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Xét \( A = 1 + \dfrac{{2014}}{2} + \dfrac{{2015}}{3} + ... + \dfrac{{4023}}{{2011}} + \dfrac{{4024}}{{2012}}\\ \)
\(\Rightarrow A - 2012 = \left( {\dfrac{{2014}}{2} - 1} \right) + \left( {\dfrac{{2015}}{3} - 1} \right) + ... + \left( {\dfrac{{4024}}{{2012}} - 1} \right)\\ \Rightarrow A - 2012 = \dfrac{{2012}}{2} + \dfrac{{2012}}{3} + ... + \dfrac{{2012}}{{2012}}\\ \Rightarrow A - 2012 = 2012\left( {\dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{{2012}}} \right)\\ \Rightarrow A = 2012\left( {1 + \dfrac{1}{2} + ... + \dfrac{1}{{2012}}} \right)\\ \Rightarrow \left( {1 + \dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{{2012}}} \right)503x = 2012\left( {1 + ... + \dfrac{1}{{2012}}} \right)\\ \Rightarrow x = \dfrac{{2012}}{{503}} = 4 \)
a) (x-5)x+2015 - (x-5)x+2014 =0
(x-5)x+2014(x-5 -1) =0
+ x -5 =0 => x =5
+ x -6 =0 => x =6
Vậy x = 5 hoặc x =6
Cộng 1 vào mỗi ps
\(\frac{x+5}{2015}+1+\frac{x+6}{2014}+1+\frac{x+7}{2013}+1=0\)
\(\Rightarrow\frac{x+2020}{2015}+\frac{x+2020}{2014}+\frac{x+2020}{2013}=0\)
\(\Rightarrow\left[x+2020\right]\left[\frac{1}{2015}+\frac{1}{2014}+\frac{1}{2013}\right]=0\)
Mà \(\frac{1}{2015}+\frac{1}{2014}+\frac{1}{2013}\ne0\Rightarrow x+2020=0\)
=> x = -2020
( x - 2 )2012 + | y2 - 9 |2014 = 0 ( 1 )
vì ( x - 2 )2012 \(\ge\)0 ; | y2 - 9 |2014 \(\ge\)0 ( 2 )
Từ ( 1 ) và ( 2 ) \(\Rightarrow\hept{\begin{cases}\left(x-2\right)^{2012}=0\\\left|y^2-9\right|^{2014}=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x-2=0\\y^2-9=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=2\\y=3\end{cases}}\)
Vậy x = 2 ; y = 3
còn lại tương tự
Vì (x -2 )2012> hoặc =0 mà |y2 -9 |2014 > hoặc =0 nên để (x -2 )2012 + | y2 -9 |2014 =0 thì (x-2)2012 =0 và |y2 -9| =0
=>( x-2)=0 và y2-9=0
=>x=0 và y2=9
=>x=o và y=3 hoặc x= -3
\(\frac{x+5}{2015}+\frac{x+6}{2014}+\frac{x+7}{2013}+\frac{x+8}{2012}+\frac{x+9}{2011}+5=0\)
\(\Rightarrow1+\frac{x+5}{2015}+1+\frac{x+6}{2014}+1+\frac{x+7}{2013}+1+\frac{x+8}{2012}+1+\frac{x+9}{2011}=0\)
\(\Rightarrow\frac{x+2020}{2015}+\frac{x+2020}{2014}+\frac{x+2020}{2013}+\frac{x+2020}{2012}+\frac{x+2020}{2011}=0\)
\(\Rightarrow\left(x+2020\right)\left(\frac{1}{2015}+\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}\right)=0\)
\(\Rightarrow x+2020=0\)
\(\Rightarrow x=-2020\)
Study well
b) \(\frac{x-99}{5}+\frac{x-97}{7}=\frac{x-95}{9}+\frac{x-93}{11}\)
\(\Leftrightarrow\left(\frac{x-99}{5}-1\right)+\left(\frac{x-97}{7}-1\right)=\left(\frac{x-95}{9}-1\right)\)\(+\left(\frac{x-93}{11}-1\right)\)
\(\Leftrightarrow\frac{x-104}{5}+\frac{x-104}{7}-\frac{x-104}{9}-\frac{x-104}{11}=0\)
\(\Leftrightarrow\left(x-104\right)\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)=0\)
Mà \(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\ne0\)
\(\Rightarrow x-104=0\)
\(\Leftrightarrow x=104\)
Vậy ....
a) \(\frac{x+1945}{45}+\frac{x+1954}{54}=\frac{x+1975}{75}+\frac{x+1969}{69}\)
\(\Leftrightarrow\left(\frac{x+1945}{45}-1\right)+\left(\frac{x+1954}{54}-1\right)=\left(\frac{x+1975}{75}-1\right)\)\(+\left(\frac{x+1969}{69}-1\right)\)
\(\Leftrightarrow\frac{x+1900}{45}+\frac{x+1900}{54}-\frac{x+1900}{75}-\frac{x+1900}{69}=0\)
\(\Leftrightarrow\left(x+1900\right)\left(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}\right)=0\)
Mà \(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}\ne0\)
\(\Rightarrow x+1900=0\)
\(\Leftrightarrow x=-1900\)
Vậy ...