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Bài giải
\(\frac{2-x}{2015}+\frac{3-x}{1007}+\frac{4-x}{671}=\frac{2005-x}{2}\)
\(( \frac{2-x}{2015}+1 )+ (\frac{3-x}{1007}+2 )+ ( \frac{4-x}{671}+3 )=\frac{2005-x}{2}+6\)
\(\frac{2017-x}{2015}+\frac{2017-x}{1007}+\frac{2017-x}{671}=\frac{2017-x}{2}\)
\(\frac{2017-x}{2015}+\frac{2017-x}{1007}+\frac{2017-x}{671}-\frac{2017-x}{2}=0\)
\((2017-x)(\frac{1}{2015}+\frac{1}{1007}+\frac{1}{671}-\frac{1}{2})=0\)
Do \(\frac{1}{2015}+\frac{1}{1007}+\frac{1}{671}-\frac{1}{2}\ne0\)
\(\Rightarrow\text{ }2017-x=0\)
\(\Rightarrow\text{ }x=2017\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1-x}{2015}+\frac{2-x}{1007}+\frac{3-x}{671}=\frac{1992-x}{4}\)
\(\Rightarrow\frac{1-x}{2015}+1+\frac{2-x}{1007}+2+\frac{3-x}{671}+3=\frac{1992-x}{4}+6\)
\(\Rightarrow\frac{2016-x}{2015}+\frac{2016-x}{1007}+\frac{2016-x}{671}=\frac{2016-x}{4}\)
\(\Rightarrow\frac{2016-x}{2015}+\frac{2016-x}{1007}+\frac{2016-x}{671}-\frac{2016-x}{4}=0\)
\(\Rightarrow\left(2016-x\right)\left(\frac{1}{2015}+\frac{1}{1007}+\frac{1}{671}-\frac{1}{4}\right)=0\)
\(\Rightarrow2016-x=0\).Do \(\frac{1}{2015}+\frac{1}{1007}+\frac{1}{671}-\frac{1}{4}\ne0\)
\(\Rightarrow x=2016\)
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\(A=\left|2x-\dfrac{1}{3}\right|+1007\)
\(\left|2x-\dfrac{1}{3}\right|\ge0\)
\(\Rightarrow\left|2x-\dfrac{1}{3}\right|+1007\ge1007\)
Dấu "=" xảy ra khi:
\(\left|2x-\dfrac{1}{3}\right|=0\Rightarrow2x=\dfrac{1}{3}\Rightarrow x=\dfrac{1}{6}\)
\(\Rightarrow MIN_A=1007\) khi \(x=\dfrac{1}{6}\)
B tương tự
\(C=\left|2018-x\right|+\left|2017-x\right|\)
\(C=\left|2018-x\right|+\left|x-2017\right|\)
Áp dụng BĐT:
\(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)
\(\Rightarrow C\ge\left|2018-x+x-2017\right|\)
\(C\ge1\)
Dấu "=" xảy ra khi:
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}2018-x\ge0\Rightarrow x\le2018\\x-2017\ge0\Rightarrow x\ge2017\end{matrix}\right.\\\left\{{}\begin{matrix}2018-x< 0\Rightarrow x>2018\\x-2017< 0\Rightarrow x< 2017\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow2017\le x\le2018\)
D tương tự
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\(5-\frac{x}{2010}+4-\frac{x}{2011}+3-\frac{x}{2012}=6-\frac{x}{2009}+1-\frac{x}{1007}.\)
\(\left(5+4+3\right)-x.\frac{1}{2010}-x.\frac{1}{2011}-x\frac{1}{2012}=\left(6+1\right)-x.\frac{1}{2009}-x\frac{1}{1007}\)
\(12-x.\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}\right)=7-x.\left(\frac{1}{2009}+\frac{1}{1007}\right)\)
\(-x.\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}\right)+x.\left(\frac{1}{2009}+\frac{1}{1007}\right)=7-12\)
\(x.\left(\frac{-1}{2010}-\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2009}+\frac{1}{1007}\right)=-5\)
\(x=\frac{-5}{\frac{-1}{2010}-\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2009}+\frac{1}{1007}}\)
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( x - \(\sqrt{3}\) )\(^{2016}\) \(\ge\) 0 với mọi x . Kí hiệu là 1
(y\(^2\) - 3 )\(^{2018}\)\(\ge\) 0 với mọi y . Kí hiệu là 2
Từ 1 và 2 suy ra ( x - \(\sqrt{3}\) )\(^{2016}\) = 0 và (y\(^2\) - 3 )\(^{2018}\) = 0 . Kí hiệu là 3
Từ 3 suy ra x - \(\sqrt{3}\) = 0 suy ra x = \(\sqrt{3}\)
y\(^2\)- 3 = 0 suy ra y\(^2\) = 0 suy ra y =..........
2. Trên tử đặt 3 ra ngoài. Dưới mẫu đặt 11 ra ngoài rồi triệt tiêu.
3. 17^18 = (17^3)^6 = 4913^6
63^12 = (63^2)^6 = 3969 ^6
Vì 4913 > 3969 nên 4913^6 > 3969^6 hay 17^18>63^12
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{x-18}{2018}+\frac{x-14}{1007}+\frac{x-13}{671}=-6\)
\(\Rightarrow\frac{x-18}{2018}+1+\frac{x-14}{1007}+2+\frac{x-13}{671}+3=-6+6\)
\(\Rightarrow\frac{x-2000}{2028}+\frac{x-2000}{1007}+\frac{x-2000}{671}=0\)
\(\Rightarrow\left(x-2000\right)\left(\frac{1}{2018}+\frac{1}{1007}+\frac{1}{671}\right)=0\)
Vì \(\frac{1}{2018}+\frac{1}{1007}+\frac{1}{671}\ne0\)
=> x - 2000 = 0
=> x = 2000