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Bài 3 :
\(\frac{x-1}{2016}+\frac{x-2}{2015}=\frac{x-3}{2014}+\frac{x-4}{2013}\)
\(\Leftrightarrow\)\(\left(\frac{x-1}{2016}-1\right)+\left(\frac{x-2}{2015}-1\right)=\left(\frac{x-3}{2014}-1\right)+\left(\frac{x-4}{2013}-1\right)\)
\(\Leftrightarrow\)\(\frac{x-1-2016}{2016}+\frac{x-2-2015}{2015}=\frac{x-3-2014}{2014}+\frac{x-4-2013}{2013}\)
\(\Leftrightarrow\)\(\frac{x-2017}{2016}+\frac{x-2017}{2015}=\frac{x-2017}{2014}+\frac{x-2017}{2013}\)
\(\Leftrightarrow\)\(\frac{x-2017}{2016}+\frac{x-2017}{2015}-\frac{x-2017}{2014}-\frac{x-2017}{2013}=0\)
\(\Leftrightarrow\)\(\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\right)=0\)
Vì \(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\ne0\)
Nên \(x-2017=0\)
\(\Rightarrow\)\(x=2017\)
Vậy \(x=2017\)
Chúc bạn học tốt ~
Bài 1 :
\(\left(8x-5\right)\left(x^2+2014\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}8x-5=0\\x^2+2014=0\end{cases}\Leftrightarrow\orbr{\begin{cases}8x=0+5\\x^2=0-2014\end{cases}}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}8x=5\\x^2=-2014\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{8}\\x=\sqrt{-2014}\left(loai\right)\end{cases}}}\)
Vậy \(x=\frac{5}{8}\)
Chúc bạn học tốt ~
Ta có:
\(\dfrac{x}{2014}+\dfrac{x+1}{2015}+\dfrac{x+2}{2016}+\dfrac{x+3}{2017}+\dfrac{x+4}{2018}=5\)
\(\Leftrightarrow\left(\dfrac{x}{2014}-1\right)+\left(\dfrac{x+1}{2015}-1\right)+\left(\dfrac{x+2}{2016}-1\right)+\left(\dfrac{x+3}{2017}-1\right)+\left(\dfrac{x+4}{2018}-1\right)=0\)\(\Leftrightarrow\dfrac{x-2014}{2014}+\dfrac{x-2014}{2015}+\dfrac{x-2014}{2016}+\dfrac{x-2014}{2017}+\dfrac{x-2014}{2018}=0\)\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}\right)=0\) (1)
Mà \(\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}>0\) (2)
Từ (1) và (2) => \(x-2014=0\) \(\Leftrightarrow x=2014\)
\(201^2=\left(200+1\right)^2=200^2+2.200.1+1^2=40000+400+1=40401\)
\(498^2=\left(500-2\right)^2=500^2-2.500.2+2^2=250000-2000+4=248004\)
b: \(=\dfrac{2014\cdot2015^2+2014\cdot2016-2016\cdot2015^2+2016\cdot2014}{2014\cdot2013^2-2014\cdot2012-2012\cdot2013^2-2012\cdot2014}\)
\(=\dfrac{2015^2\cdot\left(-2\right)+2\cdot\left(2015^2-1\right)}{2013^2\cdot\left(-2\right)-2\cdot\left(2013^2-1\right)}\)
\(=\dfrac{\left(-2\right)\cdot\left(2015^2-2015^2+1\right)}{\left(-2\right)\cdot\left(2013^2+2013^2-1\right)}=\dfrac{1}{2\cdot2013^2}\)
\(C=\dfrac{2014\left(2015^2+2016\right)-2016\left(2015^2-2014\right)}{2014\left(2013^2-2012\right)-2012\left(2013^2+2014\right)}\)
\(=\dfrac{2.2014.2016+2014.2015^2-2016.2015^2}{2014.2013^2-2012.2013^2-2.2012.2014}\)
\(=\dfrac{2.\left(2015+1\right)\left(2015-1\right)-2.2015^2}{2.2013^2-2.\left(2013+1\right)\left(2013-1\right)}\)
\(=\dfrac{2.\left(2015^2-1\right)-2.2015^2}{2.2013^2-2.\left(2013^2-1\right)}=\dfrac{-2}{2}=-1\)
\(\dfrac{x}{2012}+\dfrac{x+1}{2013}+\dfrac{x+2}{2014}+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}=5\)
\(\Leftrightarrow\dfrac{x}{2012}+\dfrac{x+1}{2013}+\dfrac{x+2}{2014}+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}-5=0\)
\(\Leftrightarrow\dfrac{x}{2012}-1+\dfrac{x+1}{2013}-1+\dfrac{x+2}{2014}-1+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}-1=0\)
\(\Leftrightarrow\dfrac{x-2012}{2012}+\dfrac{x-2012}{2013}+\dfrac{x-2012}{2014}+\dfrac{x-2012}{2015}+\dfrac{x-2012}{2016}=0\)
\(\Leftrightarrow\left(x-12\right).\left(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}\right)=0\)
\(\Leftrightarrow x-12=0\)
\(\Leftrightarrow x=12\)
A=20182+20162+20142+...+42 +22-(20172 +20152+20132+...+ 32 + 1)
A=(2018²-2017²)+(20162-20152)+(2014²-2013²)+...+(2² −1²)
A=2018+2017+2016+2015+2014+2013+...+2+1
\(A=\dfrac{2018\left(2018+1\right)}{2}=\text{2 037 171}\)
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