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\(A=\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{2\sqrt{x}-1}{\sqrt{x}-1}+\frac{x-2}{x-3\sqrt{x}+2}\)
\(A=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\frac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\) \(+\frac{x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{x-4\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\frac{2x-5\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\) \(+\frac{x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{x-4\sqrt{x}+3-2x+5\sqrt{x}-2+x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{1}{\sqrt{x}-2}\)
vậy \(A=\frac{1}{\sqrt{x}-2}\)
A có nghĩa khi \(\sqrt{x}-2>0\)
\(\Leftrightarrow\sqrt{x}=2\)
\(\Leftrightarrow x=4\)
vậy \(x=4\) thì A có nghĩa
b) theo ý a) \(A=\frac{1}{\sqrt{x}-2}\)
theo bài ra \(A>2\) \(\Leftrightarrow\frac{1}{\sqrt{x}-2}>2\)
\(\Leftrightarrow\frac{1}{\sqrt{x}-2}-2>0\)
\(\Leftrightarrow\frac{1}{\sqrt{x}-2}-\frac{2\left(\sqrt{x}-2\right)}{\sqrt{x}-2}>0\)
\(\Leftrightarrow\frac{1-2\sqrt{x}+4}{\sqrt{x}-2}>0\)
\(\Leftrightarrow\frac{5-2\sqrt{x}}{\sqrt{x}-2}>0\)
\(\Rightarrow\hept{\begin{cases}5-2\sqrt{x}>0\\\sqrt{x}-2>0\end{cases}}\) hoặc \(\hept{\begin{cases}5-2\sqrt{x}< 0\\\sqrt{x}-2< 0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}-2\sqrt{x}>-5\\\sqrt{x}>2\end{cases}}\) hoặc \(\hept{\begin{cases}-2\sqrt{x}< -5\\\sqrt{x}< 2\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x< \frac{25}{4}\\x>4\end{cases}}\)hoặc \(\hept{\begin{cases}x>\frac{25}{4}\\x< 4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}4< x< \frac{25}{4}\\x\notin\varnothing\end{cases}}\)
vậy \(4< x< \frac{25}{4}\) thì \(A>2\)
\(ĐKXĐ:\hept{\begin{cases}x\ne9\\x\ne64\end{cases}}\)
\(P=\left(\frac{\sqrt{x}}{\sqrt{x-3}}+\frac{2\sqrt{x}-24}{x-9}\right).\frac{7}{\sqrt{x}+8}\)
\(\Leftrightarrow P=\left(\frac{\sqrt{x}}{\sqrt{x}-3}+\frac{2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right).\frac{7}{\sqrt{x}+8}\)
\(\Leftrightarrow P=\frac{\sqrt{x}\left(\sqrt{x}+3\right)+2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{7}{\sqrt{x}+8}\)
\(\Leftrightarrow P=\frac{x+3\sqrt{x}+2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{7}{\sqrt{x}+8}\)
\(\Leftrightarrow P=\frac{x+5\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{7}{\sqrt{x}+8}\)
\(\Leftrightarrow P=\frac{x+8\sqrt{x}-3\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x+3}\right)}.\frac{7}{\sqrt{x}+8}\)
\(\Leftrightarrow P=\frac{\sqrt{x}\left(\sqrt{x}+8\right)-3\left(\sqrt{x}+8\right)}{\left(\sqrt{x-3}\right)\left(\sqrt{x}+3\right)}.\frac{7}{\sqrt{x}+8}\)
\(\Leftrightarrow P=\frac{\left(\sqrt{x}+8\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{7}{\sqrt{x}+8}\)
\(\Leftrightarrow P=\frac{7}{\sqrt{x}+3}\)
Để P nguyên \(\Leftrightarrow7⋮\sqrt{x}+3\) \(\left(\sqrt{x}\ge0\Rightarrow\sqrt{x}+3\ge3\right)\)
\(\Leftrightarrow\sqrt{x}+3\inƯ\left(7\right)\)
Ta có bảng sau :
| \(\sqrt{x}+3\) | \(1\) | \(-1\) | \(7\) | \(-7\) |
| \(\sqrt{x}\) | \(-2\)(ktm) | \(-4\)(ktm) | \(4\)(tm) | \(-10\)(ktm) |
| \(x\) | \(ktm\) | \(ktm\) | \(16\) | \(ktm\) |
Vậy \(x=16\Leftrightarrow P\in Z\)
Đặt \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)\rightarrow\left(x;y;z\right)\)\(\Rightarrow\)\(x^2+y^2+z^2=4\)
\(P=\frac{x^3}{x+3y}+\frac{y^3}{y+3z}+\frac{z^3}{z+3x}=\frac{x^4}{x^2+3xy}+\frac{y^4}{y^2+3yz}+\frac{z^4}{z^2+3zx}\)
\(\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^2+y^2+z^2+3\left(xy+yz+zx\right)}\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^2+y^2+z^2+3\left(x^2+y^2+z^2\right)}=\frac{4^2}{4+3.4}=1\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=\frac{2}{\sqrt{3}}\)