a: \(A=\left(2x-1\right)^2-3\left|2x-1\right|+2\)
\(=\left(\left|2x-1\right|\right)^2-2\cdot\left|2x-1\right|\cdot\frac32+\frac94-\frac14\)
\(=\left(\left|2x-1\right|-\frac32\right)^2-\frac14\ge-\frac14\forall x\)
Dấu '=' xảy ra khi \(\left|2x-1\right|-\frac32=0\)
=>\(\left|2x-1\right|=\frac32\)
=>\(\left[\begin{array}{l}2x-1=\frac32\\ 2x-1=-\frac32\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=\frac52\\ 2x=-\frac12\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac54\\ x=-\frac14\end{array}\right.\)
b: \(B=\left(0,5x^2+x\right)^2-3\cdot\left|0,5x^2+x\right|\)
\(=\left(\left|0,5x^2+x\right|\right)^2-3\left|0,5x^2+x\right|+\frac94-\frac94\)
\(=\left(\left|\frac12x^2+x\right|-\frac32\right)^2-\frac94\ge-\frac94\forall x\)
Dấu '=' xảy ra khi \(\left|\frac12x^2+x\right|-\frac32=0\)
=>\(2\cdot\left|\frac12x^2+x\right|-3=0\)
=>\(\left|x^2+2x\right|-3=0\)
=>\(\left|x^2+2x\right|=3\)
=>\(\left[\begin{array}{l}x^2+2x=3\\ x^2+2x=-3\end{array}\right.\Rightarrow\left[\begin{array}{l}x^2+2x-3=0\\ x^2+2x+3=0\left(vôlý\right)\end{array}\right.\)
=>\(x^2+2x-3=0\)
=>(x+3)(x-1)=0
=>\(\left[\begin{array}{l}x+3=0\\ x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-3\\ x=1\end{array}\right.\)
c: \(C=\left|x-7\right|+\left|x+5\right|\)
=>\(C=\left|7-x\right|+\left|x+5\right|\ge\left|7-x+x+5\right|=12\forall x\)
Dấu '=' xảy ra khi (x-7)(x+5)<=0
=>-5<=x<=7
d: \(D=\left|x^2-x+1\right|+\left|x^2-x-2\right|\)
\(=\left|x^2-x+1\right|+\left|-x^2+x+2\right|\ge\left|x^2-x+1-x^2+x+2\right|=3\forall x\)
Dấu '=' xảy ra khi \(\left(x^2-x+1\right)\left(x^2-x-2\right)\le0\)
=>\(x^2-x-2\le0\)
=>(x-2)(x+1)<=0
=>-1<=x<=2
