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a) \(ĐKXĐ:\hept{\begin{cases}x>0\\x\ne4\\x\ne9\end{cases}}\)
\(P=\left(\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}+2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right):\left(1-\frac{3\sqrt{x}-9}{x-9}\right)\)
\(=\left[\frac{-\left(\sqrt{x}-3\right)}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{\sqrt{x}+3}+\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]:\left[1-\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]\)
\(=\left[\frac{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]:\left(1-\frac{3}{\sqrt{x}+3}\right)\)
\(=\left[\frac{-x+9+x-4+x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]:\left(\frac{\sqrt{x}+3-3}{\sqrt{x}+3}\right)\)
\(=\frac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}:\frac{\sqrt{x}}{\sqrt{x}+3}\)
\(=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}.\frac{\sqrt{x}+3}{\sqrt{x}}=\frac{\sqrt{x}+2}{\sqrt{x}}\)
b) Ta có: \(P=\frac{\sqrt{x}+2}{\sqrt{x}}=1+\frac{2}{\sqrt{x}}\)
Vì \(x\inℤ\)\(\Rightarrow\)Để P nguyên thì \(\frac{2}{\sqrt{x}}\inℤ\)
\(\Rightarrow2⋮\sqrt{x}\)\(\Rightarrow\sqrt{x}\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
Vì \(\sqrt{x}>0\)\(\Rightarrow\sqrt{x}\in\left\{1;2\right\}\)
\(\Rightarrow x\in\left\{1;4\right\}\)
So sánh với ĐKXĐ ta thấy \(x=1\)thỏa mãn
\(\Rightarrow P=\frac{\sqrt{1}+2}{\sqrt{1}}=\frac{1+2}{1}=3\)
Vậy \(x=1\)khi đó \(P=3\)
\(P=\left(\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}+2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right)\div\left(1-\frac{3\sqrt{x}-9}{x-9}\right)\)
a) ĐK : \(\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)
\(=\left(\frac{3-\sqrt{x}}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{\sqrt{x}+3}-\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\div\left(1-\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right)\)
\(=\left(\frac{\left(3-\sqrt{x}\right)\left(x+\sqrt{3}\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\div\left(1-\frac{3}{\sqrt{x}+3}\right)\)
\(=\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\div\left(\frac{\sqrt{x}+3}{\sqrt{x}+3}-\frac{3}{\sqrt{x}+3}\right)\)
\(=\left(\frac{9-x+x-4-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\div\left(\frac{\sqrt{x}}{\sqrt{x}+3}\right)\)
\(=\frac{x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\times\frac{\sqrt{x}+3}{\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}+2}{\sqrt{x}}\)
b) Ta có : \(\frac{\sqrt{x}+2}{\sqrt{x}}=1+\frac{2}{\sqrt{x}}\)
Để P nguyên => \(\frac{2}{\sqrt{x}}\)nguyên
=> \(2⋮\sqrt{x}\)
=> \(\sqrt{x}\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
=> \(\sqrt{x}\in\left\{1;2\right\}\)( vì x ≥ 0 )
=> \(x\in\left\{1;4\right\}\Rightarrow x=1\)( vì x ≠ 4 )
Vậy với x = 1 thì P có giá trị nguyên
ĐK \(\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)
a, \(R=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\frac{3x-6\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}\)
b. \(R< -1\Rightarrow R+1< 0\Rightarrow\frac{3\sqrt{x}-9+\sqrt{x}+3}{\sqrt{x}+3}< 0\Rightarrow\frac{4\sqrt{x}-6}{\sqrt{x}+3}< 0\)
\(\Rightarrow0\le x< \frac{9}{4}\)
c. \(R=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}=3+\frac{-18}{\sqrt{x}+3}\)
Ta thấy \(\sqrt{x}+3\ge3\Rightarrow\frac{-18}{\sqrt{x}+3}\ge-6\Rightarrow3+\frac{-18}{\sqrt{x}+3}\ge-3\Rightarrow R\ge-3\)
Vậy \(MinR=-3\Leftrightarrow x=0\)
a) ĐKXĐ: \(\hept{\begin{cases}x-9\ne0\\\sqrt{x}\ge0\\\sqrt{x}\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne9\\x\ge0\\x\ne0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne9\\x>0\end{cases}}}\)
\(A=\left(\frac{x+3}{x-9}+\frac{1}{\sqrt{x}+3}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(\Leftrightarrow A=\left(\frac{x+3}{x-9}+\frac{\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right).\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(\Leftrightarrow A=\left(\frac{x+3}{x-9}+\frac{\sqrt{x}-3}{x-9}\right).\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(\Leftrightarrow A=\frac{x+\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(\Leftrightarrow A=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+3}.\frac{1}{\sqrt{x}}=\frac{\sqrt{x}+1}{\sqrt{x}+3}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{x-9}\)
b) \(x=\sqrt{6+4\sqrt{2}}-\sqrt{3+2\sqrt{2}}\)
\(\Leftrightarrow x=\sqrt{4+4\sqrt{2}+2}-\sqrt{2+2\sqrt{2}+1}\)
\(\Leftrightarrow x=\sqrt{\left(2+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}+1\right)^2}\)
\(\Leftrightarrow x=\left|2+\sqrt{2}\right|-\left|\sqrt{2}+1\right|\)
\(\Leftrightarrow x=2+\sqrt{2}-\sqrt{2}-1=1\left(TM\right)\)
Vậy với x= 1 thì giá trị của biểu thức \(A=\frac{\left(1+1\right)\left(1-3\right)}{1-9}=\frac{2.\left(-2\right)}{-8}=\frac{-4}{-8}=\frac{1}{2}\)
c)
Ta có :
\(\frac{x-9}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+3}{\sqrt{x}+1}=1+\frac{2}{\sqrt{x}+1}\)
+) \(\frac{1}{A}\)nguyên
\(\Leftrightarrow1+\frac{2}{\sqrt{x}+1}\)nguyên
\(\Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)\)
\(\Leftrightarrow x=1\)
Vậy ..............
Q= \(\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)+\(\frac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)= \(\frac{2\sqrt{x}-9-\left(x-9\right)+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)=\(\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)=\(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)=\(\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
b) Q <1 <=> \(\frac{\sqrt{x}-3+4}{\sqrt{x}-3}< 1< =>1+\frac{4}{\sqrt{x}-3}\)<1 <=> \(\frac{4}{\sqrt{x}-3}< 0\) <=> \(\sqrt{x}-3< 0< =>\sqrt{x}< 3\)<=> \(0\le\)x< 9
c) Q = 1 \(+\frac{4}{\sqrt{x}-3}\) là số nguyên khi 4 chia hết cho\(\sqrt{x}-3\) <=> \(\sqrt{x}-3=1;\sqrt{x}-3=-1;\sqrt{x}-3=2\);\(\sqrt{x}-3=-2;\sqrt{x}-3=4;\sqrt{x}-3=-4\)
<=> x= 16; x = 4; x = 25; x = 1 ; x = 49
Bài làm của bạn Mạnh có hai lỗi:
+) ĐKXĐ: \(\hept{\begin{cases}x-5\sqrt{x}+6\ne0;\sqrt{x}-2\ne0;3-\sqrt{x}\ne0\\x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne4;9\end{cases}}\)
+) Vì ko có điều kiện nên câu c chưa loại nghiệm. x = 4 loại nhé
Với x >= 0 ; x khác 9
\(B=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}=\frac{-3\sqrt{x}-3}{x-9}=\frac{-3\left(\sqrt{x}+1\right)}{x-9}\)
\(\frac{B}{A}=\frac{-3\left(\sqrt{x}+1\right)}{x-9}:\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{-3}{\sqrt{x}+3}+\frac{1}{2}< 0\)
\(\Leftrightarrow\frac{-6+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}< 0\Rightarrow\sqrt{x}-3< 0\Leftrightarrow x< 9\)
Kết hợp đk vậy 0 =< x < 9