Tinh\(\frac{x-1}{2010}+\frac{x-2}{2009}+.....+\frac{x-2010}{1}=2010\)
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28 tháng 4 2018
\(\frac{x+1}{2010}+\frac{x+2}{2009}+\frac{x+3}{2008}+...+\frac{x+2010}{1}=\left(-2010\right)\)
\(\Rightarrow\left(\frac{x+1}{2010}+1\right)+\left(\frac{x+2}{2009}+1\right)+...+\left(\frac{x+2010}{1}+1\right)=-2010+2010\)
\(\Rightarrow\frac{x+2011}{2010}+\frac{x+2011}{2009}+...+\frac{x+2011}{1}=0\)
\(\Rightarrow\left(x+2011\right)\left(1+\frac{1}{2}+...+\frac{1}{2009}+\frac{1}{2010}\right)=0\)
\(\Rightarrow x+2011=0\Leftrightarrow x=-2011\)
HG
1
HY
13 tháng 9 2016
\(\frac{x-1}{2010}+...+\frac{x-2010}{1}=2010\\ \Leftrightarrow\left(\frac{x-1}{2010}-1\right)+\left(\frac{x-2}{2009}-1\right)+...+\left(\frac{x-2010}{1}-1\right)=0\)
\(\Leftrightarrow\frac{x-2011}{2010}+\frac{x-2011}{2009}+...+\frac{x-2011}{1}=0\)
\(\Leftrightarrow\left(x-2011\right)\left(\frac{1}{2010}+\frac{1}{2009}+...+1\right)=0\)
\(\Leftrightarrow x-2011=0\) (Vì 1/2010 +1/2009 + ... +1 khác 0 )
\(\Leftrightarrow x=2011\)
LP
1
18 tháng 2 2017
\(\frac{x-1}{2010}+\frac{x-2}{2009}+....+\frac{x-2010}{2}=2010\)
\(\Leftrightarrow\left(\frac{x-1}{2010}-1\right)+\left(\frac{x-2}{2009}-1\right)+...+\left(\frac{x-2010}{1}-1\right)=0\)
\(\Leftrightarrow\frac{x-2011}{2010}+\frac{x-2011}{2009}+....+\frac{x-2011}{1}=0\)
\(\Leftrightarrow\left(x-2011\right)\left(\frac{1}{2010}+\frac{1}{2009}+...+1\right)=0\)
\(\Rightarrow x-2011=0\Rightarrow x=2011\)
14 tháng 7 2017
1. \(\left(2x-1\right)^3+\left(x+2\right)^3=\left(3x+1\right)^3\)
\(\Rightarrow8x^3-12x^2+6x-1+x^3+6x^2+12x+8=27x^3+27x^2+9x+1\)
\(\Rightarrow-18x^3-33x^2+9x+6=0\)\(\Rightarrow\left(x+2\right)\left(-18x^2+3x+3\right)=0\)
\(\Rightarrow\left(x+2\right)\left(2x-1\right)\left(-9x-3\right)=0\Rightarrow\orbr{\begin{cases}x=-2\\x=\frac{1}{2};x=-\frac{1}{3}\end{cases}}\)
Vậy \(x=-2;x=\frac{1}{2};x=-\frac{1}{3}\)
2. \(\frac{x-1988}{15}+\frac{x-1969}{17}+\frac{x-1946}{19}+\frac{x-1919}{21}=10\)
\(\Rightarrow\left(\frac{x-1988}{15}-1\right)+\left(\frac{x-1969}{17}-2\right)+\left(\frac{x-1946}{19}-3\right)+\left(\frac{x-1919}{21}-4\right)=0\)
\(\Rightarrow\frac{x-2003}{15}+\frac{x-2003}{17}+\frac{x-2003}{19}+\frac{x-2003}{21}=0\)
\(\Rightarrow x-2003=0\)do \(\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\ne0\)
Vậy \(x=2003\)
3. Đặt \(\hept{\begin{cases}2009-x=a\\x-2010=b\end{cases}}\)
\(\Rightarrow\frac{a^2+ab+b^2}{a^2-ab+b^2}=\frac{19}{49}\Rightarrow49a^2+49ab+49b^2=19a^2-19ab+19b^2\)
\(\Rightarrow30a^2+68ab+30b^2=0\Rightarrow\left(5a+3b\right)\left(3a+5b\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5a=-3b\\3a=-5b\end{cases}}\)
Với \(5a=-3b\Rightarrow5\left(2009-x\right)=-3\left(x-2010\right)\)
\(\Rightarrow-2x=-4015\Rightarrow x=\frac{4015}{2}\)
Với \(3a=-5b\Rightarrow3\left(2009-x\right)=-5\left(x-2010\right)\)
\(\Rightarrow2x=4023\Rightarrow x=\frac{4023}{2}\)
Vậy \(x=\frac{4023}{2}\)hoặc \(x=\frac{4015}{2}\)