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a: \(\Leftrightarrow\dfrac{3}{5}-\dfrac{8}{5}\left(\dfrac{2}{3}x-\dfrac{3}{2}\right)=\dfrac{-17}{5}\)
=>8/5(2/3x-3/2)=3/5+17/5=4
=>2/3x-3/2=4:8/5=4*5/8=5/2
=>2/3x=4
=>x=4:2/3=6
b: =>x^2-4x-5=x^2-7x
=>-4x-5=-7x
=>3x=5
=>x=5/3
![](https://rs.olm.vn/images/avt/0.png?1311)
a) | x - 1,5 | = 2
\(\Rightarrow\orbr{\begin{cases}x-1,5=2\\x-1,5=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=3,5\\x=0,5\end{cases}}\)
b) | x - 2 | = x
Nếu : x - 2 = x
\(\Leftrightarrow x-2-x=0\)
\(\Leftrightarrow x-x-2=0\)
\(\Leftrightarrow0-2=0\)( Loại )
Nếu : x - 2 = - x
\(\Leftrightarrow x-2+x=0\)
\(\Leftrightarrow x+x-2=0\)
\(\Leftrightarrow2x-2=0\)
\(\Rightarrow2x=2\)
\(\Rightarrow x=\frac{2}{2}=1\)
Vậy : x = 1
c) | x - 3,4 | + | 2,6 - x | = 0
\(\Rightarrow\hept{\begin{cases}x-3,4=0\\2,6-x=0\end{cases}}\Rightarrow\hept{\begin{cases}x=3,4\\x=2,6\end{cases}}\)
Vì : x chỉ có được 1 giá trị
\(\Rightarrow x\in\varnothing\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left|x-1,5\right|=2\\ \Rightarrow\left[{}\begin{matrix}x-1,5=2\\x-1,5=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3,5\\x=-0,5\end{matrix}\right.\)
Vậy \(x\in\left\{3,5;-0,5\right\}\)
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\(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\\ \Rightarrow\left|x+\frac{3}{4}\right|=\frac{1}{2}\\ \Rightarrow\left[{}\begin{matrix}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{2}\\x=-\frac{5}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{2};-\frac{5}{4}\right\}\)
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\(\left|x-2\right|=x\left(ĐK:x\ge0\right)\\ \Rightarrow\left[{}\begin{matrix}x-2=x\\x-2=-x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-x=2\\x+x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}0=2\left(\text{vô lý}\right)\\2x=2\end{matrix}\right.\\ \Rightarrow x=1\left(tmđk\right)\)
Vậy \(x=1\)
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\(\left|x-3,4\right|+\left|2,6-x\right|=0\\ \Rightarrow\left|x-3,4\right|=-\left|2,6-x\right|\)
Mà \(\left|2,6-x\right|\ge0\forall x\Rightarrow-\left|2,6-x\right|\le0\forall x\)
\(\Rightarrow\left|x-3,4\right|\le0\forall x\left(\text{vô lý}\right)\)
Vậy \(x\in\varnothing\)
a/ \(\left|x-1,5\right|=2\)
\(\Rightarrow\left[{}\begin{matrix}x-1,5=2\\x-1,5=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2+1,5=3,5\\x=-2+1,5=-0,5\end{matrix}\right.\)
b/ \(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)
\(\Rightarrow\left|x+\frac{3}{4}\right|=0+\frac{1}{2}=\frac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}-\frac{3}{4}=\frac{2}{4}-\frac{3}{4}=-\frac{1}{4}\\x=-\frac{1}{2}-\frac{3}{4}=\left(-\frac{2}{4}\right)+\left(-\frac{3}{4}\right)=-\frac{5}{4}\end{matrix}\right.\)
c/ \(\left|x-2\right|=x\)
\(\Rightarrow\left[{}\begin{matrix}x-2=x\\x-2=-x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-x=2\\x+x=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}0=2\left(vô-lý\right)\\2x=2\end{matrix}\right.\)
=> 2x = 2
=> x = 2 : 2 = 1
d/ \(\left|x-3,4\right|+\left|2,6-x\right|=0\)
Ta có: \(\left\{{}\begin{matrix}\left|x-3,4\right|\ge0\\\left|2,6-x\right|\ge0\end{matrix}\right.\)
=> Để \(\left|x-3,4\right|+\left|2,6-x\right|=0\) thì \(\left\{{}\begin{matrix}\left|x-3,4\right|=0\\\left|2,6-x\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-3,4=0\\2,6-x=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=0+3,4=3,4\\x=2,6-0=2,6\end{matrix}\right.\)
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x và y tỉ lệ nghịch
=>x1y1=x2y2
=>y1/x2=y2/x1
=>y1/5,6=y2/3,4=(5y1-3y2)/(5*5,6-3*3,4)=35,6/17,8=2
=>y1=11,2; y2=6,8
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