Biểu thức gì vậy bạn?

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

a: Thay \(x=\dfrac{1}{4}\) vào A, ta được:

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

b: Ta có: \(B=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}-8}{x-5\sqrt{x}+6}\)

\(=\dfrac{x-4+\sqrt{x}-8}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x+\sqrt{x}-12}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{\sqrt{x}+4}{\sqrt{x}-2}\)

c: Để B là số tự nhiên thì \(\sqrt{x}+4⋮\sqrt{x}-2\)

\(\Leftrightarrow\sqrt{x}-2\in\left\{1;2;3;6\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{3;4;5;8\right\}\)

hay \(x\in\left\{16;25;64\right\}\)

\(A+B=1+\dfrac{1}{\sqrt{x}+3}\left(đk:x\ge0\right)\)

Để A+B nhận giá trị nguyên thì: \(\sqrt{x}+3\inƯ\left(1\right)=\left\{1;-1\right\}\)

Vì \(x\ge0\)

\(\Rightarrow x\in\varnothing\)

Để A+B là số nguyên thì \(\sqrt{x}+4⋮\sqrt{x}+3\)

\(\Leftrightarrow\sqrt{x}+3\in\left\{1;-1\right\}\)(vô lý)

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

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

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

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

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

b/

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

Để B nguyên

\(x-1=\left\{-1;-2;1;2\right\}\Rightarrow x=\left[0;-1;2;3\right]\)

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

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

\(P=B:A\)

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

P nguyên

=>căn x+3 thuộc Ư(-3)

=>căn x+3 thuộc {1;-1;3;-3}

=>căn x+3=3

=>x=0

\(P\in Z\Rightarrow3P\in Z\Rightarrow\dfrac{3\sqrt{x}+15}{3\sqrt{x}+1}\in Z\)

\(\Rightarrow1+\dfrac{14}{3\sqrt{x}+1}\in Z\)

\(\Rightarrow3\sqrt{x}+1=Ư\left(14\right)=\left\{1;2;7;14\right\}\) (do \(3\sqrt{x}+1\ge1\))

\(3\sqrt{x}+1=1\Rightarrow x=0\)

\(3\sqrt{x}+1=2\Rightarrow x=\dfrac{1}{9}\notin Z\) (loại)

\(3\sqrt{x}+1=7\Rightarrow x=4\)

\(3\sqrt{x}+1=14\Rightarrow x=\dfrac{169}{9}\notin Z\) (loại)

Thế \(x=\left\{0;4\right\}\) vào P đều thỏa mãn

Vậy ....