Bài 1:
a: \(\left(\left|x-1\right|-2\right)\left(-\dfrac{1}{2}x-2\right)=0\)
=>\(\left[{}\begin{matrix}\left|x-1\right|-2=0\\-\dfrac{1}{2}x-2=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}\left|x-1\right|=2\\-\dfrac{1}{2}x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=2\\x-1=-2\\x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\\x=-4\end{matrix}\right.\)
b: \(\left|2x+5\right|=\left|x-\dfrac{1}{2}\right|\)
=>\(\left[{}\begin{matrix}2x+5=x-\dfrac{1}{2}\\2x+5=-x+\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2x-x=-\dfrac{1}{2}-5\\2x+x=\dfrac{1}{2}-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{11}{5}\\3x=-\dfrac{9}{2}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-\dfrac{11}{5}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
c: \(x^2+4\left|x\right|=0\)
=>\(\left(\left|x\right|\right)^2+4\cdot\left|x\right|=0\)
=>\(\left|x\right|\left(\left|x\right|+4\right)=0\)
mà \(\left|x\right|+4>=4>0\forall x\)
nên |x|=0
=>x=0
d: |x|+3x=5(2)
TH1: x>=0
(2) sẽ trở thành:
x+3x=5
=>4x=5
=>\(x=\dfrac{5}{4}\left(nhận\right)\)
TH2: x<0
(2) sẽ trở thành:
-x+3x=5
=>2x=5
=>\(x=\dfrac{5}{2}\left(loại\right)\)
e: \(\left|2x+1\right|+\left|x+3\right|+\left|x+4\right|=8x\)(1)
TH1: x<-4
(1) sẽ trở thành: \(-2x-1-x-3-x-4=8x\)
=>8x=-4x-8
=>12x=-8
=>\(x=-\dfrac{8}{12}=-\dfrac{2}{3}\left(loại\right)\)
TH2: \(-4< =x< -3\)
(1) sẽ trở thành:
\(-2x-1-x-3+x+4=8x\)
=>8x=-2x
=>10x=0
=>x=0(loại)
TH3: \(-3< =x< -\dfrac{1}{2}\)
(1) sẽ trở thành:
\(-2x-1+x+3+x+4=8x\)
=>8x=6
=>\(x=\dfrac{6}{8}=\dfrac{3}{4}\left(loại\right)\)
TH4: \(x>=-\dfrac{1}{2}\)
(1) sẽ trở thành:
2x+1+x+3+x+4=8x
=>8x=4x+8
=>4x=8
=>x=2(nhận)
Bài 2:
a: \(\left|2x-6\right|>=0\forall x\)
=>\(A=\left|2x-6\right|+\dfrac{5}{3}>=\dfrac{5}{3}\forall x\)
Dấu '=' xảy ra khi 2x-6=0
=>2x=6
=>x=3
b: \(\left|2y-3\right|>=0\forall y;\left|3x+y\right|>=0\forall x,y\)
Do đó: \(\left|2y-3\right|+\left|3x+y\right|>=0\forall x,y\)
=>\(B=\left|2y-3\right|+\left|3x+y\right|+\dfrac{1}{2}>=\dfrac{1}{2}\forall x,y\)
dấu '=' xảy ra khi \(\left\{{}\begin{matrix}2y-3=0\\3x+y=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2y=3\\3x=-y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{3}{2}\\x=-\dfrac{y}{3}=-\dfrac{3}{2}:3=-\dfrac{1}{2}\end{matrix}\right.\)
