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x+1/2009 + x+ 1/2010 + x + 1/2011 = x+1/2012 + x + 1/2013 + x+1/2014
= x+1/2009 + x+1/2010 + x+1/2011 - x+1/2012 - x+1/2013 - x+1/2014 = 0
= (x+1) . ( 1/2009 + 1/2010 + 1/2011 - 1/2012 - 1/2013 - 1/2014) = 0
=x+ = 0 ( Vì 1/2009 + 1/2010 + 1/2011 - 1/2012 - 1/2013 - 1/2014 ≠ 0 )
x=-1
Vậy x=-1
a) \(\left(x-1\right)=\left(x-1\right)^3\)
\(\Leftrightarrow\left(x-1\right)^3-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\\left(x-1\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x\in\left\{2;0\right\}\end{cases}}\)
Vậy \(x\in\left\{0;1;2\right\}\)
b) \(x^3+x=0\)
\(\Leftrightarrow x\left(x^2+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=-1\left(L\right)\end{cases}}\)
Vậy x = 0
(x - 3)(2x + 6) = 0
=> \(\orbr{\begin{cases}x-3=0\\2x+6=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=3\\2x=-6\end{cases}}\)
=> \(\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
Vậy ...
(x-3)(2x+6)=0
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\2x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\2x=-6\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3\\x=-3\end{cases}}}.\)
Vậy x = 3 hoặc x = -3.
\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+50\right)=1325\)
\(\Leftrightarrow50x+\left(1+2+3+...+50\right)=1325\)
\(\Leftrightarrow50x+\frac{50.\left(50+1\right)}{2}=1325\)
\(\Leftrightarrow50x+1275=1325\)
\(\Leftrightarrow50x=50\)
\(\Leftrightarrow x=1\)
Vậy x =1
Tính tổng: 1 + 2+ 3 +... +50 = (50 + 1).50 : 2 = 1275
( x +1 ) + ( x + 2 ) + ( x + 3 ) + ... + ( x + 50 ) = 1325
( chú ý vế trái của 50 hạng tử )
( x + x + x +... + x ) + ( 1+ 2 + 3 +...+ 50 ) =1325
50 . x + 1275 =1325
50 . x = 1325 - 1275
50 . x = 50
x = 1
S có số số hạng là:(2014-2):1+1=2013(số hạng)
Mà 2013=1+2X1006 nên ta nhóm như sau:
\(S=2+\left[\left(-3\right)+4\right]+\left[\left(-5\right)+6\right]+...+\left[\left(-2013\right)+2014\right]\)
\(=2+1+1+...+1=2+1006\times1=1008\)
Vậy S=1008
Ta có :\(S=\) \(2+\left(-3\right)+4+\left(-5\right)+...+\left(-2013\right)+2014\)
\(=\left[2+\left(-3\right)\right]+\left[4+\left(-5\right)\right]+...+\left[2012+\left(-2013\right)\right]+2014\)
\(=\left(-1\right)+\left(-1\right)+...+\left(-1\right)+2014\)( có 2012 só (-1 ) )
\(=\) \(\left(-1\right).2012+2014\)
\(=\left(-2012\right)+2014\)
\(=2\)
Vậy \(S=2\)
B1. 2x + 3 + 22 = 72
=> 2x + 3 + 4 = 72
=> 2x + 3 = 72 - 4
=> 2x + 3 = 68
=> ko có gtri x
B2 : Ta có : A = 1 + 2 + 22 + 23 + 24 + 25 + 26 + ... + 22001 + 22002
= (1 + 2) + (22 + 23 + 24) + (25 + 26 + 27) + ... + (22000 + 22001 + 22002)
= 3 + 22.(1 + 2 + 22) + 25.(1 + 2 + 22 ) + ... + 22000 . (1 + 2 + 22)
= 3 + 22.7 + 25.7 + ... + 22000 . 7
= 3 + (22 + 25 + .... + 22000) . 7
=> Số dư của 7 là 3
\(\Rightarrow\)\(\frac{2}{6}\)+ \(\frac{2}{12}\)+ \(\frac{2}{20}\)+...+\(\frac{2}{x\left(x+1\right)}\)= \(\frac{2011}{2013}\)
\(\Rightarrow\)\(\frac{2}{2.3}\)+ \(\frac{2}{3.4}\)+ \(\frac{2}{4.5}\)+...+ \(\frac{2}{x\left(x+1\right)}\)= \(\frac{2011}{2013}\)
\(\Rightarrow\)\(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{4}\)+...+ \(\frac{1}{x}\)- \(\frac{1}{x+1}\)= \(\frac{2011}{2013}\): 2
\(\Rightarrow\)\(\frac{1}{2}\)- \(\frac{1}{x+1}\)= \(\frac{2011}{4026}\)
\(\Rightarrow\)\(\frac{1}{x+1}\)= \(\frac{1}{2}\)- \(\frac{2011}{4026}\)= \(\frac{1}{2013}\)
\(\Rightarrow\)\(x+1=2013\)
Hình như để như này :
\(\frac{x+1}{2014}+\frac{x+2}{2013}+\frac{x+3}{2012}+3=0\)
\(\left(\frac{x+1}{2014}+1\right)+\left(\frac{x+2}{2013}+1\right)+\left(\frac{x+3}{2012}+1\right)=0\)
\(\Leftrightarrow\frac{x+2015}{2014}+\frac{x+2015}{2013}+\frac{x+2015}{2012}=0\)
\(\Leftrightarrow\left(x+2015\right)\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}\right)=0\)
Do \(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}>0\Rightarrow x+2015=0\)
\(\Leftrightarrow x=-2015\)
Vậy \(x=-2015\)
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