Câu 17.
\(\dfrac{x-3}{\sqrt{x}-\sqrt{3}}=\dfrac{\left(\sqrt{x}-\sqrt{3}\right)\left(\sqrt{x}+\sqrt{3}\right)}{\sqrt{x}-\sqrt{3}}=\sqrt{x}+3\)
⇒ Chọn D
Câu 18.
\(x^2-7=x^2-\left(\sqrt{7}\right)^2=\left(x-\sqrt{7}\right)\left(x+\sqrt{7}\right)\)
⇒ Chọn A
Câu 19.
Ta có:
\(A=-x+2\sqrt{x}+5\)
\(A=-\left(x-2\sqrt{x}-5\right)\)
\(A=-\left(x-2\sqrt{x}+1-6\right)\)
\(A=-\left[\left(\sqrt{x}-1\right)^2-6\right]\)
\(A=-\left(\sqrt{x}-1\right)^2+6\)
Mà: \(-\left(\sqrt{x}-1\right)^2\le0\forall x\) nên \(A=-\left(\sqrt{x}-1\right)^2+6\le6\)
Dấu "=" xảy ra:
\(-\left(\sqrt{x}-1\right)^2+6=6\)
\(\Leftrightarrow x=1\)
Vậy: \(A_{min}=6\) khi \(x=1\)
⇒ Chọn B
Câu 20.
Ta có:
\(B=x-4\sqrt{x}+3\)
\(B=\left(x-4\sqrt{x}+4\right)-1\)
\(B=\left(\sqrt{x}-2\right)^2-1\)
Mà: \(\left(\sqrt{x}-2\right)^2\ge0\forall x\) nên \(B=\left(\sqrt{x}-2\right)^2-1\ge-1\)
Dấu "=" xảy ra:
\(\left(\sqrt{x}-2\right)^2-1=-1\)
\(\Leftrightarrow x=4\)
Vậy: \(B_{min}=-1\) khi \(x=4\)
⇒ Chọn B