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\(\dfrac{93}{17}.\dfrac{1}{x}+\dfrac{-4}{7}.\dfrac{1}{x}+\dfrac{13}{70}=\dfrac{4}{11}\)
=>\(\dfrac{1}{x}.\left(\dfrac{93}{17}+\dfrac{-4}{7}\right)=\dfrac{137}{770}\)
=>\(\dfrac{1}{x}.\dfrac{583}{119}=\dfrac{137}{770}\)
=>\(\dfrac{1}{x}=\dfrac{2329}{64130}=>x=\dfrac{64130}{2329}\)
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a: =>|3x-5|=|x+2|
=>3x-5=x+2 hoặc 3x-5=-x-2
=>2x=7 hoặc 4x=3
=>x=7/2 hoặc x=3/4
b: \(\Leftrightarrow\left\{{}\begin{matrix}3x-5=0\\x+2=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
c: \(\Leftrightarrow\left|3x-5\right|=x-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=2\\\left(3x-5-x+2\right)\left(3x-5+x-2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=2\\\left(2x-3\right)\left(4x-7\right)=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
d: \(\dfrac{11}{2}\le\left|x\right|< \dfrac{17}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{11}{2}< =x< \dfrac{17}{2}\\-\dfrac{17}{2}< x< =-\dfrac{11}{2}\end{matrix}\right.\)
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\(\left|x+\frac{11}{67}\right|+\left|x+\frac{2}{17}\right|+\left|x+\frac{4}{17}\right|=4x\)
Có: \(\left\{{}\begin{matrix}\left|x+\frac{11}{67}\right|\\\left|x+\frac{2}{17}\right|\\\left|x+\frac{4}{17}\right|\end{matrix}\right.>0\forall x.\)
Do đó \(4x>0\Rightarrow x>0.\)
Lúc này ta có: \(\left(x+\frac{11}{67}\right)+\left(x+\frac{2}{17}\right)+\left(x+\frac{4}{17}\right)=4x\)
⇒ \(\left(x+x+x\right)+\left(\frac{11}{67}+\frac{2}{17}+\frac{4}{17}\right)=4x\)
⇒ \(3x+\frac{589}{1139}=4x\)
⇒ \(4x-3x=\frac{589}{1139}\)
⇒ \(1x=\frac{589}{1139}\)
⇒ \(x=\frac{589}{1139}:1\)
⇒ \(x=\frac{589}{1139}\left(TM\right)\)
Vậy \(x=\frac{589}{1139}.\)
Chúc bạn học tốt!
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\(\Leftrightarrow\left(5+\dfrac{8}{17}-\dfrac{4}{17}\right):x=-\dfrac{250}{1309}\)
=>89/17:x=-250/1309
hay x=-6853/250
\(\left(x-17\right)^{x+1}-\left(x-17\right)^{x+11}=0\)
\(\Rightarrow\left(x-17\right)^{x+1}\left[1-\left(x-17\right)^{10}\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-17\right)^{x+1}=0\\\left(x-17\right)^{10}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-17=0\\x-17=1\\x-17=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=17\\x=18\\z=16\end{matrix}\right.\)