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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 }
a: \(P=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}\cdot\dfrac{a+2}{a-2}\)
\(=\dfrac{2a+4}{a-2}\)
b: Để P nguyên thì \(2a-4+8⋮a-2\)
\(\Leftrightarrow a-2\in\left\{-2;-1;1;2;4;8\right\}\)
hay \(a\in\left\{0;3;4;6;10\right\}\)
a) Ta có: \(3x+2\sqrt{3x}+4=\left(\sqrt{3x}+1\right)^2+3>0;1+\sqrt{3x}>0,\forall x\ge0\), nên đk để A có nghĩa là
\(\left(\sqrt{3x}\right)^3-8-\left(\sqrt{3x}-2\right)\left(3x+2\sqrt{3x}+4\right)\ne0;x\ge0\Leftrightarrow\sqrt{3x}\ne2\Leftrightarrow0\le x\ne\frac{4}{3}\)
A=\(\left(\frac{6x+4}{\left(\sqrt{3x}\right)^3-2^3}-\frac{\sqrt{3x}}{3x+2\sqrt{3x}+4}\right)\left(\frac{1+\left(\sqrt{3x}\right)^3}{1+\sqrt{3x}}-\sqrt{3x}\right)\)
\(=\left(\frac{6x+4-\left(\sqrt{3x}-2\right)\sqrt{3x}}{\left(\sqrt{3x}-2\right)\left(3x+2\sqrt{3x}+4\right)}\right)\left(3x-\sqrt{3x}+1-\sqrt{3x}\right)\)
\(=\left(\frac{3x+4+2\sqrt{3x}}{\left(\sqrt{3x}-2\right)\left(3x+2\sqrt{3x}+4\right)}\right)\left(3x-2\sqrt{3x}+1\right)\)
\(=\frac{\left(\sqrt{3x}-1\right)^2}{\sqrt{3x}-2}\left(0\le x\ne\frac{4}{3}\right)\)
b) \(A=\frac{\left(\sqrt{3x}-1\right)^2}{\sqrt{3x}-2}=\frac{\left(\sqrt{3x}-2\right)^2+2\left(\sqrt{3x}-2\right)+1}{\sqrt{3x}-2}=\sqrt{3x}+\frac{1}{\sqrt{3x}-2}\)
Với \(x\ge0\), để A là số nguyên thì \(\sqrt{3x}-2=\pm1\Leftrightarrow\orbr{\begin{cases}\sqrt{3x}=3\\\sqrt{3x}=1\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=9\\3x=1\end{cases}\Leftrightarrow}x=3}\) (vì \(x\in Z;x\ge0\))
Khi đó A=4
Bài 1 :
a) \(P=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}}{x-2\sqrt{x}+1}\)
\(P=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right).\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
\(P=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}-1}{\sqrt{x}}\)
\(P=\frac{\sqrt{x}+1}{x}\)
b) \(P>\frac{1}{2}\)
\(\Leftrightarrow\frac{\sqrt{x}+1}{x}>\frac{1}{2}\)
\(\Leftrightarrow\frac{\sqrt{x}+1}{x}-\frac{1}{2}>0\)
\(\Leftrightarrow\frac{\sqrt{x}+1-2x}{x}>0\)
\(\Leftrightarrow\sqrt{x}-2x+1>0\left(x>0\right)\)
\(\Leftrightarrow\sqrt{x}+x^2-2x+1-x^2>0\)
\(\Leftrightarrow\sqrt{x}+x^2+\left(x-1\right)^2>0\left(\forall x>0\right)\)
Vậy P > 1/2 với mọi x> 0 ; x khác 1
Bài 2 :
a) \(K=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}+a}+\frac{2}{a-1}\right)\)
\(K=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{1}{\sqrt{a}\left(\sqrt{a}+1\right)}+\frac{2}{a-1}\right)\)
\(K=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1+2\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}\left(a-1\right)\left(\sqrt{a}+1\right)}\)
\(K=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\sqrt{a}\left(a-1\right)\left(\sqrt{a}-1\right)}{a-1+2a+2\sqrt{a}}\)
\(K=\frac{\left(a-1\right)^2}{3a+2\sqrt{a}-1}\)
b) \(a=3+2\sqrt{2}=2+2\sqrt{2}+1=\left(\sqrt{2}+1\right)^2\)( thỏa mãn ĐKXĐ )
Thay a vào biểu thức K , ta có :
\(K=\frac{\left(3+2\sqrt{2}-1\right)^2}{3\left(3+2\sqrt{2}\right)+2\sqrt{\left(\sqrt{2}+1\right)^2}-1}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{9+6\sqrt{2}+2\left|\sqrt{2}+1\right|-1}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{8+6\sqrt{2}+2\sqrt{2}+2}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{10+8\sqrt{2}}\)
\(P=\left[\frac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\right]:\frac{a+2}{a-2}\)
\(=\left(\frac{a+\sqrt{a}+1}{\sqrt{a}}-\frac{a-\sqrt{a}+1}{\sqrt{a}}\right):\frac{a+2}{a-2}\)\(=\frac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}:\frac{a+2}{a-2}\)\(=\frac{2\sqrt{a}}{\sqrt{a}}.\frac{a-2}{a+2}=\frac{2a-4}{a+2}\)