Ta có: \(P=\dfrac{4\sqrt{x}+3}{x+\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\)

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

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

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

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

Để P nguyên thì \(\sqrt{x}+3⋮\sqrt{x}\)

mà \(\sqrt{x}⋮\sqrt{x}\)

nên \(3⋮\sqrt{x}\)

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

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

mà \(\sqrt{x}>0\forall x\) thỏa mãn ĐKXĐ

nên \(\sqrt{x}\in\left\{1;3\right\}\)

\(\Leftrightarrow x\in\left\{1;9\right\}\)

Kết hợp ĐKXĐ, ta được: \(x\in\left\{1;9\right\}\)

Vậy: Để P nguyên thì \(x\in\left\{1;9\right\}\)

\(P=\frac{4\sqrt{x}+3}{x+\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+1}\)

\(P=\frac{4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}}{\sqrt{x}+1}=\frac{4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(=\frac{x+4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+1\right)}\inℤ\Leftrightarrow x+4\sqrt{x}+3⋮\sqrt{x}\)

Giải tiếp nhé sau đó thử chọn :V

\(p=\frac{4\sqrt{x}+3}{x+\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+1}\)

\(=\frac{4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

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

\(=\frac{\sqrt{x}+3}{\sqrt{x}}=1+\frac{3}{\sqrt{x}}\)

Để \(x\in Z\Rightarrow P\in Z\)

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

\(\Leftrightarrow x=9\left(t.mĐKXĐ\right)\)

Với x >0

\(P=\frac{x+4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+3}{\sqrt{x}}=1+\frac{3}{\sqrt{x}}\)

Để P nhận giá trị nguyên thì \(\frac{3}{\sqrt{x}}\in Z\Leftrightarrow\sqrt{x}\in U\left(3\right)\Leftrightarrow\sqrt{x}\in\left\{1,3\right\}\)<=> x thuộc {1, 9}

Đáp án D

x = (5*a+7)

       _____

       (a+1)

\(x^2-2xy+y^2+3x-3y-4=0\)

\(\Leftrightarrow\left(x-y\right)^2+3\left(x-y\right)-4=0\)

\(\Leftrightarrow\left(x-y\right)\left(x-y+3\right)-4=0\)

Thay y = 3 vào biểu thức trên ta được : 

\(x\left(x-3\right)-4=0\)

\(\Leftrightarrow x^2-3x-4=0\Leftrightarrow\left(x-4\right)\left(x+1\right)=0\Leftrightarrow x=4;x=-1\)

Vậy với y = 3 thì x = 4 ; x = -1 

Thay y = 3 vào bthuc ta được :

x² - 6x + 9 + 3x - 9 - 4 = 0

<=> x² - 3x - 4 = 0

<=> ( x + 1 )( x - 4 ) = 0

<=> x = -1 hoặc x = 4