a) \(A=\sqrt[]{x-4}-2\left(x\ge4\right)\)
Vì \(\sqrt[]{x-4}\ge0,\forall x\ge4\)
\(\Leftrightarrow A=\sqrt[]{x-4}-2\ge-2,\forall x\ge4\)
\(\Leftrightarrow GTNN\left(A\right)=-2\left(khi.x=4\right)\)
b) \(B=x-4\sqrt[]{x}+10\left(x\ge0\right)\)
\(\Leftrightarrow B=x-4\sqrt[]{x}+4+6\)
\(\Leftrightarrow B=\left(\sqrt[]{x}-2\right)^2+6\)
mà \(\left(\sqrt[]{x}-2\right)^2\ge0,\forall x\ge0\)
\(\Leftrightarrow B=\left(\sqrt[]{x}-2\right)^2+6\ge6,\forall x\ge0\)
\(\Leftrightarrow GTNN\left(B\right)=6,\left(khi.x=4\right)\)
c) \(C=x-\sqrt[]{x}\left(x\ge0\right)\)
\(\Leftrightarrow C=x-\sqrt[]{x}+\dfrac{1}{4}-\dfrac{1}{4}\)
\(\Leftrightarrow C=\left(\sqrt[]{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)
mà \(\left(\sqrt[]{x}-\dfrac{1}{2}\right)^2\ge0,\forall x\ge0\)
\(\Leftrightarrow C=\left(\sqrt[]{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4},\forall x\ge0\)
\(\Leftrightarrow GTNN\left(C\right)=-\dfrac{1}{4}\left(khi.x=\dfrac{1}{4}\right)\)
d) \(D=\sqrt[]{x^2-2x+4}+1\)
\(\Leftrightarrow D=\sqrt[]{x^2-2x+1+3}+1\)
\(\Leftrightarrow D=\sqrt[]{\left(x-1\right)^2+3}+1\)
mà \(\left(x-1\right)^2\ge0,\forall x\in R\)
\(\Leftrightarrow D=\sqrt[]{\left(x-1\right)^2+3}+1\ge\sqrt[]{3}+1,\forall x\in R\)
\(\Leftrightarrow GTNN\left(D\right)=\sqrt[]{3}+1\left(khi.x=1\right)\)
