Làm ko chắc lắm, có gì sai mong bạn thông cảm ._.
1/ ĐKXĐ: \(x\ge4\)
Với mọi \(x\ge4\) ta có: \(\sqrt{x-4}\ge0\Leftrightarrow\sqrt{x-4}-2\ge-2\)
Vậy min A = -2 khi x = 4
2/ ĐKXĐ: \(x\ge0\)
Ta có:
\(B=x-4\sqrt{x}+10=\left(\sqrt{x}\right)^2-2\cdot\sqrt{x}\cdot2+2^2+6\\ =\left(\sqrt{x}-2\right)^2+6\ge6\forall x\ge0\)
Vậy min B = 6 khi x = 4
3/ ĐKXĐ: \(x\ge0\)
Ta có:
\(C=x-\sqrt{x}=\left(\sqrt{x}\right)^2-2\cdot\sqrt{x}\cdot\frac{1}{2}+\left(\frac{1}{2}\right)^2-\frac{1}{4}\\ =\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\forall x\ge0\)
Vậy min C = \(-\frac{1}{4}\)khi x=\(\frac{1}{4}\)
4/ Ta có:
\(D=\sqrt{x^2-2x+4}+1\\ \Leftrightarrow\sqrt{x^2-2x+1+3}+1\\ \Leftrightarrow\sqrt{\left(x-1\right)^2+3}+1\)
Ta có:
\(\left(x-1\right)^2\ge0\Leftrightarrow\left(x-1\right)^2+3\ge3\\ \Leftrightarrow\sqrt{\left(x-1\right)^2+3}\ge\sqrt{3}\\ \Leftrightarrow\sqrt{\left(x-1\right)^2+3}+1\ge\sqrt{3}+1\forall x\)
Vậy min D = \(\sqrt{3}+1\) khi x = 1
