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

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

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

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}

1) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)

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

\(\Leftrightarrow P=\frac{\left(2+\sqrt{x}\right)^2-\left(2-\sqrt{x}\right)^2+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)

\(\Leftrightarrow P=\frac{4+4\sqrt{x}+x-4+4\sqrt{x}-x+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)

\(\Leftrightarrow P=\frac{4x+8\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)

\(\Leftrightarrow P=\frac{4\sqrt{x}}{2-\sqrt{x}}\)

2) Để \(P=2\)

\(\Leftrightarrow\frac{4\sqrt{x}}{2-\sqrt{x}}=2\)

\(\Leftrightarrow4\sqrt{x}=4-2\sqrt{x}\)

\(\Leftrightarrow6\sqrt{x}=4\)

\(\Leftrightarrow\sqrt{x}=\frac{2}{3}\)

\(\Leftrightarrow x=\frac{4}{9}\)

Vậy để \(P=2\Leftrightarrow x=\frac{4}{9}\)

3) Khi \(\left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-2=0\\2\sqrt{x}-1==0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=2\\\sqrt{x}=\frac{1}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\left(ktm\right)\\x=\frac{1}{4}\left(tm\right)\end{cases}}\)

Thay \(x=\frac{1}{4}\)vào P, ta được :

\(\Leftrightarrow P=\frac{4\sqrt{\frac{1}{4}}}{2-\sqrt{\frac{1}{4}}}=\frac{4\cdot\frac{1}{2}}{2-\frac{1}{2}}=\frac{2}{\frac{3}{2}}=\frac{4}{3}\)

4) Để \(P=\frac{\sqrt{x}+3}{2\sqrt{x}-1}\)

\(\Leftrightarrow\frac{4\sqrt{x}}{2-\sqrt{x}}=\frac{\sqrt{x}+3}{2\sqrt{x}-1}\)

\(\Leftrightarrow8x-4\sqrt{x}=-x-\sqrt{x}+6\)

\(\Leftrightarrow9x-3\sqrt{x}-6=0\)

\(\Leftrightarrow3x-\sqrt{x}-2=0\)

\(\Leftrightarrow\sqrt{x}=3x-2\)

\(\Leftrightarrow x=9x^2-12x+4\)

\(\Leftrightarrow9x^2-13x+4=0\)

\(\Leftrightarrow\left(9x-4\right)\left(x-1\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}9x-4=0\\x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{9}\\x=1\end{cases}}\)

Thử lại ta được kết quá : \(x=\frac{4}{9}\left(ktm\right)\); \(x=1\left(tm\right)\)

Vậy để \(P=\frac{\sqrt{x}+3}{2\sqrt{x}-1}\Leftrightarrow x=1\)

5) Để biểu thức nhận giá trị nguyên

\(\Leftrightarrow\frac{4\sqrt{x}}{2-\sqrt{x}}\inℤ\)

\(\Leftrightarrow4\sqrt{x}⋮2-\sqrt{x}\)

\(\Leftrightarrow-4\left(2-\sqrt{x}\right)+8⋮2-\sqrt{x}\)

\(\Leftrightarrow8⋮2-\sqrt{x}\)

\(\Leftrightarrow2-\sqrt{x}\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{1;3;0;4;-2;6;-6;10\right\}\)

Ta loại các giá trị < 0

\(\Leftrightarrow\sqrt{x}\in\left\{1;3;0;4;6;10\right\}\)

\(\Leftrightarrow x\in\left\{1;9;0;16;36;100\right\}\)

Vậy để \(P\inℤ\Leftrightarrow x\in\left\{1;9;0;16;36;100\right\}\)

\(\)

 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 \(\inℤ\)

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

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

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

Mà \(P\in Z\Rightarrow P=1\)

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

a) 

ĐKXĐ: \(x-4\ge0\text{ (1)};\text{ }x+4\sqrt{x-4}\ge0\text{ (2); }\frac{16}{x^2}-\frac{8}{x}+1>0\text{ (3)}\)

\(\left(1\right)\Leftrightarrow x\ge4\)

\(\left(2\right)\Leftrightarrow\left(\sqrt{x-4}+2\right)^2\ge0\text{ (đúng }\forall x\ge4\text{)}\)

\(\left(3\right)\Leftrightarrow\left(\frac{4}{x}-1\right)^2>0\Leftrightarrow\frac{4}{x}-1\ne0\Leftrightarrow x\ne4\)

Vậy ĐKXĐ là \(x>4\)

b)

\(A=\frac{\left|\sqrt{x-4}+2\right|+\left|\sqrt{x-4}-2\right|}{\left|\frac{4}{x}-1\right|}=\frac{\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|}{1-\frac{4}{x}}=\frac{x\left(\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|\right)}{x-4}\)

\(+\sqrt{x-4}\le2\Leftrightarrow04\)

\(A=\frac{x\left(\sqrt{x-4}+2+\sqrt{x-4}-2\right)}{x-4}=\frac{2x\sqrt{x-4}}{x-4}=\frac{2x}{\sqrt{x-4}}\)

Nếu \(\sqrt{x-4}\)là số vô tỉ thì A là số vô tỉ.

Để A là hữu tỉ thì \(\sqrt{x-4}=t\text{ }\left(t\in Z;\text{ }t>4\right)\Rightarrow x=t^2+4\)

Khi đó, \(A=\frac{2\left(t^2+4\right)}{t}=2t+\frac{8}{t}\)

A nguyên khi \(\frac{8}{t}\) nguyên hay \(t=8\text{ (do }t>4\text{)}\)

\(t=\sqrt{x-4}=8\Leftrightarrow x=8^2+4=68\)

Vậy \(x\in\left\{6;8;68\right\}\)

c/

\(+0

P nguyên <=> \(\dfrac{x-5}{\sqrt{x}+1}\) nguyên

<=> \(\dfrac{x-1}{\sqrt{x}+1}-\dfrac{4}{\sqrt{x}+1}\) nguyên

<=> \(\sqrt{x}-1-\dfrac{4}{\sqrt{x}+1}\) nguyên

=> \(\sqrt{x}+1\inƯ\left(4\right)=\left\{1;2;4\right\}\)  (vì \(\sqrt{x}+1>0\forall x\in N\))

...