ĐKXĐ: \(x\ge0;x\ne25\)

\(A=\dfrac{\sqrt{x}+2}{\sqrt{x}-5}=\dfrac{\sqrt{x}-5+7}{\sqrt{x}-5}=1+\dfrac{7}{\sqrt{x}-5}\)

Để \(A\in\mathbb{Z}\) thì: \(\dfrac{7}{\sqrt{x}-5}\) nhận giá trị nguyên

\(\Rightarrow 7\vdots\sqrt{x}-5\)

\(\Rightarrow\sqrt{x}-5\inƯ\left(7\right)\)

\(\Rightarrow\sqrt{x}-5\in\left\{1;7;-1;-7\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{6;12;4;-2\right\}\) mà \(\sqrt{x}\ge0\)

\(\Rightarrow\sqrt{x}\in\left\{4;6;12\right\}\)

\(\Rightarrow x\in\left\{16;36;144\right\}\left(tm\right)\)

Vậy \(A\in \mathbb{Z}\) khi \(x\in\left\{16;36;144\right\}\)

GIới hạn đã cho hữu hạn

\(\Rightarrow\sqrt[3]{13x^2+2x+5}-\sqrt[3]{81x^2+ax+1}=0\) có nghiệm \(x=-1\)

\(\Rightarrow a=18\)

Khi đó:

\(\lim\limits_{x\rightarrow-1}\dfrac{\sqrt{13x^2+2x+5}-\sqrt[3]{81x^2+18x+1}}{\left(x+1\right)^2}\)

\(=\lim\limits_{x\rightarrow-1}\dfrac{\left(\sqrt[]{13x^2+2x+5}-\left(1-3x\right)\right)+\left(1-3x-\sqrt[3]{81x^3+18x+1}\right)}{\left(x+1\right)^2}\)

\(=...=\dfrac{17}{16}\)

Đặt \(x+\sqrt{3}=a;\frac{1}{x}-\sqrt{3}=b\left(a,b\in Z\right)\)

=> \(a-\sqrt{3}=\frac{1}{b+\sqrt{3}}=x\)

=> \(ab-3=\sqrt{3}\left(b-a\right)\)

Do \(a,b\in Z\)

=> \(\sqrt{3}\left(b-a\right)\in Z\)

=> \(a=b\)

=> \(ab=3\)=> \(a=b=\sqrt{3}\)(Loại)

Vậy không có giá trị nào của x t/m đề bài

Câu trả lời trên sai rồi, câu trả lời đúng đây:

 Đặt \(\hept{\begin{cases}x+\sqrt{3}=a\\\frac{1}{x}-\sqrt{3}=b\end{cases}}\left(a,b\inℤ\right)\Rightarrow\hept{\begin{cases}x=a-\sqrt{3}\\\frac{1}{x}=b+\sqrt{3}\end{cases}\Rightarrow\hept{\begin{cases}x=a-\sqrt{3}\\x=\frac{1}{b+\sqrt{3}}\end{cases}\Rightarrow}a-\sqrt{3}=\frac{1}{b+\sqrt{3}}}\)

\(\Rightarrow\left(a-\sqrt{3}\right)\left(b+\sqrt{3}\right)=1\Rightarrow4-ab=\sqrt{3}\left(a-b\right)\)

TH1: \(a-b\ne0\Rightarrow\sqrt{3}\left(a-b\right)\notinℤ\)

mà\(4-ab\inℤ\)

 suy ra mâu thuẫn

TH2:\(a-b=0\Rightarrow a=b\Rightarrow4-a^2=4-b^2=0\Rightarrow a=b=2\)

 Khi đó \(x=2-\sqrt{3}\)

 Vậy........................................

trường hợp a=b=-2 đâu

1.

\(a,Q=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{x+5}{x-\sqrt{x}-2}\)

\(Q=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\\ Q=\dfrac{x-3\sqrt{x}+2-x-4\sqrt{x}-3-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\\ Q=\dfrac{-x-7\sqrt{x}-6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\\ Q=\dfrac{-\left(x+7\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\\ Q=\dfrac{-\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\dfrac{-\sqrt{x}-6}{\sqrt{x}-2}\)

\(b,Q\in Z\Leftrightarrow\dfrac{-\sqrt{x}-6}{\sqrt{x}-2}\in Z\)

\(\Leftrightarrow\dfrac{-\left(\sqrt{x}-2\right)-8}{\sqrt{x}-2}\in Z\\ \Leftrightarrow-1-\dfrac{8}{\sqrt{x}-2}\in Z\)

Mà \(-1\in Z\Leftrightarrow\dfrac{8}{\sqrt{x}-2}\in Z\)

\(\Leftrightarrow8⋮\sqrt{x}-2\\ \Leftrightarrow\sqrt{x}-2\inƯ\left(8\right)=\left\{-8,-4,-2,-1,1,2,4,8\right\}\)

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

Mà \(x\in Z\) và \(\sqrt{x}\ge0\)

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

Vậy \(x\in\left\{0;1;4\right\}\) thì \(Q\in Z\)

\(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\left(\text{đ}k\text{x}\text{đ}:x\ge3\right)\\ =\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\\ =\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{2\sqrt{x}-9-\left(x-9\right)-\left(2x-4\sqrt{x}+\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{2\sqrt{x}-9-x+9-2x+4\sqrt{x}-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ =\dfrac{5\sqrt{x}-3x+2}{x-5\sqrt{x}+6}\)

__

Để \(M\in Z\) thì \(x-5\sqrt{x}+6\) thuộc ước của \(5\sqrt{x}-3x+2\)

\(\Rightarrow x-5\sqrt{x}+6=-5\sqrt{x}-3x+2\\ \Leftrightarrow x-5\sqrt{x}+6+5\sqrt{x}+3x-2=0\\ \Leftrightarrow4x-4=0\\ \Leftrightarrow4x=4\\ \Leftrightarrow x=1\)

đk x khác 0 

\(A=4+\dfrac{6}{\sqrt{x}}\Rightarrow\sqrt{x}\inƯ\left(6\right)=\left\{1;2;3;6\right\}\)

\(a.P=\dfrac{2\sqrt{x}-5}{x-5\sqrt{x}+4}+\dfrac{2}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-4}=\dfrac{2\sqrt{x}-5+2\sqrt{x}-8-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-4\right)}=\dfrac{3}{\sqrt{x}-1}\) ( x ≥ 0 ; x # 1 ; x # 16 )

\(b.\) \(P\text{∈}Z\) ⇔ \(\dfrac{3}{\sqrt{x}-1}\text{∈}Z\) ⇔ \(\sqrt{x}-1\text{∈}\left\{1;-1;3;-3\right\}\)

+) \(\sqrt{x}-1=1\text{⇔}x=4\left(TM\right)\)

+) \(\sqrt{x}-1=-1\text{⇔}x=0\left(TM\right)\)

+) \(\sqrt{x}-1=3\text{⇔}x=16\left(KTM\right)\)

+) \(\sqrt{x}-1=-3\text{⇔}vo-nghiem\)

KL............