1) +) ta có : \(C-\dfrac{1}{3}\Leftrightarrow\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{3}=\dfrac{3\sqrt{x}-x+\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}\)

\(=\dfrac{-\left(x-4\sqrt{x}+4\right)+3}{3\left(x+\sqrt{x}+1\right)}=\dfrac{-\left(\sqrt{x}-2\right)^2+3}{3\left(x+\sqrt{x}+1\right)}\)

không thể cm được đâu bn --> xem lại đề

2) +) ta có : \(D=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}=1-\dfrac{3}{\sqrt{x}+2}\)

--> để \(D\in Z\Leftrightarrow\sqrt{x}+2\) là ước của 3 \(\Leftrightarrow\sqrt{x}+2\in\left\{\pm1;\pm3\right\}\)

\(\Leftrightarrow x=1\) vậy \(x=1\)

3) +) tương tự 2)

4) a) +) điều kiện xác định : \(x>0;x\ne4\)

ta có : \(A=\left(\dfrac{2}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}}\right):\dfrac{\sqrt{x}-2}{x+3\sqrt{x}}\)

\(\Leftrightarrow A=\left(\dfrac{2\sqrt{x}-\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}}{\sqrt{x}-2}=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\)

b) ta có : \(A=3\Leftrightarrow\dfrac{\sqrt{x}-3}{\sqrt{x}-2}=3\Leftrightarrow\sqrt{x}-3=3\sqrt{x}-6\)

\(\Leftrightarrow2\sqrt{x}=3\Leftrightarrow\sqrt{x}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{9}{4}\) vậy \(x=\dfrac{9}{4}\)

c) ta có : \(B=A.\dfrac{\sqrt{x}+3}{\sqrt{x}+2}=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}.\dfrac{\sqrt{x}+3}{\sqrt{x}+2}=\dfrac{x-9}{x-4}=1-\dfrac{5}{x-4}\)

tương tự 2 )
\(\)

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)

Ta có: \(A=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)

\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{3}{\sqrt{x}+3}\)

b: Thay x=16 vào A, ta được:

\(A=\dfrac{3}{4+3}=\dfrac{3}{7}\)

c)\(A=\dfrac{3}{\sqrt{x}+3}=\dfrac{1}{3}\)

\(\Rightarrow\sqrt{x}+3=9\\ \Rightarrow\sqrt{x}=6\\ \Rightarrow x=36\)

d) \(A=\dfrac{3}{\sqrt{x}+3}\)

Vì \(3>0;\sqrt{x}+3>0\Rightarrow\dfrac{3}{\sqrt{x}+3}>0\)

e) \(2A\in Z\Rightarrow\dfrac{6}{\sqrt{x}+3}\in Z \Rightarrow6⋮x+3\\\Rightarrow\sqrt{x}+3\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\Rightarrow x=\left\{0;9\right\}\)

\(M=\dfrac{\sqrt{a}+2}{\sqrt{a}-2}=\dfrac{\sqrt{a}-2}{\sqrt{a}-2}+\dfrac{4}{\sqrt{a}-2}=1+\dfrac{4}{\sqrt{a}-2}\in Z\)

\(\Rightarrow\sqrt{a}-2\inƯ\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)

Do \(\sqrt{a}\ge0\)

\(\Leftrightarrow\sqrt{a}\in\left\{3;1;4;0;6\right\}\)

\(\Rightarrow a\in\left\{9;1;16;0;36\right\}\)

Đề yêu cầu tìm a nguyên thì đúng hơn.

Vì yêu cầu tìm a hữu tỉ bài này sẽ có vô số số hữu tỉ thỏa mãn

Ta có \(M=\dfrac{\sqrt{a}+2}{\sqrt{a}-2}=1+\dfrac{4}{\sqrt{a}-2}\)

Để \(M\) nhận giá trị nguyên thì \(4⋮\left(\sqrt{a}-2\right)\Rightarrow\left(\sqrt{a}-2\right)\inƯ\left(4\right)\)

\(\Rightarrow\sqrt{a}-2\in\left\{1;-1;2;-2;4;-4\right\}\)

\(\Rightarrow\sqrt{a}\in\left\{3;1;4;0;6\right\}\)

\(\Rightarrow a\in\left\{\sqrt{3};1;2;0;\sqrt{6}\right\}\)

mà a là số hữu tỉ nên \(a\in\left\{1;2;0\right\}\)

a) ĐKXĐ: \(x\ge0;x\ne9;x\ne4\)

\(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)

\(M=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\)

\(M=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(M=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(M=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(M=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(M=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

b) Ta có M ϵ Z thì \(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3+4}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3}{\sqrt{x}-3}+\dfrac{4}{\sqrt{x}-3}=1+\dfrac{4}{\sqrt{x}-3}\)

Phải thuộc Z vậy:

4 ⋮ \(\sqrt{x}-3\)

\(\Rightarrow\sqrt{x}-3\inƯ\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)

Mà: \(x\ge0,x\ne4,x\ne9\) nên \(\sqrt{x}-3\in\left\{1;2;-2;4\right\}\)

\(\Rightarrow x\in\left\{16;25;1;49\right\}\)

a) ĐKXĐ: \(x\ge0;x\ne9\) . Rút gọn: \(A=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{\sqrt{x}-3}-\dfrac{x-4\sqrt{x}+7}{x-2\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{\sqrt{x}-3}-\dfrac{x-4\sqrt{x}+7}{x+\sqrt{x}-3\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{\sqrt{x}-3}-\dfrac{x-4\sqrt{x}+7}{\sqrt{x}\left(\sqrt{x}+1\right)-3\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{\sqrt{x}-3}-\dfrac{x-4\sqrt{x}+7}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)+\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)-x+4\sqrt{x}-7}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{x-3\sqrt{x}-2\sqrt{x}+6+x+\sqrt{x}+3\sqrt{x}+3-x+4\sqrt{x}-7}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{x+\sqrt{x}+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)

A>-1\(\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)>-1\(\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+1>0\Leftrightarrow\dfrac{\sqrt{x}+2+\sqrt{x}-3}{\sqrt{x}-3}>0\Leftrightarrow\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}>0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2\sqrt{x}-1>0\\\sqrt{x}-3>0\end{matrix}\right.\\\left\{{}\begin{matrix}2\sqrt{x}-1< 0\\\sqrt{x}-3< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}>0,5\\\sqrt{x}>3\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}< 0,5\\\sqrt{x}< 3\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0,25\\x>9\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0,25\\x< 9\end{matrix}\right.\end{matrix}\right.\Leftrightarrow}}\left[{}\begin{matrix}x>9\\0\le x< 0,25\end{matrix}\right.\)

ĐKXĐ: \(x\ge0;x\ne4\)

\(A=\dfrac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

b. \(x=36\Rightarrow A=\dfrac{\sqrt{36}}{\sqrt{36}-2}=\dfrac{6}{6-2}=\dfrac{3}{2}\)

c. \(A=-\dfrac{1}{3}\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}=-\dfrac{1}{3}\Rightarrow3\sqrt{x}=2-\sqrt{x}\)

\(\Rightarrow4\sqrt{x}=2\Rightarrow\sqrt{x}=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{4}\)

d. \(A>0\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}>0\Rightarrow\sqrt{x}-2>0\Rightarrow x>4\)

e. \(A=\dfrac{\sqrt{x}-2+2}{\sqrt{x}-2}=1+\dfrac{2}{\sqrt{x}-2}\in Z\Rightarrow\sqrt{x}-2=Ư\left(2\right)\)

\(\Rightarrow\sqrt{x}-2=\left\{-2;-1;1;2\right\}\)

\(\Rightarrow\sqrt{x}=\left\{0;1;3;4\right\}\Rightarrow x=\left\{0;1;9;16\right\}\)

a: Ta có: \(A=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)

\(=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

b: Thay x=36 vào A, ta được:

\(A=\dfrac{6}{6-2}=\dfrac{6}{4}=\dfrac{3}{2}\)

c: Để \(A=-\dfrac{1}{3}\) thì \(3\sqrt{x}=-\sqrt{x}+2\)

\(\Leftrightarrow4\sqrt{x}=2\)

hay \(x=\dfrac{1}{4}\)

d: Để A>0 thì \(\sqrt{x}-2>0\)

hay x>4

e: Để A nguyên thì \(\sqrt{x}⋮\sqrt{x}-2\)

\(\Leftrightarrow\sqrt{x}-2\in\left\{-1;1;2;-2\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{1;3;4;0\right\}\)

hay \(x\in\left\{1;9;16;0\right\}\)