a, \(ĐKXĐ:a;b>0;a\ne2b\\ \)

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

b, Tự lm nhé.

\(a,dkxd:x\ge0,x\ne4\)

\(b,B=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{x-2\sqrt{x}}\right)\dfrac{1}{\sqrt{x}-2}\\ =\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\dfrac{1}{\sqrt{x}-2}\\ =\dfrac{\sqrt{x^2}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}.\dfrac{1}{\sqrt{x}-2}\\ =\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)^2}\\ =\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(c,x=16\left(tm\right)\Rightarrow B=\dfrac{\sqrt{16}+2}{\sqrt{16}\left(\sqrt{16}-2\right)}=\dfrac{4+2}{4\left(4-2\right)}=\dfrac{6}{8}=\dfrac{3}{4}\)

\(d,B>0\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}>0\Leftrightarrow\sqrt{x}+2>0\Leftrightarrow\sqrt{x}>-2\left(ktm\right)\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)< 0\Leftrightarrow\sqrt{x}< 2\Leftrightarrow x< 4\)

Kết hợp với \(dk:x\ge0\) ta kết luận \(0\le x< 4\) thì \(B>0\).

a) Điều kiện xác định:

\(\left\{{}\begin{matrix}x-2\sqrt{x}\ne0\\x\ge0\end{matrix}\right.\)\(\Leftrightarrow x>0,x\ne4\)

Vậy...

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

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

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

c) Tại x=16 ( thỏa mãn đk) thay vào B đã rút gọn ta được:

\(B=\dfrac{\sqrt{16}+2}{\sqrt{16}\left(\sqrt{16}-2\right)}=\dfrac{3}{4}\)

d) \(B>0\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}>0\)

\(\Leftrightarrow\sqrt{x}-2>0\)\(\Leftrightarrow\sqrt{x}>2\Leftrightarrow x>4\)

Vậy x>4 thì B>0

a: ĐKXĐ: a>=0; b>=0; ab<>0; a<>1\(M=\dfrac{3\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)-3a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\cdot\dfrac{2\left(a+\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(a-1\right)}\)

\(=\dfrac{3a-3\sqrt{ab}-3a+a+\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{1}{\left(\sqrt{a}-\sqrt{b}\right)\left(a-1\right)}\)

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

b: M nguyên khi a-1 thuộc {1;-1}

=>a thuộc {2;0}

ĐKXĐ: \(\left\{{}\begin{matrix}a^2-b^2>0\\b\ne0\end{matrix}\right.\)

a: ĐKXĐ: a>=0; a<>1; a<>4

b: \(A=\dfrac{\sqrt{a}+1+\sqrt{a}-1}{a-1}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)

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

c: Để A>0 thì căn a-2>0

=>a>4

\(a.b.A=\left(2-\dfrac{a-3\sqrt{a}}{\sqrt{a}-3}\right)\left(2-\dfrac{5\sqrt{a}-\sqrt{ab}}{\sqrt{b}-5}\right)=\left(2-\dfrac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}-3}\right)\left(2+\dfrac{\sqrt{a}\left(5-\sqrt{b}\right)}{5-\sqrt{b}}\right)=\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)=4-a\) ( a ; b ≥ 0 ; a # 9 ; b # 25 )

ĐKXĐ : \(\left\{{}\begin{matrix}a^2-b^2>0\\a-\sqrt{a^2-b^2}\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2>b^2\\a^2-b^2\ne a^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2>b^2\\b^2\ne0\end{matrix}\right.\)

Lời giải:
ĐKXĐ: \(\left\{\begin{matrix} a^2-b^2>0\\ a-\sqrt{a^2-b^2}\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a^2>b^2\\ a\neq \sqrt{a^2-b^2}\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} a^2> b^2\\ a\neq \sqrt{a^2-b^2}\end{matrix}\right.\)