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ĐK \(\hept{\begin{cases}x\ge0\\x\ne4;x\ne9\end{cases}}\)
a. P=\(\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-3}\right):\frac{2\sqrt{x}+2-\sqrt{x}}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}+2+\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}:\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}+2+x-9-x+4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}+1}{\sqrt{x}+2}=\frac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(=\frac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
b. Với \(x=4-2\sqrt{3}\Rightarrow P=\frac{\sqrt{4-2\sqrt{3}}+1}{4-2\sqrt{3}-4}=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}+1}{-2\sqrt{3}}\)
\(=\frac{\sqrt{3}-1+1}{-2\sqrt{3}}=-\frac{1}{2}\)
c. Để \(\frac{1}{P}\le\frac{-5}{2}\Leftrightarrow\frac{x-4}{\sqrt{x}+1}+\frac{5}{2}\le0\Leftrightarrow\frac{2x-8+5\sqrt{x}+5}{2\left(\sqrt{x}+1\right)}\le0\)
\(\Leftrightarrow\frac{2x+5\sqrt{x}-3}{2\left(\sqrt{x}+1\right)}\le0\Leftrightarrow2x+5\sqrt{x}-3\le0\)vì \(2\left(\sqrt{x}+1\right)\ge0\forall x\ge0\)
\(\Leftrightarrow\left(\sqrt{x}+3\right)\left(2\sqrt{x}-1\right)\le0\Leftrightarrow2\sqrt{x}-1\le0\Leftrightarrow0\le x\le\frac{1}{4}\left(tm\right)\)
Vậy với \(0\le x\le\frac{1}{4}\)thì \(\frac{1}{P}\le-\frac{5}{2}\)
d. Ta có \(B=P\left(\sqrt{x}-2\right)=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}+1}{\sqrt{x}+2}=1-\frac{1}{\sqrt{x}+2}\)
Gỉa sử \(B\in Z\Leftrightarrow\sqrt{x}+2\inƯ\left(1\right)\Leftrightarrow\sqrt{x}+2\in\left\{-1;1\right\}\Leftrightarrow x\in\left\{\phi\right\}\)
Vậy B không nhận giá trị nguyên với mọi x để P có nghĩa
ĐK: \(x-9\ne0\Rightarrow x\ne9\)
\(\sqrt{x}\ge0\Rightarrow x\ge0\)
\(x+\sqrt{x}-6\ne0\Rightarrow x+3\sqrt{x}-2\sqrt{x}-6\ne0\Rightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)\ne0\)
\(\Rightarrow\sqrt{x}-2\ne0\Rightarrow\sqrt{x}\ne2\Rightarrow x\ne4\)
ĐKXĐ: \(x\ge0;x\ne4;x\ne9\)
\(A=\left(\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{1}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\left(\frac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\frac{\sqrt{x}}{\sqrt{x}+3}:\left(\frac{1+\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\)
\(=\frac{\sqrt{x}}{\sqrt{x}+3}:\frac{1+x-9-x+4\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}+3}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{4\sqrt{x}-12}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{4\left(\sqrt{x}-3\right)}\)
2, Với \(x=\frac{25}{16}\)\(\Rightarrow\sqrt{x}=\sqrt{\frac{25}{16}}=\frac{5}{4}\)
\(A=\frac{\frac{5}{4}\left(\frac{5}{4}-2\right)}{4\left(\frac{5}{4}-3\right)}=\frac{5}{4}.\left(-\frac{3}{4}\right):4\left(-\frac{7}{4}\right)=-\frac{15}{16}:-7=\frac{15}{112}\)
\(\orbr{\begin{cases}\orbr{\begin{cases}\\\end{cases}}\\\end{cases}}\)\(\orbr{\begin{cases}\orbr{\begin{cases}\sqrt{x}-2< 0\\\sqrt{x}-3>0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}< 2\\\sqrt{x}>3\end{cases}}\Rightarrow\orbr{\begin{cases}x< 4\\x>9\end{cases}}}\\\orbr{\begin{cases}\sqrt{x}-2>0\\\sqrt{x}-3< 0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}>2\\\sqrt{x}< 3\end{cases}\Rightarrow\orbr{\begin{cases}x>4\\x< 9\end{cases}}}}\end{cases}}\)
a: \(P=\dfrac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{x-1-x+4}\)
\(=\dfrac{1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}-2}{3}=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
b: P=1/4
=>\(\dfrac{\sqrt{x}-2}{3\sqrt{x}}=\dfrac{1}{4}\)
=>\(4\left(\sqrt{x}-2\right)=3\sqrt{x}\)
=>\(4\sqrt{x}-8-3\sqrt{x}=0\)
=>\(\sqrt{x}=8\)
=>x=64
c: Khi \(x=4+2\sqrt{3}\) thì \(P=\dfrac{\sqrt{4+2\sqrt{3}}-2}{3\cdot\sqrt{4+2\sqrt{3}}}\)
\(=\dfrac{\sqrt{3}+1-2}{3\left(\sqrt{3}+1\right)}=\dfrac{\sqrt{3}-1}{3\sqrt{3}+3}=\dfrac{2-\sqrt{3}}{3}\)
ĐK: \(x\ge0,x\ne9\).
\(P=\frac{x+2\sqrt{x}-10}{x-\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}+2}+\frac{\sqrt{x}-2}{3-\sqrt{x}}\)
\(=\frac{x+2\sqrt{x}-10}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}+\frac{\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+2\sqrt{x}-10+x-6\sqrt{x}+9-x+4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x-4\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
Tại \(x=1-\frac{\sqrt{3}}{2}\): \(\sqrt{x}=\sqrt{1-\frac{\sqrt{3}}{2}}=\frac{1}{2}\sqrt{4-2\sqrt{3}}=\frac{1}{2}\sqrt{\left(\sqrt{3}-1\right)^2}=\frac{1}{2}\left(\sqrt{3}-1\right)\)
\(P=\frac{\sqrt{3}-1-2}{\sqrt{3}-1+4}=\frac{\sqrt{3}-3}{\sqrt{3}+3}=-2+\sqrt{3}\).
\(P=\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac{\sqrt{x}+2-3}{\sqrt{x}+2}=1-\frac{3}{\sqrt{x}+2}\)
Dễ thấy \(P< 1\).
\(\sqrt{x}+2\ge2\Rightarrow\frac{3}{\sqrt{x}+2}\le\frac{3}{2}\Rightarrow P\ge1-\frac{3}{2}=-\frac{1}{2}\)
Suy ra \(-\frac{1}{2}\le P< 1\)do đó \(P\)chỉ có thể nhận \(1\)giá trị nguyên duy nhất là \(P=0\).
Với \(P=0\Rightarrow x=1\).
Do đó ta có đpcm.