Ta có : \(K=\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)

4 \(⋮\)(\(\sqrt{x}-3\))

=> \(\sqrt{x}-3\inƯ\left(4\right)\)

=> \(\sqrt{x}-3\in\left\{\pm1;\pm2;\pm4\right\}\)

Lập bảng :

Vậy : ...

\(\dfrac{\sqrt{x}-5}{\sqrt{x-3}}=1-\dfrac{2}{\sqrt{x}-3}=P\)

Để P nguyên thì \(2⋮\sqrt{x}-3\Leftrightarrow\sqrt{x}-3\inƯ\left(2\right)=\left\{\pm1,\pm2\right\}\)

\(\begin{matrix}\sqrt{x}-3&-1&-2&1&2\\\sqrt{x}&-2\left(L\right)&1&4&5\\x&&1\left(tm\right)&16\left(tm\right)&25\left(tm\right)\end{matrix}\) 

Mà x nguyên lớn nhất \(\Rightarrow x=25\)

Để P là số nguyên thì

căn x-3-2 chia hết cho căn x-3

=>căn x-3 thuộc Ư(-2)

mà x nguyên lớn nhất

nên căn x-3=2

=>x=25

\(P=\dfrac{\sqrt x-5}{\sqrt x -3}=\dfrac{\sqrt x-3-2}{\sqrt x -3}=1-\dfrac{2}{\sqrt x -3}\)

Để \(P \in Z \Leftrightarrow 2\vdots \sqrt x -3 \Rightarrow \sqrt x -3 \in \text{Ư(2)={1;-1;2;-2}}\)

\(\Rightarrow \sqrt x \in \text{{4;2;5;1}} \Rightarrow x \in \text{{16;4;25;1}}\)

\(\Rightarrow x_{max}=25\)

a) (Tự giải) ĐKXĐ: \(x\ge0;x\ne4;x\ne9\)

b) \(Q=\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)

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

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

        \(=\frac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

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

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

c) Để Q là 1 số nguyên => \(1-\frac{4}{\sqrt{x}-3}\in Z\) 

                                    Mà \(1\in Z\Rightarrow\frac{4}{\sqrt{x}-3}\in Z\)

                                     => \(4⋮\sqrt{x}-3\)

Hay \(\sqrt{x}-3\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

ta lập bảng

Vậy x={1;16;25;49}
 

a) Để \(\frac{11}{\sqrt{x}-5}\)nhận giá trị nguyên thì \(\sqrt{\text{x}}-5\inƯ\left(11\right)\)(DK : \(0\le x\ne25\))

Vì \(\sqrt{\text{x}}-5\ge-5\)nên ta có : 

\(\sqrt{x}-5\in\left\{-1;1;11\right\}\)\(\Rightarrow\sqrt{x}\in\left\{4;6;16\right\}\Rightarrow x\in\left\{16;36;256\right\}\)

b) \(B=\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)(DK : \(0\le x\ne9\))

Để B nhận giá trị nguyên thì \(\sqrt{x}-3\inƯ\left(4\right)\)

Vì \(\sqrt{\text{x}}-3\ge-3\)nên ta có : 

\(\sqrt{\text{x}}-3\in\left\{-2;-1;1;2;4\right\}\)\(\Rightarrow\sqrt{x}\in\left\{1;2;4;5;7\right\}\Rightarrow x\in\left\{1;4;16;25;49\right\}\)

1. Ta có: A = \(\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)

Để A \(\in\)Z <=> \(4⋮\sqrt{x}-3\) <=> \(\sqrt{x}-3\inƯ\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)

Lập bảng:

Vậy ....

2. Ta có: B = \(\frac{x^2+15}{x^2+3}=\frac{\left(x^2+3\right)+12}{x^2+3}=1+\frac{12}{x^2+3}\)

Do x² + 3 \(\ge\)3  \(\forall\)x => \(\frac{12}{x^2+3}\le4\forall x\)

=> \(1+\frac{12}{x^2+3}\le5\forall x\)

Dấu "=" xảy ra <=> x = 0

Vậy Max B = 5 khi x = 0