a) ĐKXĐ: \(x\ge0;x\ne9\)

\(B=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{3-11\sqrt{x}}{9-x}\)

\(B=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}-\frac{3-11\sqrt{x}}{x-9}\)

\(B=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)-3+11\sqrt{x}}{x-9}\)

\(B=\frac{2x-6+x+4\sqrt{x}+3-3+11\sqrt{x}}{x-9}\)

\(B=\frac{3x-6+15\sqrt{x}}{x-9}\)

a: ĐKXĐ: x>=0; x<>1

\(A=\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{-\left(5\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)

b: \(A-\dfrac{2}{3}=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{2}{3}\)

\(=\dfrac{-15\sqrt{x}+6-2\sqrt{x}-6}{3\left(\sqrt{x}+3\right)}=\dfrac{-17\sqrt{x}}{3\left(\sqrt{x}+3\right)}< =0\)

Do đó: A<=2/3

a)ĐKXĐ:x>=0;x khác 9

A=[\(\frac{\sqrt{x}}{\sqrt{x}-3}\) - \(\frac{3\sqrt{x}+9}{x-9}\)+ \(\frac{2\sqrt{x}}{\sqrt{x}+3}\)] \(\div\) [\(\frac{2\sqrt{x}-2}{\sqrt{x}-3}\)-1]

 A=[\(\frac{\sqrt{x}\left(\sqrt{x}-3\right)-3\sqrt{x}-9+2\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}\)] \(\div\) [\(\frac{\left(2\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-x+9}{x-9}\)]

A=[\(\frac{3x-12\sqrt{x}-9}{x-9}\)].[\(\frac{x-9}{x-4\sqrt{x}+3}\)]

A=\(\frac{3x-12\sqrt{x}-9}{x-4\sqrt{x}+3}\)

a) DK : x > 0; x khác 1

 \(P=\sqrt{x}\left(\sqrt{x}-1\right)-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\)

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

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

<=> \(xQ-\left(Q+2\right)\sqrt{x}+Q=0\)(1)

TH1: Q = 0 => x = 0 loại

TH2: Q khác 0

(1) là phương trình bậc 2 với tham số Q ẩn x.

(1) có nghiệm <=> \(\left(Q+2\right)^2-4Q^2\ge0\)

<=> \(-3Q^2+4Q+4\ge0\)

<=> \(-\frac{2}{3}\le Q\le2\)

Vì Q nguyên và khác 0 nên Q =  1 hoặc Q = 2

Với Q = 1 => \(x-3\sqrt{x}+1=0\)

<=> \(\sqrt{x}=\frac{3}{2}\pm\frac{\sqrt{5}}{2}\)----> Tìm được x 

Với Q = 2 => \(2x-4\sqrt{x}+1=0\Leftrightarrow\sqrt{x}=1\pm\frac{1}{\sqrt{2}}\)-----> tìm đc x.

Tự làm tiếp nhé! Kiểm tra lại đề bài câu b.