\(A=\frac{x-2}{x+3}\in Z\)

=> (x- 2) \(⋮\)(x+ 3)

=> (x- 2)-( x+3) \(⋮\)(x +3)

=> -5 \(⋮\)(x+ 3)

Ta có bảng sau:

Để A thuộc Z thì x= { -4;-8; -2; 2}

a, x= -1;1

b, x= 0 thì A nguyên

a) bài 1

để \(x\in Z\)thì \(3x-1⋮x-1\)

mà \(x-1⋮x-1\)

\(\Rightarrow3\left(x-1\right)⋮x-1\)

\(\Rightarrow\left(3x-1\right)-\left[3x-3\right]⋮x-1\)

\(\Rightarrow2⋮x-1\)

\(\Rightarrow x-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

ta có bảng

vậy \(x\in\left\{2;0;3;-1\right\}\)

còn nữa mà bạn

câu c bài 2:

x chỉ nhận giá trị là 2

thật vậy

\(2^2=4\)

\(2.2=4\)

a) \(ĐKXĐ:x\ne4;x\ne9\)

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

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

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

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

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

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

\(\Rightarrow\sqrt{x}-3\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\) (ĐK: x thuộc Z)

Vậy để A thuộc Z khi x = {2;\(\sqrt{2};\sqrt{5};\sqrt{1};\sqrt{7}\) }

A = \(\frac{x-2}{x+3}\)=\(\frac{x+3-3-2}{x+3}\)= 1 +\(\frac{-5}{x+3}\)

suy ra x + 3 ∈ Ư(5) = { -5; -1 ; 1 ; 1}

Vây x ∈ { -8; -4; -2; 2}

Câu a:

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

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

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

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

A nguyên khi và chỉ khi \(3⋮\left(\sqrt{x}+1\right)\)

Câu b : \(\frac{m\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)}=\sqrt{x}-2\Leftrightarrow2m\sqrt{x}-m-x+\sqrt{x}+2=0\)

\(\Leftrightarrow x-\left(2m+1\right)\sqrt{x}+m-2=0\)phương trình có hai nghiệm phân biệt khi 

\(\Delta>0\)hay \(\Delta=\left(2m+1\right)^2-\left(m-2\right)4=m^2+9>0\forall m\)

Câu C: để \(A=2-\frac{3}{\sqrt{x}+1}\ge2-\frac{3}{0+1}=-1\)\(\Rightarrow A_{Min}=-1\)khi \(x=0\)