b tìm x,y,z biết 3(x -1 ) = 2(y-2) , 4 ( y -2 ) = 3( x-3) và 2x + 3y - z = 50
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Đề hình như hơi sai sai \(\left|x+2017\right|^{20}\)hay \(\left(x+2017\right)^{20}\)hay \(\left|x+2017\right|\)
Theo mk đề là: \(\left|x+2017\right|+\left|x+2018\right|=1\)
\(\left|x+2017\right|+\left|-x-2018\right|=1\)
+)Ta có: \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)nên
\(\left|x+2017\right|+\left|-x-2018\right|\ge\left|x+2017-x-2018\right|\)
\(\Rightarrow\left|x+2017\right|+\left|-x-2018\right|\ge\left|-1\right|\)
\(\Rightarrow\left|x+2017\right|+\left|-x-2018\right|\ge1\)
+)Dấu "=" xảy ra khi
\(\left(x+2017\right).\left(-x-2018\right)\ge0\)
\(\Leftrightarrow\hept{\begin{cases}x+2017\ge0\\-x-2018\ge0\end{cases}hoac\hept{\begin{cases}x+2017< 0\\-x-2018< 0\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ge-2017\\-x\ge2018\end{cases}hoac\hept{\begin{cases}x< -2017\\-x< 2018\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ge-2017\\x\le-2018\end{cases}hoac\hept{\begin{cases}x< -2017\\x>-2018\end{cases}}}\)
Vậy \(-2018< x< -2017\)(tm)
Chúc bạn học tốt
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Bài 2 :
a,\(\frac{x-1}{3}=2-\frac{x}{-2}\)
\(\Leftrightarrow\frac{x-1}{3}=\frac{-4-x}{-2}\Leftrightarrow-2x+2=-12-3x\Leftrightarrow x=-14\)
b, \(\frac{x-1}{x+5}=\frac{6}{7}\Leftrightarrow7x-7=6x+30\Leftrightarrow x=37\)
c, \(\frac{2x-1}{4}=\frac{4}{2x-1}\Leftrightarrow\left(2x-1\right)^2=16\)
\(\Leftrightarrow\left(2x-1\right)^2-4^2=0\Leftrightarrow\left(2x-5\right)\left(2x+3\right)=0\Leftrightarrow x=\frac{5}{2};-\frac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(3\left(x+\frac{1}{2}\right)-\frac{x^2}{3}\left(x+\frac{1}{2}\right)=0\)
=> \(\left(3-\frac{x^2}{3}\right)\left(x+\frac{1}{2}\right)=0\)
=> \(\orbr{\begin{cases}3-\frac{x^2}{3}=0\\x+\frac{1}{2}=0\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{x^2}{3}=3\\x=-\frac{1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x^2=9\\x=-\frac{1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\pm3\\x=-\frac{1}{2}\end{cases}}\)
Vậy \(x\in\left\{\pm3;-\frac{1}{2}\right\}\)là giá trị cần tìm
\(3\left(x+\frac{1}{2}\right)-\frac{x^2}{3}\left(x+\frac{1}{2}\right)=0\)
\(\left(x+\frac{1}{2}\right)\times\left(3-\frac{x^2}{3}\right)=0\)
\(\Rightarrow\hept{\begin{cases}x+\frac{1}{2}=0\\3-\frac{x^2}{3}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{1}{2}\\\frac{x^2}{3}=3\end{cases}\Rightarrow\hept{\begin{cases}x=-\frac{1}{2}\\x=3\end{cases}}}\)
vậy \(x=-\frac{1}{2}\)hoặc \(x=3\)