\(\left(\frac{1}{7}x-\frac{2}{7}\right).\left(\frac{-1}{5}x+\frac{3}{5}\right).\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)

\(\hept{\begin{cases}\frac{1}{7}x-\frac{2}{7}=0\\\frac{-1}{5}x+\frac{3}{5}=0\\\frac{1}{3}x+\frac{4}{3}=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\x=3\\x=-4\end{cases}}}\)

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

lang nhang qua

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 \(\left|x-\frac{3}{4}\right|+\left|\frac{2}{5}-y\right|+\left|x-y+z\right|=0\)

\(\Rightarrow\hept{\begin{cases}x-\frac{3}{4}=0\\\frac{2}{5}-y=0\\x-y+z=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{3}{4}\\y=\frac{2}{5}\\\frac{3}{4}-\frac{2}{5}+z=0\Rightarrow z=\frac{-7}{20}\end{cases}}\)

\(4\left(x-7\right)-2\left(3x-8\right)=\left|-4\right|\cdot3-\left(-22\right)\)

\(4x-28-6x+16=34\)

\(-2x-12=34\)

\(-2x=46\)

\(x=-23\)

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

\(\left(x+20\right)^{100}\ge0,\left|y+4\right|\ge0\)

mà \(\left(x+20\right)^{100}+\left|y-4\right|=0\)

dấu "=" xảy ra khi \(\hept{\begin{cases}\left(x+20\right)^{100}=0\Rightarrow x=-20\\\left|y+4\right|=0\Rightarrow y+4=0\Rightarrow y=-4\end{cases}}\)

vậy x=-20, y=-4

\(\left|x+\frac{1}{2}\right|+\left|y-\frac{3}{4}\right|+\left|z+1\right|=0\)

\(\Rightarrow\hept{\begin{cases}\left|x+\frac{1}{2}\right|=0\\\left|y-\frac{3}{4}\right|=0\\\left|z+1\right|=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=0-\frac{1}{2}\\y=0+\frac{3}{4}\\z=0-1\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{-1}{2}\\y=\frac{3}{4}\\z=-1\end{cases}}\)