Bài 1:

1: \(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)

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

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

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

2: Thay x=1/9 vào A, ta được:

\(A=\left(-5\cdot\sqrt{\frac19}+2\right):\left(\sqrt{\frac19}+3\right)=\left(-5\cdot\frac13+2\right):\left(\frac13+3\right)\)

\(=\frac13:\frac{10}{3}=\frac{1}{10}\)

Thay \(x=4-2\sqrt3=\left(\sqrt3-1\right)^2\) vào A, ta được:

\(A=\frac{-5\cdot\sqrt{\left(\sqrt3-1\right)^2}+2}{\sqrt{\left(\sqrt3-1\right)^2}+3}=\frac{-5\left(\sqrt3-1\right)+2}{\sqrt3-1+3}\)

\(=\frac{-5\sqrt3+7}{2+\sqrt3}=\left(7-5\sqrt3\right)\left(2-\sqrt3\right)=14-7\sqrt3-10\sqrt3+15=29-17\sqrt3\)

3: \(A=\frac12\)

=>\(\frac{-5\sqrt{x}+2}{\sqrt{x}+3}=\frac12\)

=>\(-10\sqrt{x}+4=\sqrt{x}+3\)

=>\(-11\sqrt{x}=3-4=-1\)

=>\(\sqrt{x}=\frac{1}{11}\)

=>\(x=\frac{1}{121}\) (nhận)

4: A>-1

=>A+1>0

=>\(\frac{-5\sqrt{x}+2+\sqrt{x}+3}{\sqrt{x}+3}>0\)

=>\(-4\sqrt{x}+5>0\)

=>\(-4\sqrt{x}>-5\)

=>\(\sqrt{x}<\frac54\)

=>0<=x<25/16

Kết hợp DKXĐ, ta được: 0<=x<25/16 và x<>1

5: A nguyên khi \(-5\sqrt{x}+2\) ⋮\(\sqrt{x}+3\)

=>\(-5\sqrt{x}-15+17\) ⋮\(\sqrt{x}+3\)

=>17⋮\(\sqrt{x}+3\)

mà \(\sqrt{x}+3\ge3\forall x\) thỏa mãn ĐKXĐ

nên \(\sqrt{x}+3=17\)

=>\(\sqrt{x}=14\)

=>x=196(nhận)

6: \(A+5=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}+5=\frac{-5\sqrt{x}+2+5\sqrt{x}+15}{\sqrt{x}+3}=\frac{17}{\sqrt{x}+3}>0\)

=>A>-5

Bài 2:

ĐKXĐ: x>=0; x<>4

1: \(B=\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{x+5}{x-\sqrt{x}-2}\)

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

\(=\frac{x-3\sqrt{x}+2-x-4\sqrt{x}-3-x-5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}=\frac{-x-7\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{-\left(\sqrt{x}+6\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}=\frac{-\sqrt{x}-6}{\sqrt{x}-2}\)

2: Khi x=9 thì \(B=\frac{-3-6}{3-2}=-9\)

Khi \(x=9-4\sqrt5=\left(\sqrt5-2\right)^2\) vào B, ta được:

\(B=\frac{-\sqrt{\left(\sqrt5-2\right)^2}-6}{\sqrt{\left(\sqrt5-2\right)^2}-2}=\frac{-\sqrt5+2-6}{\sqrt5-2-2}=\frac{-\sqrt5-4}{\sqrt5-4}\)

\(=\frac{4+\sqrt5}{4-\sqrt5}=\frac{\left(4+\sqrt5\right)^2}{\left(4-\sqrt5\right)\left(4+\sqrt5\right)}=\frac{16+8\sqrt5+5}{16-5}=\frac{21+8\sqrt5}{11}\)

3: B=-1

=>\(-\sqrt{x}-6=-1\left(\sqrt{x}-2\right)=-\sqrt{x}+2\)

=>-6=2(loại)

4: B<-1

=>B+1<0

=>\(\frac{-\sqrt{x}-6+\sqrt{x}-2}{\sqrt{x}-2}<0\)

=>\(-\frac{8}{\sqrt{x}-2}<0\)

=>\(\sqrt{x}-2>0\)

=>\(\sqrt{x}>2\)

=>x>4