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

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

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

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

c: A/B>4/3

=>\(\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}>\dfrac{4}{3}\)

=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{4}{3}>0\)

=>\(\dfrac{3\left(\sqrt{x}+1\right)-4\sqrt{x}}{3\sqrt{x}}>0\)

=>\(3\left(\sqrt{x}+1\right)-4\sqrt{x}>0\)

=>\(3\sqrt{x}+3-4\sqrt{x}>0\)

=>\(-\sqrt{x}>-3\)

=>\(\sqrt{x}< 3\)

=>0<=x<9

Kết hợp ĐKXĐ, ta được: 0<x<9

a) Vì \(x=\dfrac{1}{4}\) thỏa mãn ĐKXĐ

nên Thay \(x=\dfrac{1}{4}\) vào biểu thức \(A=\dfrac{x-4}{\sqrt{x}+2}\), ta được:

\(A=\dfrac{\dfrac{1}{4}-4}{\sqrt{\dfrac{1}{4}}+2}=\left(\dfrac{1}{4}-\dfrac{16}{4}\right):\left(\dfrac{1}{2}+2\right)=\dfrac{-15}{4}:\dfrac{5}{2}\)

\(\Leftrightarrow A=\dfrac{-15}{4}\cdot\dfrac{2}{5}=\dfrac{-30}{20}=\dfrac{-3}{2}\)

Vậy: Khi \(x=\dfrac{1}{4}\) thì \(A=\dfrac{-3}{2}\)

b) Ta có: \(B=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{\sqrt{x}-1}{2-\sqrt{x}}-\dfrac{9-x}{4-x}\)

\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{9-x}{x-4}\)

\(=\dfrac{x-2\sqrt{x}+\sqrt{x}-2+x+2\sqrt{x}-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{2x-4+9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

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

Thay x = \(\dfrac{1}{4}\)vào bt A ta có: A= \(\dfrac{\dfrac{1}{4}-4}{\sqrt{\dfrac{1}{4}}+2}=\dfrac{-15}{4}:\dfrac{5}{2}=\dfrac{-3}{2}\)

Vậy x = \(\dfrac{1}{4}\)vào bt A nhận giá trị là -3/2

b)

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:

\(0< x< 4\)

a) \(ĐKXĐ:\left\{{}\begin{matrix}x>0\\x\ne1\\x\ne4\end{matrix}\right.\)

\(\Leftrightarrow B=\dfrac{\sqrt{x}-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)

\(\Leftrightarrow B=\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}\)

\(\Leftrightarrow B=\dfrac{2-\sqrt{x}}{3\sqrt{x}}\)

b) \(x=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\Rightarrow\sqrt{x}=\sqrt{3}+1\)  (*)

Thay (*) vào B , ta được : \(B=\dfrac{2-\sqrt{3}-1}{3\sqrt{3}+3}=\dfrac{-\sqrt{3}+1}{3\sqrt{3}+3}\)

Bạn santa làm sai r nha!

a, ĐKXĐ: x \(\ge\) 0; x \(\ne\) 4; x \(\ne\) 0

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

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

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

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

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

B = \(\dfrac{2-\sqrt{x}}{3\sqrt{x}}\) (Đoạn này bạn kia viết sai đề mà vẫn đúng kết quả được?)

Vậy ...

b, Ta có: x = 4 + 2\(\sqrt{3}\) = (\(\sqrt{3}\) + 1)² (TMĐK)

\(\Rightarrow\) \(\sqrt{x}\) = \(\sqrt{3}+1\) (1)

Thay (1) vào B ta được:

B = \(\dfrac{2-\sqrt{3}-1}{3\left(\sqrt{3}-1\right)}\) = \(\dfrac{1-\sqrt{3}}{-3\left(1-\sqrt{3}\right)}\) = \(\dfrac{-1}{3}\)

Vậy ...

Chúc bn học tốt!

mình làm lại nhé :

đkxđ : \(\left\{{}\begin{matrix}x>0\\x\ne1\\x\ne4\end{matrix}\right.\)

\(\Leftrightarrow B=\dfrac{\sqrt{x}-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}:\dfrac{x-1-x+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

\(\Leftrightarrow B=\dfrac{-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{3}\)

\(\Leftrightarrow B=\dfrac{2-\sqrt{x}}{3\sqrt{x}}\)

câu b làm như kia là oke rồi nhé <3

1:

\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)

2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)

\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)

\(=\dfrac{20-6}{2}=7\)

\(a,P=\dfrac{\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{2-\sqrt{x}}{\sqrt{x}}=\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{-2}{\sqrt{x}+2}\\ P=-\dfrac{3}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\\ \Leftrightarrow3\sqrt{x}+6=10\Leftrightarrow\sqrt{x}=\dfrac{4}{3}\Leftrightarrow x=\dfrac{16}{9}\left(tm\right)\)

a) Thay x=25 vào B ta có:

\(B=\dfrac{\sqrt{25}+2}{\sqrt{25}-2}=\dfrac{7}{3}\)

b) \(A=\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{2\sqrt{x}-1}{x-5\sqrt{x}+6}\)

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

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

\(A=\dfrac{x-9-x+4+2\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

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

\(A=\dfrac{2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(A=\dfrac{2}{\sqrt{x}-2}\)

c) Ta có: \(A>B\) Khi:

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

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}-\sqrt{x}< 0\\\sqrt{x}-2< 0\end{matrix}\right.\\\left\{{}\begin{matrix}-\sqrt{x}>0\\\sqrt{x}-2>0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x< 4\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x>4\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow0< x< 4\) 

`a)A=[2\sqrt{3}+2-2\sqrt{3}+2]/[(2\sqrt{3}-2)(2\sqrt{3}+2)]`

   `A=4/[12-4]=1/2`

Với `x > 0,x ne 1` có:

`B=[x-2\sqrt{x}+1]/[\sqrt{x}(\sqrt{x}-1)]`

`B=[(\sqrt{x}-1)^2]/[\sqrt{x}(\sqrt{x}-1)]=[\sqrt{x}-1]/\sqrt{x}`

`b)B=2/5A`

`=>[\sqrt{x}-1]/\sqrt{x}=2/5 . 1/2`

`<=>5\sqrt{x}-5=\sqrt{x}`

`<=>\sqrt{x}=5/4`

`<=>x=25/16` (t/m)