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

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

Biểu thức gì vậy bạn?

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

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

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

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

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

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

b: \(A-5=\dfrac{2x-4\sqrt{x}+2}{\sqrt{x}}=\dfrac{2\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>=0\)

=>A>=5

a = 9;0

a: \(P=\dfrac{x-\sqrt{x}-1-\sqrt{x}+1}{x-1}\cdot\dfrac{4\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)^2}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)\cdot4\left(\sqrt{x}-2\right)}{\sqrt{x}\left(x-1\right)}=\dfrac{4}{x-1}\)

Để P nguyên dương thì x-1 thuộc {1;4;2}

=>x thuộc {2;5;3}

b: x+y+z=0

=>x=-y-z; y=-x-z; z=-x-y

\(P=\dfrac{x^2}{y^2+z^2-\left(y+z\right)^2}+\dfrac{y^2}{z^2+x^2-\left(x+z\right)^2}+\dfrac{z^2}{x^2+y^2-\left(x+y\right)^2}\)

\(=\dfrac{x^2}{-2yz}+\dfrac{y^2}{-2xz}+\dfrac{z^2}{-2xy}\)

\(=\dfrac{x^3+y^3+z^3}{2xyz}\cdot\left(-1\right)\)

\(=-\dfrac{\left(x+y\right)^3+z^3-3xy\left(x+y\right)}{2xyz}\)

\(=-\dfrac{\left(-z\right)^3+z^3-3xy\cdot\left(-z\right)}{2xyz}=-\dfrac{3}{2}\)

ĐKXĐ: \(x>0;x\ne9\)

\(P=\left(\dfrac{x+7}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}-\dfrac{4\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\right)\left(\dfrac{\sqrt{x}+6}{\sqrt{x}}\right)\)

\(=\left(\dfrac{x+7-4\sqrt{x}-4+\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\right)\left(\dfrac{\sqrt{x}+6}{\sqrt{x}}\right)\)

\(=\left(\dfrac{x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\right).\left(\dfrac{\sqrt{x}+6}{\sqrt{x}}\right)\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}.\dfrac{\left(\sqrt{x}+6\right)}{\sqrt{x}}\)

\(=\dfrac{\sqrt{x}+6}{\sqrt{x}+1}\)

b.

Ta có \(P=\dfrac{\sqrt{x}+1+5}{\sqrt{x}+1}=1+\dfrac{5}{\sqrt{x}+1}\)

Do \(\sqrt{x}+1>0\Rightarrow\dfrac{5}{\sqrt{x}+1}>0\Rightarrow P>1\)

\(P=\dfrac{6\left(\sqrt{x}+1\right)-5\sqrt{x}}{\sqrt{x}+1}=6-\dfrac{5\sqrt{x}}{\sqrt{x}+1}\)

Do \(\left\{{}\begin{matrix}5\sqrt{x}>0\\\sqrt{x}+1>0\end{matrix}\right.\) ;\(\forall x>0\Rightarrow\dfrac{5\sqrt{x}}{\sqrt{x}+1}>0\)

\(\Rightarrow P< 6\Rightarrow1< P< 6\)

Mà P nguyên \(\Rightarrow P=\left\{2;3;4;5\right\}\)

- Để \(P=2\Rightarrow\dfrac{\sqrt{x}+6}{\sqrt{x}+1}=2\Rightarrow\sqrt{x}+6=2\sqrt{x}+2\Rightarrow x=16\)

- Để \(P=3\Rightarrow\dfrac{\sqrt{x}+6}{\sqrt{x}+1}=3\Rightarrow\sqrt{x}+6=3\sqrt{x}+3\Rightarrow\sqrt{x}=\dfrac{3}{2}\Rightarrow x=\dfrac{9}{4}\)

- Để \(P=4\Rightarrow\dfrac{\sqrt{x}+6}{\sqrt{x}+1}=4\Rightarrow\sqrt{x}+6=4\sqrt{x}+4\Rightarrow\sqrt{x}=\dfrac{2}{3}\Rightarrow x=\dfrac{4}{9}\)

- Để \(P=5\Rightarrow\dfrac{\sqrt{x}+6}{\sqrt{x}+1}=5\Rightarrow\sqrt{x}+6=5\sqrt{x}+5\Rightarrow\sqrt{x}=\dfrac{1}{4}\Rightarrow x=\dfrac{1}{16}\)

ĐKXĐ: x>0; x<>1

a: \(P=\frac{x}{x-\sqrt{x}}+\frac{2}{x+2\sqrt{x}}+\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+2\sqrt{x}\right)}\)

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

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

\(=\frac{x\sqrt{x}+2x+2\sqrt{x}-2+x+2}{\sqrt{x}\cdot\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\frac{x\sqrt{x}+3x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)\cdot\left(\sqrt{x}-1\right)}\)

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

b: \(x=3+2\sqrt2=\left(\sqrt2+1\right)^2\)

\(P=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(=\frac{\sqrt{\left(\sqrt2+1\right)^2}+1}{\sqrt{\left(\sqrt2+1\right)^2}-1}=\frac{\sqrt2+1+1}{\sqrt2+1-1}\)

\(=\frac{2+\sqrt2}{\sqrt2}=\sqrt2+1\)

c: Để P nguyên thì \(\sqrt{x}+1\) ⋮\(\sqrt{x}-1\)

=>\(\sqrt{x}-1+2\) ⋮\(\sqrt{x}-1\)

=>2⋮\(\sqrt{x}-1\)

=>\(\sqrt{x}-1\in\left\lbrace1;-1;2;-2\right\rbrace\)

=>\(\sqrt{x}\in\left\lbrace2;0;3\right\rbrace\)

=>x∈{0;4;9}

Kết hợp ĐKXĐ, ta được: x∈{4;9}