Bài 6:
1) \(A=\left(x-2\right)^2-5\)
Ta có \(\left(x-2\right)^2\ge0\forall x\Rightarrow\left(x-2\right)^2-5\ge-5\forall x\)
Dấu "=" xảy ra khi \(x=2\)
Vậy \(A_{min}=-5\) khi \(x=2\)
2) \(B=\left|x+\dfrac{1}{2}\right|+2\)
Ta có \(\left|x+\dfrac{1}{2}\right|\ge0\forall x\Rightarrow\left|x+\dfrac{1}{2}\right|+2\ge2\forall x\)
Dấu "=" xáy ra khi \(x=-\dfrac{1}{2}\)
Vậy \(B_{min}=2\) khi \(x=-\dfrac{1}{2}\)
Bài 6:
1: Ta có: \(\left(x-2\right)^2\ge0\forall x\)
\(\Leftrightarrow\left(x-2\right)^2-5\ge-5\forall x\)
Dấu '=' xảy ra khi x=2
2: Ta có: \(\left|x+\dfrac{1}{2}\right|\ge0\forall x\)
\(\Leftrightarrow\left|x+\dfrac{1}{2}\right|+2\ge2\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{1}{2}\)
Bài 5:
1: Để A nguyên thì \(x+3⋮x-1\)
\(\Leftrightarrow x-1\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(x\in\left\{2;0;3;-1;5;-3\right\}\)
2: Để B nguyên thì \(2x+1⋮x-2\)
\(\Leftrightarrow x-2\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{3;1;7;-3\right\}\)
