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a, Q=\(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\left(x\ge0,x\ne4,x\ne9\right)\)
=\(\frac{2\sqrt{x}-9}{x-2\sqrt{x}-3\sqrt{x}+6}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}\)
=\(\frac{2\sqrt{x}-9}{\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)}-\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
= \(\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
=\(\frac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
=\(\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{x-2\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
=\(\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
b, Để Q<1 <=> \(\frac{\sqrt{x}+1}{\sqrt{x}-3}< 1\)
<=> \(\frac{\sqrt{x}+1}{\sqrt{x}-3}-1< 0\) <=> \(\frac{\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}-3}< 0\) <=> \(\frac{4}{\sqrt{x}-3}< 0\)
<=> \(\sqrt{x}-3< 0\) <=> \(\sqrt{x}< 3\) <=> x<9. Kết hợp vs đk => \(0\le x< 9\) và \(x\ne2\)
Vậy Q<1 <=> \(0\le x< 9\) và \(x\ne2\)
c, Có \(Q=\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)
Để Q\(\in Z\) <=> \(\frac{4}{\sqrt{x}-3}\in Z\)
Vs \(x\in Z\) => \(\left\{{}\begin{matrix}\sqrt{x}\in Z\\\sqrt{x}\notin Z\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\sqrt{x}-3\in Z\\\sqrt{x}-3\notin Z\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\frac{4}{\sqrt{x}-3}\in Z\left(tm\right)\\\frac{4}{\sqrt{x}-3}\notin Z\left(ktm\right)\end{matrix}\right.\)
=> \(\sqrt{x}-3\inƯ\left(4\right)=\left\{\pm1,\pm2,\pm4\right\}\)
<=> \(\sqrt{x}\in\left\{4,2,1,5,-1,7\right\}\)
mà \(\sqrt{x}\ge0,\sqrt{x}\ne2\)
=> \(\sqrt{x}\in\left\{1,4,5,7\right\}\)
<=> x\(\in\left\{1,16,25,49\right\}\)
Vậy x\(\in\left\{1,16,25,49\right\}\) thì Q\(\in Z\)
\(M=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
b/ \(x=11-6\sqrt{2}=\left(3-\sqrt{2}\right)^2\Rightarrow\sqrt{x}=3-\sqrt{2}\)
\(\Rightarrow M=\frac{3-\sqrt{2}+1}{3-\sqrt{2}-3}=1-2\sqrt{2}\)
c/ \(M=2\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}-3}=2\Leftrightarrow\sqrt{x}+1=2\sqrt{x}-6\)
\(\Leftrightarrow\sqrt{x}=7\Rightarrow x=49\)
d/ \(M< 1\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}-3}< 1\Leftrightarrow\frac{\sqrt{x}-3+4}{\sqrt{x}-3}< 1\Leftrightarrow1+\frac{4}{\sqrt{x}+3}< 1\)
\(\Leftrightarrow\frac{4}{\sqrt{x}+3}< 0\) \(\Rightarrow\) không tồn tại m thỏa mãn
e/ \(M=\frac{\sqrt{x}+1}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)
Để M nguyên \(\Leftrightarrow\frac{4}{\sqrt{x}-3}\) nguyên
\(\Rightarrow4⋮\sqrt{x}-3\Rightarrow\sqrt{x}-3=Ư\left(4\right)=\left\{-2;-1;1;2;4\right\}\)
\(\Rightarrow\sqrt{x}=\left\{1;2;4;5;7\right\}\)
\(\Rightarrow x=\left\{1;4;16;25;49\right\}\)
Kết hợp ĐKXĐ ban đầu ta được \(x=\left\{1;16;25;49\right\}\)
1: Ta có: \(P=\left(1-\frac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)
\(=\left(\frac{1+\sqrt{x}-\sqrt{x}}{1+\sqrt{x}}\right):\left(\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{1}{1+\sqrt{x}}:\frac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{1}{1+\sqrt{x}}:\frac{\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{1}{\sqrt{x}+1}\cdot\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-1}\)
\(=\frac{x-5\sqrt{x}+6}{x-1}\)
Ta có: \(x=\sqrt{4+2\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{1+2\cdot1\cdot\sqrt{3}+3}+\sqrt{3-2\cdot\sqrt{3}\cdot2+4}\)
\(=\sqrt{\left(1+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\left|1+\sqrt{3}\right|+\left|\sqrt{3}-2\right|\)
\(=1+\sqrt{3}+2-\sqrt{3}=3\)
Thay x=3 vào biểu thức \(P=\frac{x-5\sqrt{x}+6}{x-1}\), ta được:
\(P=\frac{3-5\cdot\sqrt{3}+6}{3-2}=\frac{9-5\sqrt{3}}{1}=9-5\sqrt{3}\)
Vậy: Khi x=3 thì giá trị của biểu thức P là \(9-5\sqrt{3}\)
2: Để P<0 thì \(\frac{x-5\sqrt{x}+6}{x-1}< 0\)
Trường hợp 1:
\(\left\{{}\begin{matrix}x-5\sqrt{x}+6>0\\x-1< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)>0\\x-1< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}-2>0\\\sqrt{x}-3>0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}-2< 0\\\sqrt{x}-3< 0\end{matrix}\right.\end{matrix}\right.\\x< 1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}>2\\\sqrt{x}>3\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}< 2\\\sqrt{x}< 3\end{matrix}\right.\end{matrix}\right.\\x< 1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}x>4\left(ktm\right)\\x>9\left(ktm\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x< 4\\x< 9\end{matrix}\right.\end{matrix}\right.\\x< 1\end{matrix}\right.\Leftrightarrow x< 1\)
Vậy: Khi x<1 thì P<0
Lời giải:
1. \(A=\frac{x-3\sqrt{x}-x+9}{x-9}=\frac{9-3\sqrt{x}}{x-9}=\frac{3(3-\sqrt{x})}{(\sqrt{x}-3)(\sqrt{x}+3)}=\frac{-3}{\sqrt{x}+3}\)
\(x=57-24\sqrt{3}=48+9-2\sqrt{48.9}=(\sqrt{48}-\sqrt{9})^2\)
\(\Rightarrow \sqrt{x}=\sqrt{48}-\sqrt{9}=4\sqrt{3}-3\)
\(\Rightarrow A=\frac{-3}{4\sqrt{3}}=\frac{-\sqrt{3}}{4}\)
2. \(B=\frac{(3-\sqrt{x})(3+\sqrt{x})}{(\sqrt{x}+3)(\sqrt{x}-2)}+\frac{3-\sqrt{x}}{2-\sqrt{x}}-\frac{\sqrt{x}+2}{\sqrt{x}+3}=\frac{3-\sqrt{x}}{\sqrt{x}-2}+\frac{3-\sqrt{x}}{2-\sqrt{x}}-\frac{\sqrt{x}+2}{\sqrt{x}+3}\)
\(=-\frac{\sqrt{x}+2}{\sqrt{x}+3}\)
3.
\(P=A:B=\frac{-3}{\sqrt{x}+3}:\frac{-(\sqrt{x}+2)}{\sqrt{x}+3}=\frac{3}{\sqrt{x}+2}\)
$P$ nguyên $\Leftrightarrow 3\vdots \sqrt{x}+2$
Mà $\sqrt{x}+2\geq 2$ nên $\sqrt{x}+2=3$
$\Rightarrow x=1$
\(P=\left(\frac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\left(\frac{2\sqrt{x}-2-\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\right)\)
\(=\left(\frac{3x+3-2x+6\sqrt{x}-x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-3}\right)\)
\(=\frac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}=\frac{3}{\sqrt{x}+3}\)
\(x=20-6\sqrt{11}=\left(\sqrt{11}-3\right)^2\Rightarrow\sqrt{x}=\sqrt{11}-3\)
\(\Rightarrow P=\frac{3}{\sqrt{11}-3+3}=\frac{3\sqrt{11}}{11}\)
\(P>\frac{1}{2}\Rightarrow\frac{3}{\sqrt{x}+3}>\frac{1}{2}\Rightarrow\sqrt{x}+3< 6\Rightarrow x< 9\)
Kết hợp ĐKXD \(\Rightarrow0\le x< 9\)
\(Q=\frac{2\sqrt{x}}{3}.\frac{3}{\left(\sqrt{x}+3\right)}=\frac{2\sqrt{x}}{\sqrt{x}+3}=2-\frac{6}{\sqrt{x}+3}\)
Để Q nguyên \(\Rightarrow\sqrt{x}+3=Ư\left(6\right)\)
Mà \(\sqrt{x}+3\ge3\Rightarrow\sqrt{x}+3=\left\{3;6\right\}\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+3=3\\\sqrt{x}+3=6\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=9\left(l\right)\end{matrix}\right.\)
Bài 2:
a) Ta có: \(M=\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{3x+9}{x-9}\)
\(=\frac{\sqrt{x}\cdot\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{2\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{3x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{3}{\sqrt{x}+3}\)
bạn ơi đây là x thuộc R không là x nguyên. Hình như bạn làm nhầm rồi.
1: Ta có: \(A=\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}-1}{3-\sqrt{x}}\)
\(=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{2\sqrt{x}-9-\left(x-9\right)+2x-5\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{-3\sqrt{x}-7+2x-x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x-3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}-3}\)
2: Để A<1 thì \(\frac{\sqrt{x}-1}{\sqrt{x}-3}< 1\)
\(\Leftrightarrow\sqrt{x}-1< \sqrt{x}-3\)
\(\Leftrightarrow\sqrt{x}-1-\sqrt{x}+3< 0\)
\(\Leftrightarrow2< 0\)(vô lý)
Vậy: Không có giá trị nào của x thỏa mãn A<1
3: Để A nhận giá trị nguyên thì \(\sqrt{x}-1⋮\sqrt{x}-3\)
\(\Leftrightarrow\sqrt{x}-3+2⋮\sqrt{x}-3\)
mà \(\sqrt{x}-3⋮\sqrt{x}-3\)
nên \(2⋮\sqrt{x}-3\)
\(\Leftrightarrow\sqrt{x}-3\inƯ\left(2\right)\)
\(\Leftrightarrow\sqrt{x}-3\in\left\{1;-1;2;-2\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{4;2;5;1\right\}\)
\(\Leftrightarrow x\in\left\{16;4;25;1\right\}\)
mà \(x\ge0\) và \(x\notin\left\{4;9\right\}\)(ĐKXĐ)
nên \(x\in\left\{16;25;1\right\}\)
Vậy: để A nhận giá trị nguyên thì \(x\in\left\{16;25;1\right\}\)