a) Ta có: \(B=\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)

\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\)

=1-a

b) Ta có: \(2x^2-5x+2=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{1}{2};2\right\}\)

a) 

\(P=\left(\dfrac{b-a}{\sqrt{b}-\sqrt{a}}-\dfrac{a\sqrt{a}-b\sqrt{b}}{a-b}\right):\dfrac{\left(\sqrt{b}-\sqrt{a}\right)^2+\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\)

\(=\left[\sqrt{b}+\sqrt{a}-\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\right]:\dfrac{b-\sqrt{ab}+a}{\sqrt{a}+\sqrt{b}}\)

\(=\left(\sqrt{b}+\sqrt{a}-\dfrac{a+\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}\right).\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\)

\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-a-\sqrt{ab}-b}{\sqrt{a}+\sqrt{b}}.\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\)

\(=\dfrac{\sqrt{ab}}{\sqrt{a}+\sqrt{b}}.\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\)\(=\dfrac{\sqrt{ab}}{a-\sqrt{ab}+b}\)

b) \(P=\dfrac{\sqrt{ab}}{a-\sqrt{ab}+b}=\dfrac{\sqrt{ab}}{\left(\sqrt{a}-\dfrac{1}{2}\sqrt{b}\right)^2+\dfrac{3}{4}b}\)

Vì \(\left(\sqrt{a}-\dfrac{1}{2}\sqrt{b}\right)^2+\dfrac{3}{4}b>0;\forall a\ge0;b\ge0;a\ne b\)

\(\sqrt{ab}\ge0\)\(\forall a\ge0;b\ge0\)

\(\Rightarrow P=\dfrac{\sqrt{ab}}{\left(\sqrt{a}-\dfrac{1}{2}\sqrt{b}\right)^2+\dfrac{3}{4}b}\ge0\)

Vậy...

cảm ơn tất cả mọi người

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

\(=\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)\)

=a-b

Với `a > 0,b > 0,a \ne b` có:

   `[a+b-2\sqrt{ab}]/[\sqrt{a}-\sqrt{b}]:1/[\sqrt{a}+\sqrt{b}]`

`=[(\sqrt{a}-\sqrt{b})^2]/[\sqrt{a}-\sqrt{b}]. (\sqrt{a}+\sqrt{b})`

`=(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})`

`=a-b`

\(=\dfrac{\left(\sqrt{a}-\sqrt{b}^2\right)}{\left(\sqrt{a-\sqrt{b}}\right)}:\left(\sqrt{a+\sqrt{b}}\right)\)

\(=\left(\sqrt{a-\sqrt{b}}\right).\left(\sqrt{a+\sqrt{b}}\right)\)

ĐA:\(a-b\)

\(a,=\left|2-\sqrt{3}\right|=2-\sqrt{3}\\ b,=\left|3-\sqrt{11}\right|=\sqrt{11}-3\\ c,=2\left|a\right|=2a\\ d,=3\left|a-2\right|=3\left(2-a\right)\left(a< 0\Leftrightarrow a-2< 0\right)\)

a: \(P=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)=a-b\)

\(a.\sqrt{8}-2\sqrt{50}+\sqrt{18}=2\sqrt{2}-10\sqrt{2}+3\sqrt{2}=\sqrt{2}\left(2-10+3\right)=-5\sqrt{2}\)

\(b.\left(\dfrac{\sqrt{a}-a}{1-\sqrt{a}}+\sqrt{a}\right):\dfrac{2\sqrt{a}}{1+\sqrt{a}}\left(đk:a\ge0;a\ne1\right)\)

\(=\left(\sqrt{a}+\sqrt{a}\right).\dfrac{1+\sqrt{a}}{2\sqrt{a}}\)

\(=2\sqrt{a}.\dfrac{1+\sqrt{a}}{2\sqrt{a}}\)

\(=1+\sqrt{a}\)

(Chỗ điều kiện bài b mik thấy a = 0 cũng có thể là nghiệm nên mik sửa lại nhé)

b. \(=\left(\dfrac{\sqrt{a}-a+a\left(1-\sqrt{a}\right)}{1-\sqrt{a}}\right):\left(\dfrac{2\sqrt{a}}{1+\sqrt{a}}\right)\)

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

\(=\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)\)

\(=1-a\)

a) \(A=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{a-\sqrt{a}}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{a-1}\right)\)

\(A=\left[\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right]:\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{a-1}\right)\)

\(A=\left[\dfrac{a}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right]:\left[\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{a-1}\right]\)

\(A=\dfrac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\left[\dfrac{\sqrt{a}-1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}+\dfrac{2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right]\)

\(A=\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\sqrt{a}-1+2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)

\(A=\dfrac{\sqrt{a}+1}{\sqrt{a}}:\dfrac{\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)

\(A=\dfrac{\sqrt{a}+1}{\sqrt{a}}\cdot\left(\sqrt{a}-1\right)\)

\(A=\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}}\)

\(A=\dfrac{a-1}{\sqrt{a}}\)

b) Ta có:

\(a=4+2\sqrt{3}=\left(\sqrt{3}\right)^2+2\sqrt{3}\cdot1+1^2=\left(\sqrt{3}+1\right)^2\)

Thay vào A ta có:

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

c) \(A< 0\) khi:

\(\dfrac{a-1}{\sqrt{a}}< 0\)

Mà: \(\sqrt{a}\ge0\forall x\) (xác định) 

\(\Leftrightarrow a-1< 0\)

\(\Leftrightarrow a< 1\)

Kết hợp với đk:

\(0< a< 1\)

Với `a > 0,b >= 0` có:

`Bth=[a\sqrt{b}+b]/[a-b] . \sqrt{[b(a+b-2\sqrt{ab})]/[a^2+2a\sqrt{b}+b]} . (\sqrt{a}+\sqrt{b})`

 `=[\sqrt{b}(a+\sqrt{b})]/[a-b].\sqrt{[b(\sqrt{a}-\sqrt{b})^2]/[(a+\sqrt{b})^2]}.(\sqrt{a}+\sqrt{b})`

`=[\sqrt{b}(a+\sqrt{b})|\sqrt{a}-\sqrt{b}|.\sqrt{b}.(\sqrt{a}+\sqrt{b})]/[(a-b)(a+\sqrt{b})]`

`=[b|\sqrt{a}-\sqrt{b}|]/[\sqrt{a}-\sqrt{b}]`

`={(b\text{ nếu }\sqrt{a} >= \sqrt{b}),(-b\text{ nếu }\sqrt{a} < \sqrt{b}):}`