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 \(x\in\left\{1;9\right\}\)

ĐKXĐ : \(x\ne2\)

Ta có HĐT sau (a - b)(a + b) = a² - ab + ab - b² = a² - b² 

Áp dụng vào bài toán ta có:

 x⁴ + 3 = (x⁴ - 16) + 19

= [(x²)² - 4²] + 19

= (x² - 4)(x² + 4) + 19

= (x - 2)(x + 2)(x² + 4) + 19

Từ đó \(A=\dfrac{x^2+3}{x-2}=\dfrac{\left(x-2\right).\left(x+2\right).\left(x^2+4\right)+19}{x-2}\)

\(=\left(x+2\right).\left(x^2+4\right)+\dfrac{19}{x-2}\)

Do \(x\inℤ\) nên \(A\inℤ\Leftrightarrow19⋮x-2\)

\(\Leftrightarrow x-2\inƯ\left(19\right)=\left\{1;-1;19;-19\right\}\)

hay \(x\in\left\{3;1;21;-17\right\}\)

Để A là số nguyên nhỏ nhất thì x+3=-1

hay x=-4

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

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

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

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}

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

\(\sqrt{x}+\sqrt{2-x}\ge\sqrt{x+2-x}=\sqrt{2}\)

\(\Rightarrow\dfrac{2}{2}\le P\le\dfrac{2}{\sqrt{2}}\Rightarrow1\le P\le\sqrt{2}\)

Mà \(P\in Z\Rightarrow P=1\)

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

Ta có : B = 2x+1/x-3 = (2x-6)+7/x-3 = 2+ 7/x-3 

Để B nhận giá trị nguyên thì x-3 thuộc Ư(7) = (+-1;+-7)

suy ra : x-3=-1 => x=2                               x-3=1 => x=4

             x-3=-7 => x=-4                               x-3=7 => x=10

Vậy x =(-4;2;4;10) thì B nhận giá trị nguyên

 P = A.B = \(\dfrac{x-7}{\sqrt{x}+2}=\dfrac{\left(x-4\right)-3}{\sqrt{x}+2}=\dfrac{\left(\sqrt{x}-2\right).\left(\sqrt{x}+2\right)-3}{\sqrt{x}+2}\)

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

\(P\inℤ\) <=> x là số chính phương và \(\dfrac{3}{\sqrt{x}+2}\inℤ\)

mà \(\sqrt{x}+2\ge2\Rightarrow\dfrac{3}{\sqrt{x}+2}\inℤ\Leftrightarrow\sqrt{x}+2=3\)

\(\Leftrightarrow x=1\) (thỏa)

Vậy x = 1 thì P \(\inℤ\)

a) \(\left(\frac{x+3}{x-2}+\frac{x+2}{3-x}+\frac{x+2}{x^2-5x+6}\right):\left(\frac{1-x}{x+1}\right)\)

= \(\left(\frac{x+3}{x-2}-\frac{x+2}{x-3}+\frac{x+2}{x^2-2x-3x+6}\right):\left(\frac{1-x}{x+1}\right)\)

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

= \(\left(\frac{x^2-9-x^2+4+x+2}{\left(x-2\right)\left(x-3\right)}\right).\frac{x+1}{1-x}\)

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

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

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

b) Để A >1 \(\Leftrightarrow\frac{x+1}{\left(x-2\right)\left(1-x\right)}>1\)

\(\Leftrightarrow\frac{-\left(1-x\right)\left(3-x\right)}{\left(x-2\right)\left(1-x\right)}\)

\(\Leftrightarrow\frac{x-3}{x-2}>0\)

\(\Rightarrow\orbr{\begin{cases}x-3\ge0\\x-2>0\end{cases}\Leftrightarrow\orbr{\begin{cases}x\ge3\\x>2\end{cases}\Leftrightarrow}x\ge3}\)

\(\Rightarrow\orbr{\begin{cases}x-3< 0\\x-2< 0\end{cases}\Leftrightarrow\orbr{\begin{cases}x< 3\\x< 2\end{cases}\Leftrightarrow}x< 2}\)

Vậy ...