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a, ĐK: \(x\ge0;x\ne9\)
\(P=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{3x+9}{9-x}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{2x-6\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{-3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=-\dfrac{3}{\sqrt{x}-3}\)
b, \(P>0\Leftrightarrow-\dfrac{3}{\sqrt{x}-3}>0\)
\(\Leftrightarrow\sqrt{x}-3>0\)
\(\Leftrightarrow x>9\)
c, \(P=-\dfrac{3}{\sqrt{x}-3}\in Z\)
\(\Leftrightarrow\sqrt{x}-3\inƯ_3=\left\{\pm1;\pm3\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{0;2;4;6\right\}\)
\(\Leftrightarrow x\in\left\{0;4;16;36\right\}\)
Bài 1:
a: ĐKXĐ: 2x+3>=0 và x-3>0
=>x>3
b: ĐKXĐ:(2x+3)/(x-3)>=0
=>x>3 hoặc x<-3/2
c: ĐKXĐ: x+2<0
hay x<-2
d: ĐKXĐ: -x>=0 và x+3<>0
=>x<=0 và x<>-3
a: \(A=\dfrac{\sqrt{x}-1+\sqrt{x}}{\sqrt{x}\left(1-\sqrt{x}\right)}:\left(\dfrac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)}+\dfrac{\sqrt{x}\left(2x+\sqrt{x}-1\right)}{1+x\sqrt{x}}\right)\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(1-\sqrt{x}\right)}:\left(2x+\sqrt{x}-1\right)\cdot\left(\dfrac{1}{1-x}+\dfrac{\sqrt{x}}{1+x\sqrt{x}}\right)\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(1-\sqrt{x}\right)}:\left[\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)\cdot\dfrac{1+x\sqrt{x}+\sqrt{x}-x\sqrt{x}}{\left(1-x\right)\left(1+x\sqrt{x}\right)}\right]\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(1-\sqrt{x}\right)}:\left[\dfrac{\left(2\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}{\left(1-\sqrt{x}\right)\left(1+x\sqrt{x}\right)}\right]\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(1-\sqrt{x}\right)}\cdot\dfrac{\left(1-\sqrt{x}\right)\left(x+\sqrt{x}+1\right)}{\left(2\sqrt{x}-1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
b: Khi x=17-12 căn 2 thì \(A=\dfrac{17-12\sqrt{2}+3-2\sqrt{2}+1}{3-2\sqrt{2}}=7\)
a: ĐKXĐ: \(x>0\)
b: Ta có: \(A=\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+1\)
\(=x+\sqrt{x}-2\sqrt{x}-1+1\)
\(=x-\sqrt{x}\)