a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

b: Thay x=9 vào A, ta được:

\(A=\dfrac{3-1}{3+1}=\dfrac{1}{2}\)

c: Ta có: P=AB

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\left(\dfrac{\sqrt{x}+3}{\sqrt{x}+1}+\dfrac{4}{\sqrt{x}-1}+\dfrac{5-x}{x-1}\right)\)

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

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\cdot\dfrac{6\sqrt{x}+6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{6}{\sqrt{x}+1}\)

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

Ta có: \(A=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)

\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

Thay \(x=6-2\sqrt{5}\) vào A, ta được:

\(A=\dfrac{\sqrt{5}-1-1}{\sqrt{5}-1+1}=\dfrac{\sqrt{5}-2}{\sqrt{5}}=\dfrac{5-2\sqrt{5}}{5}\)

b: Để \(A< \dfrac{1}{2}\) thì \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{1}{2}< 0\)

\(\Leftrightarrow2\sqrt{x}-2-\sqrt{x}-1< 0\)

\(\Leftrightarrow x< 9\)

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne1\end{matrix}\right.\)

1) ĐKXĐ: \(x\notin\left\{0;1\right\}\)

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

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

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

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

a) ĐKXĐ: \(x\ge0,x\ne1\)

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

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

c) \(A=2\sqrt{x}-1< -1\Leftrightarrow2\sqrt{x}< 0\)(vô lý do \(2\sqrt{x}\ge0\forall x\))

Vậy \(S=\varnothing\)

Bài 3:

\(A=\dfrac{x+1-2\sqrt{x}}{\sqrt{x}-1}+\dfrac{x+\sqrt{x}}{\sqrt[]{x}+1}\\ DKXD:x\ne1;x\ge0\\ A=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\\ A=\sqrt{x}-1+\sqrt{x}\\ A=2\sqrt{x}+1\)

\(C.A< -1\Leftrightarrow2\sqrt{x}-1< -1\\ \Leftrightarrow2\sqrt{x}< 0\\ \Leftrightarrow x< 0\left(ktmdk\right)\\ =>BPTVN:S=\varnothing\)

`A)đk:x>=0,x ne 25`

`A=9=>A=(3+2)/(3-5)=-5/2`

`B)B=(3sqrtx-15+20-2sqrtx)/(x-25)`

`=(sqrtx+5)/(x-25)`

`=1/(sqrtx-5)`

`A=B.|x-4|`

`<=>A/B=|x-4|`

`<=>\sqrtx+2=|x-4|`

`<=>\sqrtx+2=(sqrtx+2)|sqrtx-2|`

`<=>|sqrtx-2|=1`

`+)sqrtx-2=1<=>x=9(tm)`

`+)sqrtx-2=-1<=>x=1(tm)`

Vậy `S={1,9}`

a, Thay x=9 vào biểu thức A ta có

\(A=\dfrac{\sqrt{9}+2}{\sqrt{9}-5}\)

\(A=\dfrac{3+2}{3-5}=\dfrac{5}{-2}=-2,5\)

Vậy A =-2,5 khi x=9

a. A=\(\dfrac{\sqrt{x}+2}{\sqrt{x}-5}\)

=\(\dfrac{\sqrt{9}+2}{\sqrt{9}-5}=\dfrac{-5}{2}\)

a. B = \(\dfrac{\sqrt{36}}{\sqrt{36}-3}=\dfrac{6}{6-3}=2\)

a: Thay x=36 vào B, ta được:

\(B=\dfrac{6}{6-3}=\dfrac{6}{3}=2\)

a: Sửa đề: \(A=\dfrac{\sqrt{x}}{\sqrt{x}+6}\)

Khi x=4 thì \(A=\dfrac{\sqrt{4}}{\sqrt{4}+6}=\dfrac{2}{2+6}=\dfrac{2}{8}=\dfrac{1}{4}\)

b: \(B=\dfrac{4}{x-1}+\dfrac{\sqrt{x}+3}{\sqrt{x}+1}-\dfrac{5}{1-\sqrt{x}}\)

\(=\dfrac{4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+3}{\sqrt{x}+1}+\dfrac{5}{\sqrt{x}-1}\)

\(=\dfrac{4+\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)+5\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

\(=\dfrac{x+7\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}+6}{\sqrt{x}-1}\)

c: \(P=A\cdot B=\dfrac{\sqrt{x}+6}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+6}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)

Để P<0 thì \(\dfrac{\sqrt{x}}{\sqrt{x}-1}< 0\)

mà \(\sqrt{x}>0\)

nên \(\sqrt{x}-1< 0\)

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

=>0<=x<1

1: Khi x=64 thì \(A=\dfrac{8+2}{8}=\dfrac{10}{8}=\dfrac{5}{4}\)

2: \(B=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

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

\(=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)

3: A/B>3/2

=>\(\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{3}{2}>0\)

=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{3}{2}>0\)

=>\(\dfrac{2\sqrt{x}+2-3\sqrt{x}}{\sqrt{x}\cdot2}>0\)

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

=>-căn x>-2

=>căn x<2

=>0<x<4

1) Thay x=64 vào A ta có:

\(A=\dfrac{2+\sqrt{64}}{\sqrt{64}}=\dfrac{2+8}{8}=\dfrac{5}{4}\)

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

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

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

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

\(B=\dfrac{x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(B=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(B=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)

3) Ta có:

\(\dfrac{A}{B}>\dfrac{3}{2}\) khi

\(\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}>\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}>\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}}>\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{3}{2}>0\)

\(\Leftrightarrow\dfrac{2\sqrt{x}+2-3\sqrt{x}}{2\sqrt{x}}>0\)

\(\Leftrightarrow\dfrac{2-\sqrt{x}}{2\sqrt{x}}>0\)

Mà: \(2\sqrt{x}\ge0\forall x\)

\(\Leftrightarrow2-\sqrt{x}>0\)

\(\Leftrightarrow\sqrt{x}< 2\)

\(\Leftrightarrow x< 4\)

Kết hợp với đk:

\(0< x< 4\)