ĐKXĐ:\(x\ge0\)

Để \(\dfrac{2\sqrt{x}}{\sqrt{x}+3}\) nhận giá trị nguyên thì \(2\sqrt{x}⋮\sqrt{x}+3\)

                                                      \(\Leftrightarrow2\left(\sqrt{x}+3\right)-6⋮\sqrt{x}+3\)

                                                     \(\Leftrightarrow-6⋮\sqrt{x}+3hay\sqrt{x}+3\inƯ_{\left(-6\right)}\)

Vì \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+3\ge3\)

TH1.\(\sqrt{x}+3=3\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\left(tmĐKXĐ\right)\)

TH2.\(\sqrt{x}+3=6\Leftrightarrow\sqrt{x}=3\Leftrightarrow x=9\left(tmĐKXĐ\right)\)

Vậy,x={0;9}                                                                                                                                                                                                                                                                                               

Để A là số nguyên dương thì \(\left\{{}\begin{matrix}3\sqrt{x}+6-7⋮\sqrt{x}+2\\x>\dfrac{1}{9}\end{matrix}\right.\Leftrightarrow\sqrt{x}+2=7\)

hay x=25

Ta có: \(A=\dfrac{\sqrt{x}-4}{\sqrt{x}+3}=\dfrac{\sqrt{x}+3-7}{\sqrt{x}+3}=1-\dfrac{7}{\sqrt{x}+3}\) (ĐKXĐ: \(x\ge0\))

Để \(A\in Z\) thì \(\sqrt{x}+3\inƯ\left(7\right)=\left\{1;-1;7;-7\right\}\)

\(\Rightarrow x=16\) (TMĐK)

Vậy \(x=16\) thì \(A\in Z\)

\(A=\dfrac{\sqrt{x}-4}{\sqrt{x}+3}\)

\(A=1-\dfrac{7}{\sqrt{x}+3}\)

Để A nguyên thì \(\sqrt{x}+3\) phải là ước của 7 . 

\(\sqrt{x}+3=1;-1;7;-7\)

\(\Rightarrow16\)

\(A=\dfrac{\sqrt{x}-4}{\sqrt{x}+3}=\dfrac{\sqrt{x}+3-7}{\sqrt{x}+3}=1-\dfrac{7}{\sqrt{x}+3}\)

\(\Rightarrow\sqrt{x}+3\inƯ\left(-7\right)=\left\{1;7\right\}\)

Ta có :

\(K=\frac{2\sqrt{x}+3}{\sqrt{x}-5}=\frac{2\sqrt{x}-10}{\sqrt{x}-5}+\frac{13}{\sqrt{x}-5}=2+\frac{13}{\sqrt{x}-5}\)là số nguyên dương 

<=> 13 chia hết cho \(\sqrt{x}-5\)

<=> \(\sqrt{x}-5\inƯ\left(13\right)=\left\{-13;-1;1;13\right\}\)

<=> \(\sqrt{x}\in\left\{-12;4;6;18\right\}\)

<=> \(x\in\left\{16;36;324\right\}\) (vì \(\sqrt{x}\ge0\))

Do x nguyên và x có GTLN nên x = 324

ĐKXĐ: x>=0; x<>4

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

M nguyên khi \(x-2\sqrt{x}+4\sqrt{x}-8+12⋮\sqrt{x}-2\)

=>\(\sqrt{x}-2\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)

=>\(\sqrt{x}\in\left\{3;1;4;0;5;6;8;14\right\}\)

=>\(x\in\left\{9;1;16;0;25;36;64;196\right\}\)

Để M là số nguyên thì \(12\sqrt{x}+5⋮3\sqrt{x}-1\)

=>\(12\sqrt{x}-4+9⋮3\sqrt{x}-1\)

=>\(3\sqrt{x}-1\in\left\{1;-1;3;-3;9;-9\right\}\)

=>\(3\sqrt{x}\in\left\{2;0;4;10\right\}\)

=>\(\sqrt{x}\in\left\{0;\dfrac{2}{3};\dfrac{4}{3};\dfrac{10}{3}\right\}\)

mà x là số chính phương

nên x=0

\(M=\dfrac{12\sqrt{x}+5}{3\sqrt{x}-1}\)

\(M=\dfrac{12\sqrt{x}-4+9}{3\sqrt{x}-1}\)

\(M=\dfrac{4\left(3\sqrt{x}-1\right)+9}{3\sqrt{x}-1}\)

\(M=\dfrac{4\left(3\sqrt{x}-1\right)}{3\sqrt{x}-1}+\dfrac{9}{3\sqrt{x}-1}\)

\(M=4+\dfrac{9}{3\sqrt{x}-1}\)

M nguyên khi: 

\(9\) ⋮ \(3\sqrt{x}-1\)

Mà: \(3\sqrt{x}-1\ge-1\)

\(\Rightarrow3\sqrt{x}-1\in\left\{1;-1;3;9\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{\dfrac{2}{3};0;\dfrac{4}{3};\dfrac{10}{3}\right\}\)

\(\Rightarrow x\in\left\{\dfrac{4}{9};0;\dfrac{16}{9};\dfrac{100}{9}\right\}\)

Mà: x là số chính phương nên:

x = 0

\(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\left(đk:x\ge0,x\ne1\right)\)

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

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

\(=\dfrac{\left(\sqrt{x}-1\right)^2.2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}=\dfrac{2}{x+\sqrt{x}+1}\)

Để A nguyên thì: \(x+\sqrt{x}+1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)

Mà \(x+\sqrt{x}+1=\left(x+\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)

\(\Rightarrow x+\sqrt{x}+1\in\left\{1;2\right\}\)

+ Với \(x+\sqrt{x}+1=1\)

\(\Leftrightarrow\sqrt[]{x}\left(\sqrt{x}+1\right)=0\)

\(\Leftrightarrow x=0\left(tm\right)\left(do.\sqrt{x}+1\ge1>0\right)\)

+ Với \(x+\sqrt{x}+1=2\)

\(\Leftrightarrow\left(x+\sqrt{x}+\dfrac{1}{4}\right)=\dfrac{5}{4}\)

\(\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{2}\right)^2=\dfrac{5}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+\dfrac{1}{2}=\dfrac{\sqrt{5}}{2}\\\sqrt{x}+\dfrac{1}{2}=-\dfrac{\sqrt{5}}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{\sqrt{5}-1}{2}\\\sqrt{x}=-\dfrac{\sqrt{5}+1}{2}\left(VLý\right)\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{3-\sqrt{5}}{2}\left(tm\right)\)

Vậy \(S=\left\{1;\dfrac{3-\sqrt{5}}{2}\right\}\)

Biểu thức gì vậy bạn?

tìm các giá trị nguyên của x để biểu thức P=A.B  nhận giá trị nguyên

\(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\left(x\ge0;x\ne9\right)=\dfrac{\sqrt{x}+3-2}{\sqrt{x}+3}=1-\dfrac{2}{\sqrt{x}+3}\)

Để \(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\in Z\Leftrightarrow\dfrac{2}{\sqrt{x}+3}\in Z\)

\(\Leftrightarrow2⋮\sqrt{x}+3\\ \Leftrightarrow\sqrt{x}+3\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{-5;-4;-2;-1\right\}\\ \Leftrightarrow x\in\left\{1;4;16;25\right\}\)

Vậy \(x\in\left\{1;4;16;25\right\}\) thì \(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\in Z\)

Tick plz

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}+3\ne0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne-3\left(loại\right)\end{matrix}\right.\)\(\Rightarrow x\ge0\)

\(x\in Z\Rightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\in Z\Rightarrow\left(\sqrt{x}+1\right)⋮\left(\sqrt{x}+3\right)\)

\(\Rightarrow\left(\sqrt{x}+3-2\right)⋮\left(\sqrt{x}+3\right)\)

Vì \(\Rightarrow\left(\sqrt{x}+3\right)⋮\left(\sqrt{x}+3\right)\)

\(\Rightarrow2⋮\left(\sqrt{x}+3\right)\Rightarrow\sqrt{x}+3\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

Ta có bảng:

Vậy không có x nguyên thỏa mãn đề bài

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

Để \(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\in Z\) thì \(2⋮\sqrt{x}+3\Rightarrow\sqrt{x}+3\in\) Ư(2)\(=\left\{1;-1;2;-2\right\}\)

Vì \(\sqrt{x}\ge0\Rightarrow x\in\varnothing\)

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

Để A nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-3\)

\(\Leftrightarrow\sqrt{x}-3+4⋮\sqrt{x}-3\)

mà \(\sqrt{x}-3⋮\sqrt{x}-3\)

nên \(4⋮\sqrt{x}-3\)

\(\Leftrightarrow\sqrt{x}-3\inƯ\left(4\right)\)

\(\Leftrightarrow\sqrt{x}-3\in\left\{1;-1;2;-2;4;-4\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{4;2;5;1;7;-1\right\}\)

mà \(\sqrt{x}\ge0\forall x\) thỏa mãn ĐKXĐ

nên \(\sqrt{x}\in\left\{1;2;4;5;7\right\}\)

hay \(x\in\left\{1;4;16;25;49\right\}\)(nhận)

Vậy: Để A nguyên thì \(x\in\left\{1;4;16;25;49\right\}\)