a) P = \(\left(\frac{3\sqrt{a}}{a+\sqrt{a}+b}-\frac{3a}{a\sqrt{a}-b\sqrt{b}}+\frac{1}{\sqrt{a}-\sqrt{b}}\right):\frac{\left(a-1\right).\left(\sqrt{a}-\sqrt{b}\right)}{\left(2.a+2.\sqrt{ab}+2.b\right)}\)

        = \(\left(\frac{3\sqrt{a}.\left(\sqrt{a}-\sqrt{b}\right)-3.a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right).\left(a+\sqrt{ab}+b\right)}\right).\frac{2.\left(a+\sqrt{ab}+b\right)}{\left(a-1\right).\left(\sqrt{a}-\sqrt{b}\right)}\)

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

          = \(\frac{2}{a-1}\)

b) P nguyên <=> \(\frac{2}{a-1}\)nguyên => 2 \(⋮\)a - 1 

=> ( a- 1 ) = { \(\pm\)1 ; \(\pm\) 2} => a = { -1 ; 0 ; 2 ;3 } 

Giao luu:

\(x-\sqrt{x^2-1}\ne0\Rightarrow A.xacdinh.moi.x\)

\(0\le\left(\sqrt[4]{\frac{a}{b}}-\sqrt[4]{\frac{b}{a}}\right)^2=\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}-2\Rightarrow\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\ge2\)

\(\Rightarrow x\ge1\Rightarrow\hept{\begin{cases}x-1\ge0\\x^2-1\ge0\end{cases}\left(1\right)}\)

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

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

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

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

\(2\left(\frac{A}{2b}+1\right)=\left[\sqrt{\left(\sqrt[4]{\frac{a}{b}}-\sqrt[4]{\frac{b}{a}}\right)^2}+\sqrt{\left(\sqrt[4]{\frac{a}{b}}+\sqrt[4]{\frac{b}{a}}\right)^2}\right]^{^2}\)

\(2\left(\frac{A}{2b}+1\right)=\left[\left(\sqrt[4]{\frac{a}{b}}-\sqrt[4]{\frac{b}{a}}\right)+\left(\sqrt[4]{\frac{a}{b}}+\sqrt[4]{\frac{b}{a}}\right)\right]^{^2}=4\sqrt{\frac{a}{b}}\)

\(\frac{A}{2b}+1=2\sqrt{\frac{a}{b}}\)

\(A=4b\sqrt{\frac{a}{b}}-2b=4\sqrt{ab}-2b\)(hoa hết mắt  có khi (+-,*/,) nhầm vì số liệu chưa đẹp...hihi)

Ta có:

\(x=\frac{1}{2}\left(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\right)=\frac{a+b}{2\sqrt{ab}}\)

\(x^2-1=\left(\frac{a+b}{2\sqrt{ab}}\right)^2-1=\frac{a^2+2ab+b^2-4ab}{4ab}\)

\(=\frac{\left(a-b\right)^2}{4ab}\)

Từ đây ta có

\(A=\frac{2b\sqrt{x^2-1}}{x-\sqrt{x^2-1}}=\frac{2b\sqrt{\frac{\left(a-b\right)^2}{4ab}}}{\frac{a+b}{2\sqrt{ab}}-\sqrt{\frac{\left(a-b\right)^2}{4ab}}}\)

\(=\frac{2b.\frac{a-b}{2\sqrt{ab}}}{\frac{a+b}{2\sqrt{ab}}-\frac{a-b}{2\sqrt{ab}}}=\frac{2ab-2b^2}{2b}=a-b\)

Hoa hế mắt

a: ĐKXĐ: \(\left[{}\begin{matrix}x\ge6\\x\le2\end{matrix}\right.\)

b: ĐKXĐ: \(-1\le x\le1\)

c: ĐKXĐ: \(x\le-2\)

a. \(\sqrt{\left(x-2\right)\left(x-6\right)}\) có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}x-2\ge0\\x-6\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\ge6\end{matrix}\right.\) \(\Leftrightarrow x\ge6\)

b. \(\sqrt{1-x^2}\) có nghĩa \(\Leftrightarrow1-x^2\ge0\) \(\Leftrightarrow\left\{{}\begin{matrix}1-x\ge0\\x+1\ge0\end{matrix}\right.\) \(\Leftrightarrow-1\le x\le1\)

\(\sqrt{-5x-10}\) có nghĩa \(\Leftrightarrow-5x-10\ge0\Leftrightarrow-5x\ge10\Leftrightarrow x\ge-2\)

P = (\(\dfrac{1}{\sqrt{x}-1}\) - \(\dfrac{1}{\sqrt{x}}\)) : (\(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\) - \(\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\)) với  0 < \(x\) ≠ 1; 4

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

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

P = \(\dfrac{1}{\sqrt{x}.\left(\sqrt{x}-1\right)}\) : \(\dfrac{3}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-1\right)}\)

P = \(\dfrac{1}{\sqrt{x}.\left(\sqrt{x}-1\right)}\) \(\times\) \(\dfrac{\left(\sqrt{x}-2\right).\left(\sqrt{x}-1\right)}{3}\)

P = \(\dfrac{\sqrt{x}-2}{3.\sqrt{x}}\)

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

b, P = \(\dfrac{1}{4}\)

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

⇒4\(x\) - 8\(\sqrt{x}\) = 3\(x\)

⇒ 4\(x\) - 8\(\sqrt{x}\) - 3\(x\) = 0

     \(x\) - 8\(\sqrt{x}\)   = 0

      \(\sqrt{x}\).(\(\sqrt{x}\) - 8) = 0

       \(\left[{}\begin{matrix}x=0\\\sqrt{x}=8\end{matrix}\right.\)

      \(\left[{}\begin{matrix}x=0\\x=64\end{matrix}\right.\)

      \(x=0\) (loại)

      \(x\) = 64

Lời giải:

a. \(P=\frac{\sqrt{x}-(\sqrt{x}-1)}{\sqrt{x}(\sqrt{x}-1)}: \frac{(\sqrt{x}+1)(\sqrt{x}-1)-(\sqrt{x}-2)(\sqrt{x}+2)}{(\sqrt{x}-2)(\sqrt{x}-1)}\)

\(=\frac{1}{\sqrt{x}(\sqrt{x}-1)}: \frac{x-1-(x-4)}{(\sqrt{x}-2)(\sqrt{x}-1)}=\frac{1}{\sqrt{x}(\sqrt{x}-1)}:\frac{3}{(\sqrt{x}-1)(\sqrt{x}-2)}\\ =\frac{1}{\sqrt{x}(\sqrt{x}-1)}.\frac{(\sqrt{x}-1)(\sqrt{x}-2)}{3}=\frac{\sqrt{x}-2}{3\sqrt{x}}\)

b.

\(P=\frac{\sqrt{x}-2}{3\sqrt{x}}=\frac{1}{4}\\ \Rightarrow 4(\sqrt{x}-2)=3\sqrt{x}\\ \Leftrightarrow \sqrt{x}=8\Leftrightarrow x=64\) 

(thỏa mãn) 

c.

Tại $x=4+2\sqrt{3}=(\sqrt{3}+1)^2\Rightarrow \sqrt{x}=\sqrt{3}+1$
Khi đó:

$P=\frac{\sqrt{3}+1-2}{3(\sqrt{3}+1)}=\frac{2-\sqrt{3}}{3}$