a: \(\left|x-\dfrac{1}{2}\right|>=0\forall x\)
=>\(2\left|x-\dfrac{1}{2}\right|>=0\forall x\)
=>\(A=2\left|x-\dfrac{1}{2}\right|+\dfrac{1}{2}>=\dfrac{1}{2}\forall x\)
Dấu '=' xảy ra khi \(x-\dfrac{1}{2}=0\)
=>\(x=\dfrac{1}{2}\)
b: \(x^2>=0\forall x;\left|y-2\right|>=0\forall y\)
Do đó: \(x^2+\left|y-2\right|>=0\forall x,y\)
=>\(B=x^2+\left|y-2\right|-5>=-5\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y-2=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=0\\y=2\end{matrix}\right.\)
c: \(C=\left|3x-5\right|+\left|3x-9\right|=\left|3x-5\right|+\left|9-3x\right|\)
=>\(C>=\left|3x-5+9-3x\right|=4\)
Dấu '=' xảy ra khi (3x-5)(3x-9)<=0
=>\(\dfrac{5}{3}< =x< =\dfrac{9}{3}\)
=>\(\dfrac{5}{3}< =x< =3\)
a.
Do \(2\left|x-\dfrac{1}{2}\right|\ge0;\forall x\)
Dấu "=" xảy ra khi \(x=\dfrac{1}{2}\)
Nên \(A\ge\dfrac{1}{2}\)
Vậy \(A_{min}=\dfrac{1}{2}\) khi \(x=\dfrac{1}{2}\)
b.
Do \(\left\{{}\begin{matrix}x^2\ge0\\\left|y-2\right|\ge0\end{matrix}\right.\) ;\(\forall x;y\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=2\end{matrix}\right.\)
Nên \(B\ge-5\)
Vậy \(B_{min}=-5\) khi \(\left\{{}\begin{matrix}x=0\\y=2\end{matrix}\right.\)
c.
\(C=\left|3x-5\right|+\left|3x-9\right|=\left|3x-5\right|+\left|9-3x\right|\ge\left|3x-5+9-3x\right|=4\)
Dấu "=" xảy ra khi \(\left(3x-5\right)\left(9-3x\right)\ge0\Rightarrow\dfrac{5}{3}\le x\le3\)
Vậy \(C_{min}=4\) khi \(\dfrac{5}{3}\le x\le3\)
