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\(A=\left(1-\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\right):\left(\frac{\left(\sqrt{a}-2\right)^2-\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)-\left(9-a\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\right)\)
\(=\left(\frac{\sqrt{a}+3-\sqrt{a}}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\right):\left(\frac{\left(\sqrt{a}-2\right)^2-a+9-9+a}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\right)\)
\(=\frac{3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}:\left(\frac{\left(\sqrt{a}-2\right)^2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\right)\)
\(=\frac{3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}.\frac{\left(\sqrt{a}+3\right)}{\left(\sqrt{a}-2\right)}=\frac{3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)
Để \(A+\left|A\right|\ne0\Rightarrow\left|A\right|\ne-A\Rightarrow A>0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{a}< 2\\\sqrt{a}>3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}a< 4\\a>9\end{matrix}\right.\)
Kết hợp điều kiện \(\Rightarrow\left[{}\begin{matrix}0\le a< 4\\a>9\end{matrix}\right.\)
\(B=\frac{9-x}{\sqrt{x}+3}-\frac{x-6\sqrt{x}+9}{\sqrt{x}-3}-6\)(đk: x ≥ 0 và x ≠ 9)
\(B=\frac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}{\sqrt{x}+3}-\frac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}-3}-6\)
\(B=\left(3-\sqrt{x}\right)-\left(\sqrt{x}-3\right)-6\)
\(B=3-\sqrt{x}-\sqrt{x}+3-6\)
\(B=-2\sqrt{x}\)
\(A=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}+\frac{x}{36-x}\)(đk: x ≥ 0 và x ≠ 36)
\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+6\right)-3\left(\sqrt{x-6}\right)-x}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{x+6\sqrt{x}-3\sqrt{x}+18-x}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3\sqrt{x}+18}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3(\sqrt{x}+6)}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3}{\sqrt{x}-6}\)
a) \(\left(2-\frac{a-3.\sqrt{a}}{\sqrt{a}-3}\right).\left(2-\frac{5.\sqrt{a}+\sqrt{a}.b}{\sqrt{b}-5}\right)\)
=\(\left(2-\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}-3}\right)\left(2+\frac{\sqrt{a}\left(5-\sqrt{b}\right)}{5-\sqrt{b}}\right)\)
=\(\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)\)
=4-a ( Bạn xem lại đề bài giúp mình )
b)\(\frac{9-a}{\sqrt{a}+3}-\frac{9-6\sqrt{a}+a}{\sqrt{a}-3}\) -6
=\(\frac{\left(3-\sqrt{a}\right)\left(3+\sqrt{a}\right)}{\sqrt{a}+3}+\frac{\left(3-\sqrt{a}\right)^2}{3-\sqrt{a}}-6\)
=\(3-\sqrt{a}+3-\sqrt{a}-6\)
=-2\(\sqrt{a}\)
ĐKXĐ:
a/ \(A=\left[\frac{\left(\sqrt{a}+3\right)^2-\left(\sqrt{a}-3\right)^2}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\right].\frac{\sqrt{a}-3}{3\sqrt{a}}=\left(\frac{a+6\sqrt{a}+9-a+6\sqrt{a}-9}{\sqrt{a}+3}\right).\frac{1}{3\sqrt{a}}\)
\(=\frac{12\sqrt{a}}{\sqrt{a}+3}.\frac{1}{3\sqrt{a}}=\frac{4}{\sqrt{a}+3}\)
b/ Để \(A\in Z\)thì \(\sqrt{a}+3\inƯ\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Với \(\sqrt{a}+3=1\Rightarrow\sqrt{a}=-2\) (vô nghiệm)
Với \(\sqrt{a}+3=-1\Rightarrow\sqrt{a}=-4\)(vô nghiệm)
Với \(\sqrt{a}+3=2\Rightarrow\sqrt{a}=-1\) (vô nghiệm)
Với \(\sqrt{a}+3=-2\Rightarrow\sqrt{a}=-5\)(vô nghiệm)
Với \(\sqrt{a}+3=4\Rightarrow\sqrt{a}=1\Rightarrow a=1\)(nhận)
Với \(\sqrt{a}+3=-4\Rightarrow\sqrt{a}=-7\)(vô nghiệm)
Vậy a = 1 thì \(A\in Z\)