Tìm x :
I x-4I -x +2=0
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\(\left(\frac{1}{7}x-\frac{2}{7}\right).\left(\frac{-1}{5}x+\frac{3}{5}\right).\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
\(\hept{\begin{cases}\frac{1}{7}x-\frac{2}{7}=0\\\frac{-1}{5}x+\frac{3}{5}=0\\\frac{1}{3}x+\frac{4}{3}=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\x=3\\x=-4\end{cases}}}\)
KL
b, \(\left|\frac{5}{3}x\right|=\left|\frac{-1}{6}\right|\)
\(\left|\frac{5}{3}x\right|=\frac{1}{6}\)
\(\Rightarrow\orbr{\begin{cases}\frac{5}{3}x=\frac{1}{6}\\\frac{5}{3}x=\frac{-1}{6}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{10}\\x=\frac{-1}{10}\end{cases}}}\)
KL
c, \(\left|\frac{3}{4}x-\frac{3}{4}\right|-\frac{3}{4}=\left|\frac{-3}{4}\right|\)
\(\left|\frac{3}{4}x-\frac{3}{4}\right|-\frac{3}{4}=\frac{3}{4}\)
\(\Rightarrow\left|\frac{3}{4}x-\frac{3}{4}\right|=\frac{3}{2}\)
\(\Rightarrow\orbr{\begin{cases}\frac{3}{4}x-\frac{3}{4}=\frac{3}{2}\\\frac{3}{4}x-\frac{3}{4}=\frac{-3}{2}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{9}{4}\\x=\frac{-3}{4}\end{cases}}}\)
KL
Ta có : \(\hept{\begin{cases}\left|x-\frac{3}{4}\right|\ge0\forall x\\\left|\frac{2}{5}-y\right|\ge0\forall y\\\left|x-y+z\right|\ge0\forall x;y;z\end{cases}}\Leftrightarrow\left|x-\frac{3}{4}\right|+\left|\frac{2}{5}-y\right|+\left|x-y+z\right|\ge0\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-\frac{3}{4}=0\\\frac{2}{5}-y=0\\x-y+z=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{3}{4}\\y=\frac{2}{5}\\z=-\frac{7}{20}\end{cases}}\)
Vậy x = 3/4 ; y = 2/5 ; z = -7/20
\(\left|x-\frac{3}{4}\right|+\left|\frac{2}{5}-y\right|+\left|x-y+z\right|=0\)
Ta có: \(\left|x-\frac{3}{4}\right|;\left|\frac{2}{5}-y\right|;\left|x-y+z\right|\ge0\Rightarrow\left|x-\frac{3}{4}\right|+\left|\frac{2}{5}-y\right|+\left|x-y+z\right|\ge0\)
Mà \(\left|x-\frac{3}{4}\right|+\left|\frac{2}{5}-y\right|+\left|x-y+z\right|=0\)
\(\Rightarrow\hept{\begin{cases}x-\frac{3}{4}=0\\\frac{2}{5}-y=0\\x-y+z=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{3}{4}\\y=\frac{2}{5}\\\frac{3}{4}-\frac{2}{5}+z=0\Rightarrow z=\frac{-7}{20}\end{cases}}\)
Ta có:\(\left|x+5\right|+\left|x-4\right|=\left|x+5\right|+\left|4-x\right|>\left|x+5+4-x\right|\)-x| =9
Dấu ''='' xảy ra <=>(x+5)(4-x)>0
<=>-5<=x<=4
Vậy min(A)=9<=>-5<=x<=4
\(\left(x+20\right)^{100}+\left|y+4\right|=0.\)
\(Nx:\left(x+20\right)^{100}\ge0;\left|y+4\right|\ge0\)
\(\Rightarrow VT=0\Leftrightarrow\left(x+20\right)^{100}=0,\left|y+4\right|=0\)
\(\left(x+20\right)^{100}=0\Leftrightarrow x+20=0\Leftrightarrow x=-20\)
\(\left|y+4\right|=0\Leftrightarrow y+4=0\Leftrightarrow y=-4\)
Vậy x = -20 và y = -4
\(\left|x+\frac{1}{2}\right|+\left|y-\frac{3}{4}\right|+\left|z+1\right|=0\)
\(\Rightarrow\hept{\begin{cases}\left|x+\frac{1}{2}\right|=0\\\left|y-\frac{3}{4}\right|=0\\\left|z+1\right|=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=0-\frac{1}{2}\\y=0+\frac{3}{4}\\z=0-1\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{-1}{2}\\y=\frac{3}{4}\\z=-1\end{cases}}\)
|x - 4| - x + 2 = 0
=> |x - 4| = x - 2
+ Với x < 4 thì |x - 4| = 4 - x
Ta có: 4 - x = x - 2
=> -x - x = -2 - 4
=> -2x = -6
=> x = -6 : (-2) = 3, thỏa mãn x < 4
+ Với \(x\ge4\) thì |x - 4| = x - 4
Ta có: x - 4 = x - 2
=> x - x = -2 + 4
=> 0 = 2, vô lý
Vậy x = 3
|x-4|-x+2=0
\(\Leftrightarrow\left|x-4\right|=x-2\)
\(\Leftrightarrow x-4=\pm\left(x-2\right)\)
Nếu \(x-4=x-2\)
\(\Leftrightarrow-4=-2\left(loai\right)\)
Nếu \(x-4=-\left(x-2\right)\)
\(\Leftrightarrow x-4=-x+2\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\left(tm\right)\)
Vậy x=3