\(A=\frac{x+16}{\sqrt{x+3}}\)

Đặt \(\sqrt{x+3}=a>0\)

Ta có \(A=\frac{a^2+13}{a}\)

\(A=a+\frac{13}{a}\ge2\sqrt{a\cdot\frac{13}{a}}=2\sqrt{13}\)

Dấu "=" xảy ra \(\Leftrightarrow a=\sqrt{13}\Leftrightarrow\sqrt{x+3}=\sqrt{13}\Leftrightarrow x=10\)

\(\frac{1}{2x-\sqrt{x+3}}\) đề đây

bạn viết rõ đề đi

a: ĐKXĐ: x>=0; x<>1

\(P=\left(\frac{3}{x-1}+\frac{1}{\sqrt{x}+1}\right):\frac{1}{\sqrt{x}+1}\)

\(=\frac{3+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\left(\sqrt{x}+1\right)\)

\(=\frac{\sqrt{x}+2}{\sqrt{x}-1}\)

b: Thay \(x=3+2\sqrt2=\left(\sqrt2+1\right)^2\) vào P, ta được:

\(P=\frac{\sqrt{\left(\sqrt2+1\right)^2}+2}{\sqrt{\left(\sqrt2+1\right)^2}-1}=\frac{\sqrt2+1+2}{\sqrt2+1-1}=\frac{3+\sqrt2}{\sqrt2}=\frac{3\sqrt2+2}{2}\)

c: P<0

=>\(\frac{\sqrt{x}+2}{\sqrt{x}-1}<0\)

=>\(\sqrt{x}-1<0\)

=>\(\sqrt{x}<1\)

=>0<=x<1

Đề là: \(A=\sqrt{x-2\sqrt{x-3}}\) đúng ko em?

ĐKXĐ: \(x\ge3\)

\(A=\sqrt{x-3-2\sqrt{x-3}+1+2}=\sqrt{\left(\sqrt{x-3}-1\right)^2+2}\ge\sqrt{2}\)

\(A_{min}=\sqrt{2}\) khi \(x=4\)

ĐKXĐ: x>=3

\(A=\sqrt{x-2\sqrt{x-3}}\)

\(=\sqrt{x-3-2\sqrt{x-3}+1+2}\)

\(=\sqrt{\left(\sqrt{x-3}-1\right)^2+2}\ge\sqrt2\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi x-3=1

=>x=4