
\(2010+2009+...+x=2011-2011\)
\(2010+2009+...+x=0\)
\(\rightarrow\) Từ 2010->x có: x số
\(\rightarrow\) \(\frac{\left(x+2010\right).x}{2}=0\)
\(\left(x+2010\right).x=0.2\)
\(\left(x+2010\right).x=0\)
\(\Rightarrow x+2010=0\)
\(x=0+2010\)
\(x=-2010\)
\(V\text{ậy}\) \(x=-2010\)
Tâm Trần Hiếu sai rùi vì 0-2010 mới bằng -2010