TL:

ĐKXĐ:\(\sqrt{x^2-1}>0\) 

\(\Leftrightarrow x^2-1>0\Leftrightarrow x^2>1\Leftrightarrow x>1\) 

Vậy...

DKXD :  X > 1

a: ĐKXĐ: x>=0; x<>1

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

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

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

\(=\frac{15\sqrt{x}-11-\left(3x+7\sqrt{x}-6\right)-\left(2x+\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

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

b: Khi \(x=4-2\sqrt3\) thì \(A=\frac{-5\cdot\sqrt{4-2\sqrt3}+2}{\sqrt{4-2\sqrt3}+3}=\frac{-5\left(\sqrt3-1\right)+2}{\sqrt3-1+3}\)

\(=\frac{-5\sqrt3+5+2}{\sqrt3+2}=\frac{-5\sqrt3+7}{\sqrt3+2}=\left(-5\sqrt3+7\right)\left(2-\sqrt3\right)\)

\(=-10\sqrt3+15+14-7\sqrt3=-17\sqrt3+29\)

c: \(A=\frac12\)

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

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

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

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

=>x=1/121(nhận)

e: \(A+5=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}+5=\frac{-5\sqrt{x}+2+5\sqrt{x}+15}{\sqrt{x}+3}=\frac{17}{\sqrt{x}+3}>0\forall x\) thỏa mãn ĐKXĐ

=>A>-5∀x thỏa mãn ĐKXĐ

a: ĐKXĐ: \(x\notin\left\{-1;3\right\}\)

b: \(A=\dfrac{3x\left(x+1\right)}{\left(x+1\right)\left(2x-6\right)}=\dfrac{3x}{2x-6}\)

Để A=0 thì 3x=0

hay x=0

Ta có

\(\sqrt{x^2-3x+7}\)

\(=\sqrt{x^2-2.x.\frac{3}{2}+\frac{9}{4}+\frac{19}{4}}\)

\(=\sqrt{\left(x-\frac{3}{2}\right)^2+\frac{19}{4}}\)

Vì \(\begin{cases}\left(x-\frac{3}{2}\right)^2\ge0\\\frac{19}{4}>0\end{cases}\)\(\Rightarrow\sqrt{\left(x-\frac{3}{2}\right)^2+\frac{19}{4}}>0\)

Vậy biểu thức có ngĩa với mọi x

Để phân thức A được xác định thì x khác -2 x khác 3 

Mk có tâm rút gọn hộ bạn luôn rồi nè =)) 

a, ĐK : \(x\ne-2;3\)

b, \(A=\frac{8-x}{\left(x+2\right)\left(x-3\right)}+\frac{2}{x+2}\)

\(=\frac{8-x}{\left(x+2\right)\left(x-3\right)}+\frac{2\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}=\frac{8-x+2x-6}{\left(x+2\right)\left(x-3\right)}\)

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

`a)ĐK:(x+1)(2x-6) ne 0`

`<=>(x+1)(x-3) ne 0`

`<=> x ne -1,x ne 3`

`b)C=(3x^2+3x)/((x+1)(2x-6))`

`=(3x(x+1))/((x+1)(2x-6))`

`=(3x)/(2x-6)`

`C=1`

`=>3x=2x-6`

`<=>x=-6(tm)`

Vậy `x=-6`