8.
a) \(\left|-3x\right|=4\Leftrightarrow\left[{}\begin{matrix}-3x=4\\-3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)
b) \(3+\left|x^2+1\right|=5\)
\(\Leftrightarrow\left|x^2+1\right|=2\)
\(\Leftrightarrow x^2+1=2\) (vì \(x^2+1>0;\forall x\))
\(\Leftrightarrow x^2=1\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
c) \(\left|x-\dfrac{1}{2}\right|-2=5-\left|x-\dfrac{1}{2}\right|\)
\(\Leftrightarrow2\left|x-\dfrac{1}{2}\right|=7\)
\(\Leftrightarrow\left|x-\dfrac{1}{2}\right|=\dfrac{7}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{7}{2}\\x-\dfrac{1}{2}=-\dfrac{7}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
d) \(\dfrac{\left|4-5x\right|+4}{5}=2\)
\(\Leftrightarrow\left|4-5x\right|+4=10\)
\(\Leftrightarrow\left|4-5x\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}4-5x=6\\4-5x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-2\\5x=10\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{5}\\x=2\end{matrix}\right.\)
9.
a) \(\left|2x-1\right|=\left|2x-5\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=2x-5\\2x-1=5-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-1=-5\left(\text{vô lí}\right)\\4x=6\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{3}{2}\)
b) \(\left|7-x\right|-\left|2-3x\right|=0\)
\(\Leftrightarrow\left|7-x\right|=\left|2-3x\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}7-x=2-3x\\7-x=3x-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x=-5\\4x=9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=\dfrac{9}{4}\end{matrix}\right.\)
c) \(\left|x-4\right|+\left|x^2-5x+4\right|=0\)
\(\Leftrightarrow\left|x-4\right|+\left|\left(x-1\right)\left(x-4\right)\right|=0\)
\(\Leftrightarrow\left|x-4\right|\left(1+\left|x-1\right|\right)=0\)
\(\Leftrightarrow\left|x-4\right|=0\) (vì \(1+\left|x-1\right|>0;\forall x\))
\(\Leftrightarrow x-4=0\)
\(\Leftrightarrow x=4\)
d) \(\left|\dfrac{x^2-x-2}{x+1}\right|-\left|x\right|=0;\left(x\ne-1\right)\)
\(\Leftrightarrow\left|\dfrac{x^2-x-2}{x+1}\right|=\left|x\right|\)
\(\Leftrightarrow\left|\dfrac{\left(x+1\right)\left(x-2\right)}{x+1}\right|=\left|x\right|\)
\(\Leftrightarrow\left|x-2\right|=\left|x\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=x\\x-2=-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-2=0\left(\text{vô lí}\right)\\2x=2\end{matrix}\right.\)
\(\Leftrightarrow x=1\left(tm\right)\)
10.
a) \(\left|x-6\right|=-5x+9\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=-5x+9\left(dk:x\ge6\right)\\6-x=-5x+9\left(dk:x< 6\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}6x=15\\4x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\left(ktm\right)\\x=\dfrac{3}{4}\left(tm\right)\end{matrix}\right.\)
b) \(\left|x+1\right|=x^2+x\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=x^2+x\left(dk:x\ge-1\right)\\-x-1=x^2+x\left(dk:x< -1\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2=1\\x^2+2x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pm1\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)
c) \(\left|x^2-2x\right|+4=2x\)
\(\Leftrightarrow\left|x^2-2x\right|=2x-4\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x=2x-4\left(dk:x\le0\text{ hoặc }x\ge2\right)\\2x-x^2=2x-4\left(dk:0< x< 2\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x+4=0\\x^2=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\\left[{}\begin{matrix}x=2\left(ktm\right)\\x=-2\left(tm\right)\end{matrix}\right.\end{matrix}\right.\)
d) \(\left|\dfrac{x^2+x+6}{x-1}\right|=x-2\left(dk:x\ne1\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x^2+x+6}{x-1}=x-2\left(dk:x>1\right)\\\dfrac{x^2+x+6}{1-x}=x-2\left(dk:x< 1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+6=x^2-3x+2\\x^2+x+6=-x^2+3x-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-4\\2x^2-2x+8=0\left(\text{vô lí vì }2x^2-2x+8>0;\forall x\right)\end{matrix}\right.\)
\(\Rightarrow x=-1\left(ktm\right)\)
$\text{#}Toru$
