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\(A=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\) ĐKXĐ : x > 0 , x khác 9
\(A=\left(\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
\(A=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x+3}{x-9}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(A=\frac{-3\sqrt{x}}{\sqrt{x}+3}.\frac{1}{\sqrt{x}+1}\)
\(A=\frac{-3\sqrt{x}}{x+4\sqrt{x}+4}\)
\(A=\frac{-3\sqrt{x}}{\left(\sqrt{x}+2\right)^2}\)
a) ĐKXĐ : x>hoặc = 0 ; x khác 9
Còn câu b,c,d để vài bữa mình làm tiếp cho bây giờ mình đi ngủ đã buồn ngủ quá !
----------------- -Học tốt-----------------
a)\(P=\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{3x-9}{x-9}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)+2\sqrt{x}\left(\sqrt{x}+3\right)-3x-9}{(\sqrt{x}+3)\left(\sqrt{x}-3\right)}\)
\(=\frac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{(\sqrt{x}+3)\left(\sqrt{x}-3\right)}\)
\(=\frac{3x+3\sqrt{x}-3x-9}{(\sqrt{x}+3)\left(\sqrt{x}-3\right)}\)
\(=\frac{3\sqrt{x}-9}{(\sqrt{x}+3)\left(\sqrt{x}-3\right)}\)
\(=\frac{3(\sqrt{x}-3)}{(\sqrt{x}+3)\left(\sqrt{x}-3\right)}\)
\(=\frac{3}{\sqrt{x}+3}\)
\(x\ge0;x\ne9\)
\(P=\left(\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\left(\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
\(=\left(\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right).\left(\frac{\sqrt{x}-3}{\sqrt{x}+1}\right)\)
\(=\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)}=\frac{-3}{\sqrt{x}+3}\)
Do \(-3< 0\Rightarrow P_{min}\) khi \(\sqrt{x}+3\) nhỏ nhất
Mà \(\sqrt{x}+3\ge3\Rightarrow P_{min}=\frac{-3}{3}=-1\) khi \(x=0\)
a) đk: \(\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)
b) Ta có:
\(P=\frac{\sqrt{x}+2}{\sqrt{x}-3}+\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{3x-8\sqrt{x}+27}{9-x}\)
\(P=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)+2\sqrt{x}\cdot\left(\sqrt{x}-3\right)-3x+8\sqrt{x}-27}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{x+5\sqrt{x}+6+2x-6\sqrt{x}-3x+8\sqrt{x}-27}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{7\sqrt{x}-21}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{7\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{7}{\sqrt{x}+3}\)
c) Nếu x không là số chính phương => P vô tỉ (loại)
=> x là số chính phương khi đó để P nguyên thì:
\(\left(\sqrt{x}+3\right)\inƯ\left(7\right)\) , mà \(\sqrt{x}+3\ge3\left(\forall x\ge0\right)\)
\(\Rightarrow\sqrt{x}+3=7\Leftrightarrow\sqrt{x}=4\Rightarrow x=16\)
Vậy x = 16 thì P nguyên
\(A\sqrt{3}=\sqrt{3x+6\sqrt{3x-9}}+\sqrt{3x-6\sqrt{3x-9}}\)
\(=\sqrt{3x-9+6\sqrt{3x-9}+9}+\sqrt{3x-9-6\sqrt{3x-9}+9}\)
\(=\sqrt{\left(\sqrt{3x-9}+3\right)^2}+\sqrt{\left(\sqrt{3x-9}-3\right)^2}\)
\(=\left|\sqrt{3x-9}+3\right|+\left|\sqrt{3x-9}-3\right|\)
Do \(x\ge6\Rightarrow\sqrt{3x-9}-3\ge0\)
\(\Rightarrow A\sqrt{3}=\sqrt{3x-9}+3+\sqrt{3x-9}-3=2\sqrt{3x-9}\ge6\)
\(\Rightarrow A\ge\frac{6}{\sqrt{3}}=2\sqrt{3}\)
Dấu "=" xảy ra khi \(x=6\)