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 } 

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

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

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

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

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

\(=2\sqrt{b}\)

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

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

\(D=\frac{\left[\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}\right].\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right).\sqrt{b}}-\frac{\left(\sqrt{a}.\sqrt{b}-b\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}.\left(\sqrt{a}+\sqrt{b}\right)}\)

\(D=\frac{\left[\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}\right]-\left(\sqrt{a}.\sqrt{b}-b\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}.\left(\sqrt{a}+\sqrt{b}\right)}\)

\(D=\frac{2b.\sqrt{a}+2b.\sqrt{b}}{\sqrt{b}.\left(\sqrt{a}+\sqrt{b}\right)}\)

\(D=\frac{2b.\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}\)

\(D=2\sqrt{b}\)

a)  ĐK:  a > 0;  b > 0

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

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

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

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

\(=2\sqrt{b}-b\)

b)  \(A=1\)\(\Rightarrow\)\(2\sqrt{b}-b=1\)

                    \(\Leftrightarrow\)\(b-2\sqrt{b}+1=0\)

                   \(\Leftrightarrow\) \(\left(\sqrt{b}-1\right)^2=0\)

                   \(\Leftrightarrow\)\(\sqrt{b}-1=0\)

                   \(\Leftrightarrow\)\(\sqrt{b}=1\)

                   \(\Leftrightarrow\)\(b=1\)   (t/m ĐKXĐ)

Vậy  b=1

cau c í mk thấy bn chép sai đề nên mk sửa lại đề rồi bạn xem lại đề rồi so với bài làm của mk nha có j ko hiểu thì ib mk nha

\(a)VT = \dfrac{{{{\left( {\sqrt a + 1} \right)}^2} - 4\sqrt a }}{{\sqrt a - 1}} + \dfrac{{a + \sqrt a }}{{\sqrt a }}\\ = \dfrac{{a + 2\sqrt a + 1 - 4\sqrt a }}{{\sqrt a - 1}} + \dfrac{{\sqrt a \left( {\sqrt a + 1} \right)}}{{\sqrt a }}\\ = \dfrac{{a - 2\sqrt a + 1}}{{\left( {\sqrt a - 1} \right)}} + \sqrt a + 1\\ = \dfrac{{{{\left( {\sqrt a - 1} \right)}^2}}}{{\sqrt a - 1}} + \sqrt a + 1\\ = \sqrt a - 1 + \sqrt a + 1\\ = 2\sqrt a = VP (đpcm) \)

\(b)VT = \dfrac{{x\sqrt x + y\sqrt y }}{{\sqrt x + \sqrt y }} - {\left( {\sqrt x - \sqrt y } \right)^2}\\ = \dfrac{{\left( {\sqrt x + \sqrt y } \right)\left( {x - \sqrt {xy} + y} \right)}}{{\sqrt x + \sqrt y }} - \left( {x - 2\sqrt {xy} + y} \right)\\ = x - \sqrt {xy} + y - x + 2\sqrt {xy} - y\\ = \sqrt {xy} (đpcm)\\ c)VT = \dfrac{{a\sqrt b - b\sqrt a }}{{\sqrt {ab} }}:\dfrac{{a - b}}{{\sqrt a + \sqrt b }}\\ = \dfrac{{\sqrt {ab} \left( {\sqrt a - \sqrt b } \right)}}{{\sqrt {ab} }}.\dfrac{{\sqrt a + \sqrt b }}{{a - b}}\\ = \sqrt a - \sqrt b .\dfrac{{\sqrt a + \sqrt b }}{{a - b}}\\ = \dfrac{{\left( {\sqrt a - \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)}}{{a - b}}\\ = \dfrac{{a - b}}{{a - b}} = 1 (đpcm)\\ d)VT = \left[ {\dfrac{{{{\left( {\sqrt a - \sqrt b } \right)}^2} + 4\sqrt {ab} }}{{\sqrt a + \sqrt b }} - \dfrac{{a\sqrt b - b\sqrt a }}{{\sqrt {ab} }}} \right]:\sqrt b \\ = \dfrac{{a - 2\sqrt {ab} + b + 4\sqrt {ab} }}{{\sqrt a + \sqrt b }} - \dfrac{{\sqrt {ab} \left( {\sqrt a - \sqrt b } \right)}}{{\sqrt {ab} }}:\sqrt b \\ = \dfrac{{{{\left( {\sqrt a + \sqrt b } \right)}^2}}}{{\sqrt a + \sqrt b }} - \left( {\sqrt a - \sqrt b } \right):\sqrt b \\ = \sqrt a + \sqrt b - \sqrt a + \sqrt b :\sqrt b \\ = \dfrac{{2\sqrt b }}{{\sqrt b }} = 2 (đpcm) \)

Câu c đề sai (đã sửa)

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}