\(A=\left(x-2\right)^2+2\)
Có: \(\left(x-2\right)^2\ge0với\forall x\\ \Rightarrow\left(x-2\right)^2+2\ge0\\ \Leftrightarrow A\ge0\)
Dấu "=" xảy ra khi \(\left(x-2\right)^2=0\Leftrightarrow x=2\)
Vậy....
\(B=\left(2x+1\right)^4-1\)
Có: \(\left(2x+1\right)^4\ge0với\forall x\\ \Rightarrow\left(2x+1\right)^4-1\ge-1\\ \Leftrightarrow B\ge-1\)
Dấu "=" xảy ra khi \(\left(2x+1\right)^4=0\Leftrightarrow x=-\frac{1}{2}\)
VẬy...
\(C=\left(x^2-16\right)^2+\left|y-3\right|-2\)
Có: \(\left(x^2-16\right)^2\ge0với\forall x\\ \left|y-3\right|\ge0với\forall x\\ \Rightarrow\left(x^2-16\right)^2+\left|y-3\right|-2\ge2\\ \Leftrightarrow C\ge2\)
Dấu "=" xảy ra khi \(\left(x^2-16\right)^2=0\Leftrightarrow x\in\left\{\pm16\right\}\); \(\left|y-3\right|=0\Leftrightarrow y=3\)
Vậy...
\(D=\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2-10\)
Có: \(\left(x+2\right)^2\ge0với\forall x\\ \left(y-\frac{1}{5}\right)^2\ge0với\forall x\\ \Rightarrow\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2-10\ge-10\\ \Leftrightarrow D\ge-10\)
Dấu "=" xảy ra khi \(\left(x+2\right)^2=0\Leftrightarrow x=-2\);\(\left(y-\frac{1}{5}\right)^2=0\Leftrightarrow x=\frac{1}{5}\)
Vậy...
