a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)

b: Ta có: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)

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

\(=\dfrac{-3}{\sqrt{x}+3}\)

c: Thay \(x=4-2\sqrt{3}\) vào P, ta được:

\(P=\dfrac{-3}{\sqrt{3}-1+3}=\dfrac{-3}{2+\sqrt{3}}=-6+3\sqrt{3}\)

a: Để P nguyên thì \(-3⋮\sqrt{x}+3\)

\(\Leftrightarrow\sqrt{x}+3=3\)

hay x=0

\(A=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{x-\sqrt{x}}\right):\frac{1}{\sqrt{x}-1}\)ĐKXĐ : \(x>1\)

\(A=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\frac{1}{\sqrt{x}-1}\)

\(A=\frac{x+2}{\sqrt{x}\left(\sqrt{x}-1\right)}.\left(\sqrt{x}-1\right)\)

\(A=\frac{x+2}{\sqrt{x}}\)

\(a)\)\(A=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{x-\sqrt{x}}\right):\frac{1}{\sqrt{x}-1}\)

\(A=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\frac{1}{\sqrt{x}-1}\)

\(A=\left(\frac{\sqrt{x}}{\sqrt{x}-1}:\frac{1}{\sqrt{x}-1}\right)+\left(\frac{2}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{1}{\sqrt{x}-1}\right)\)

\(A=\sqrt{x}+\frac{2}{\sqrt{x}}\)

\(b)\) Áp dụng Cosi với hai số dương ta có : 

\(A=\sqrt{x}+\frac{2}{\sqrt{x}}\ge2\sqrt{\sqrt{x}.\frac{2}{\sqrt{x}}}=2\sqrt{2}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\sqrt{x}=\frac{2}{\sqrt{x}}\)

\(\Leftrightarrow\)\(x=2\)

Vậy GTNN của \(A\) là \(2\sqrt{2}\) khi \(x=2\)

Chúc bạn học tốt ~ 

PS : mới lớp 8 ko chắc nhé :v 

Theo câu a) thì  \(A=\frac{x+2}{\sqrt{x}}\)

Áp dụng BĐT AM-GM cho 2 số không âm\(x+2\ge2\sqrt{2x}\)

\(\Rightarrow A\ge\frac{2\sqrt{2x}}{\sqrt{x}}=2\sqrt{2}\). Vậy Min \(A=2\sqrt{2}\). Dấu "=" xảy ra <=> \(x=2\).

a: \(P=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2=x-\sqrt{x}+1\)

b: \(P=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)

Dấu '=' xảy ra khi x=1/4

a) ĐKXĐ: \(x\ne9\)

\(P=\frac{x\sqrt{x}+5\sqrt{x}-12-2\left(\sqrt{x}-3\right)^2-\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)

\(P=\frac{x\sqrt{x}+5\sqrt{x}-12-2x+12\sqrt{x}-18-x-5\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)

\(P=\frac{x\sqrt{x}-3x+12\sqrt{x}-36}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)

\(P=\frac{\left(\sqrt{x}-3\right)\left(x+12\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)

\(P=\frac{x+12}{\sqrt{x}+2}\)

b) Ta có: \(P=\frac{x+12}{\sqrt{x}+2}=\frac{x-4+16}{\sqrt{x}+2}=\sqrt{x}-2+\frac{16}{\sqrt{x}+2}\)

\(=\left(\sqrt{x}+2\right)+\frac{16}{\sqrt{x}+2}-4\)

\(\ge2\sqrt{\left(\sqrt{x}+2\right).\frac{16}{\sqrt{x}+2}}-4=4\)

P = 4 thì \(\left(\sqrt{x}+2\right)^2=16\Rightarrow\sqrt{x}=2\Rightarrow x=4\)

Vậy GTNN của P là 4 khi x = 4.