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\(\left(x-17\right)^{x+1}-\left(x-17\right)^{x+11}=0\)
\(\Rightarrow\left(x-17\right)^{x+1}\left[1-\left(x-17\right)^{10}\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-17\right)^{x+1}=0\\\left(x-17\right)^{10}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-17=0\\x-17=1\\x-17=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=17\\x=18\\z=16\end{matrix}\right.\)
\(\left|x+\frac{11}{67}\right|+\left|x+\frac{2}{17}\right|+\left|x+\frac{4}{17}\right|=4x\)
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!
\(\left|x+\dfrac{11}{17}\right|\ge0\)
\(\left|x+\dfrac{2}{17}\right|\ge0\)
\(\left|x+\dfrac{4}{17}\right|\ge0\)
\(\Leftrightarrow\left|x+\dfrac{11}{17}\right|+\left|x+\dfrac{2}{17}\right|+\left|x+\dfrac{4}{17}\right|\ge0\)
\(\Leftrightarrow x+\dfrac{11}{17}+x+\dfrac{2}{17}+x+\dfrac{4}{17}=4x\)
\(3x+\left(\dfrac{11}{17}+\dfrac{2}{17}+\dfrac{4}{17}\right)=4x\)
\(3x+1=4x\Leftrightarrow4x-3x=1\Leftrightarrow x=1\)
Vậy...
Ta có: \(\left\{{}\begin{matrix}\left|x+\dfrac{11}{17}\right|\ge0\\\left|x+\dfrac{2}{17}\right|\ge0\\\left|x+\dfrac{4}{17}\right|\ge0\end{matrix}\right.\Leftrightarrow\left|x+\dfrac{11}{17}\right|+\left|x+\dfrac{2}{17}\right|+\left|x+\dfrac{4}{17}\right|\ge0\)
\(\Leftrightarrow4x\ge0\Leftrightarrow x\ge0\)
\(\Leftrightarrow x+\dfrac{11}{17}+x+\dfrac{2}{17}+x+\dfrac{4}{17}=4x\)
\(\Leftrightarrow3x+1=4x\)
\(\Leftrightarrow x=1\)
Vậy x = 1
\(\dfrac{93}{17}.\dfrac{1}{x}+\dfrac{-4}{7}.\dfrac{1}{x}+\dfrac{13}{70}=\dfrac{4}{11}\)
=>\(\dfrac{1}{x}.\left(\dfrac{93}{17}+\dfrac{-4}{7}\right)=\dfrac{137}{770}\)
=>\(\dfrac{1}{x}.\dfrac{583}{119}=\dfrac{137}{770}\)
=>\(\dfrac{1}{x}=\dfrac{2329}{64130}=>x=\dfrac{64130}{2329}\)
\(\Leftrightarrow\left(5+\dfrac{8}{17}-\dfrac{4}{17}\right):x=-\dfrac{250}{1309}\)
=>89/17:x=-250/1309
hay x=-6853/250
x + 17 = 17 + 11
x + 17 = 28
x = 28 - 17
x = 11
x+17=28
x =28-27
x =1