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a) \(2017-\left|x-2017\right|=x\)
\(\left|x-2017\right|=2017-x\)
Mà \(\left|x-2017\right|\ge0\Leftrightarrow2017-x\ge0\)
\(\Leftrightarrow x-2017=2017-x\)
\(x+x=2017+2017\)
\(\Leftrightarrow x=2017\)
a)\(2017-\left|x-2017\right|=x\)
\(x-2017=2017-x\)
\(\Rightarrow x=2017\)
b)\(\left|2x-3\right|+\left|2x+4\right|=7\)
\(\left|2x-3\right|+\left|2x+4\right|=\left|2x-3\right|+2\left|x+2\right|\)
Giả sử \(\left|2x-3\right|\ge0\)thì dấu ''='' chỉ có thể xảy ra khi \(x=\frac{3}{2}\)
Vậy \(x=\frac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left|x+2\right|+\left|x-3\right|=7\)
Lập bảng xét dấu:
x | -2 3 |
x + 2 | - 0 + \(|\) + |
x - 3 | - \(|\) - 0 + |
* Nếu \(x< -2\) thì pttt:
\(-x-2-x+3=7\)
\(\Leftrightarrow-2x+1=7\)
\(\Leftrightarrow-2x=6\)
\(\Leftrightarrow x=-3\left(tm\right)\)
* Nếu \(-2\le x\le3\) thì pttt:
\(x+2-x+3=7\)
\(\Leftrightarrow5=7\) ( vô lí )
* Nếu \(x>3\) thì pttt:
\(x+2+x-3=7\)
\(\Leftrightarrow2x-1=7\)
\(\Leftrightarrow2x=8\)
\(\Leftrightarrow x=4\left(tm\right)\)
Vậy phương trình có tập nghiệm \(S=\left\{-3;4\right\}\)
b) \(\left|x+2\right|-6x=1\)
* Nếu \(x+2>0\Leftrightarrow x>2\) thì pttt:
\(x+2-6x=1\)
\(\Leftrightarrow-6x=-1\)
\(\Leftrightarrow x=1\left(ktm\right)\)
* Nếu \(x+2< 0\Leftrightarrow x< 2\) thì pttt:
\(-x-2-6x=1\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=-\dfrac{3}{7}\left(tm\right)\)
Vậy pt có tập nghiệm \(S=\left\{\dfrac{-3}{7}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Ta có :
\(A=\left|2x-2\right|+\left|2x-2017\right|=\left|2x-2\right|+\left|2017-2x\right|\ge\left|2x-2+2017-2x\right|=2015\)
Dấu "=" xảy ra \(\Leftrightarrow\left(2x-2\right)\left(2017-2x\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-2\ge0\\2017-2x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-2\le0\\2017-2x\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x\ge2\\2017\ge2x\end{matrix}\right.\\\left\{{}\begin{matrix}2x\le2\\2017\le2x\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge1\\\dfrac{2017}{2}\ge x\end{matrix}\right.\\\left\{{}\begin{matrix}x\le1\\\dfrac{2017}{2}\le x\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}1\le x\le\dfrac{2017}{2}\\x\in\varnothing\end{matrix}\right.\)
Vậy ...
b, Tương tự
c, \(\left|x+3\right|+\left|x+7\right|=4x\)
Mà \(\left\{{}\begin{matrix}\left|x+3\right|\ge0\\\left|x+7\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left|x-3\right|+\left|x+7\right|\ge0\)
\(\Leftrightarrow4x\ge0\)
\(\Leftrightarrow x\ge0\)
Với \(x\ge0\) ta có :
+) \(\left|x+3\right|=x+3\)
\(\left|x+7\right|=x+7\)
\(\Leftrightarrow\left|x+3\right|+\left|x+7\right|=x+3+x+7=4x\)
\(\Leftrightarrow2x+10=4x\)
\(\Leftrightarrow10=2x\)
\(\Leftrightarrow x=5\)
Vậy ..
B1b)
Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(B=\left|x-2\right|+\left|x-8\right|\)
\(B\ge\left|x-2\right|+\left|8-x\right|=6\)
Dấu "=" xảy ra khi \(\left(x-2\right)\left(8-x\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2\le0\\8-x\le0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2\ge0\\8-x\ge0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\le2\\x\ge8\end{matrix}\right.\left(C\right)}\\\left\{{}\begin{matrix}x\ge2\\x\le8\end{matrix}\right.\left(L\right)}\end{matrix}\right.\)
TH1: chọn, TH2: loại.
Vậy \(MIN_B=6\Leftrightarrow2\le x\le8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a ) \(\left(x-3\right).\left(x+2\right)+\left(x-1\right).\left(x+1\right)-\left(2x-1\right)x\)
\(=x.\left(x+2\right)-3.\left(x+2\right)+x.\left(x+1\right)-1.\left(x+1\right)-\left(2x-1\right)x\)
\(=x^2+2x-3x-6+x^2+x-x-1-2x^2+x\)
\(=-6\)
\(\RightarrowĐPCM\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a)\) \(\left|\left|3x-3\right|2x+\left(-1\right)^{2016}\right|=3x+2017^0\)
\(\Leftrightarrow\)\(\left|\left|3x-3\right|2x+1\right|=3x+1\)
Mà \(\left|\left|3x-3\right|2x+1\right|\ge0\) nên \(3x+1\ge0\)\(\Rightarrow\)\(x\ge1\)
\(\Leftrightarrow\)\(\left|3x-3\right|2x+1=3x+1\)
\(\Leftrightarrow\)\(\left|3x-3\right|=\frac{3x}{2x}\)
\(\Leftrightarrow\)\(\left|3x-3\right|=\frac{3}{2}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}3x-3=\frac{3}{2}\\3x-3=\frac{-3}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=\frac{9}{2}\\3x=\frac{3}{2}\end{cases}}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{9}{2}:3\\x=\frac{3}{2}:3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\left(tmx\ge1\right)\\x=\frac{1}{2}\left(loai\right)\end{cases}}}\)
Vậy \(x=\frac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, 2017-Ix-2017I=x
\(TH1:x>0\)
\(\Rightarrow2017-\left(x-2017\right)=x\)
\(\Rightarrow2017-x+2017\)
\(\Rightarrow4034-x=x\) ( loại )
\(TH2:x\le0\)
\(\Rightarrow2017-\left[-\left(x-2017\right)\right]=x\)
\(\Rightarrow2017-\left(-x+2017\right)=x\)
\(\Rightarrow2017+x-2017=x\)
\(\Rightarrow x+0=x\)
\(\Rightarrow x=x\) \(\left(x\in Z\right)\)