A∈Z⇒\(\dfrac{2\left(x+1\right)}{x+3}\in Z\Rightarrow\left(2x+2\right)⋮\left(x+3\right)\)

\(\Rightarrow\left(2x+6-4\right)⋮\left(x+3\right)\\ \Rightarrow\left[2\left(x+3\right)-4\right]⋮\left(x+3\right)\)

 \(\text{Mà}2\left(x+3\right)⋮\left(x+3\right)\\ \Rightarrow-4⋮\left(x+3\right)\\ \Rightarrow x+3\inƯ\left(-4\right)=\left\{-4;-2;-1;1;2;4\right\}\\ \Rightarrow x\in\left\{-7;-5;-4;-2;-1;1\right\}\)

- Bạn ơi lớp 6 cũng làm được nhé :)

x ∈{0;-6;-2;-4}

\(A=\dfrac{2\left(x+1\right)}{x+3}=\dfrac{2\left(x+3\right)-4}{x+3}=2-\dfrac{4}{x+3}\)

Để A nguyên \(\Rightarrow\dfrac{4}{x+3}\) nguyên

\(\Rightarrow x+3=Ư\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\)

\(\Rightarrow x=\left\{-7;-5;-4;-2;-1;1\right\}\)

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

Để M nguyên \(\Leftrightarrow\text{ }7\text{ }⋮\text{ }\left(\sqrt{x}-2\right)\)

=> \(\sqrt{x}-2\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{1;3;9\right\}\)

\(\Rightarrow x\in\left\{1;9;81\right\}\)

Tham Khảo

tích nha

\(A=\dfrac{2x+2}{x+3}.\left(x\ne-3\right).\)

\(A=2+\dfrac{-4}{x+3}.\)

Để \(A\in Z.\Leftrightarrow2+\dfrac{-4}{x+3}\in Z.\Leftrightarrow x+3\inƯ\left(-4\right)=\left\{1;-1;2;-2;4;-4\right\}.\)

\(\Rightarrow x\in\left\{-2;-4;-1;-5;1;-7\right\}.\)

ĐKXĐ: \(x\ne\left\{0;1\right\}\)

Rút gọn được \(P=x-\sqrt{x}+1\)

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

Do \(\left\{{}\begin{matrix}2\sqrt{x}\ge0\\x-\sqrt{x}+1=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\end{matrix}\right.\) \(\Rightarrow Q\ge0\)

\(Q=\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}=\dfrac{2\left(x-\sqrt{x}+1\right)-2x+4\sqrt{x}-2}{x-\sqrt{x}+1}=2-\dfrac{2\left(\sqrt{x}-1\right)^2}{x-\sqrt{x}+1}\le2\)

\(\Rightarrow0\le Q\le2\)

Mà \(Q\in Z\Rightarrow Q=\left\{0;1;2\right\}\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}=0\\\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}=1\\\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}2\sqrt{x}=0\\x-3\sqrt{x}+1=0\\x-2\sqrt{x}+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}=\dfrac{3+\sqrt{5}}{2}\\\sqrt{x}=\dfrac{3-\sqrt{5}}{2}\\\sqrt{x}=1\end{matrix}\right.\) \(\Rightarrow x=\left\{0;\dfrac{7+3\sqrt{5}}{2};\dfrac{7-3\sqrt{5}}{2};1\right\}\)

a: Thay x=2/3 vào A, ta được:

\(A=\dfrac{3\cdot\dfrac{2}{3}+2}{\dfrac{2}{3}}=\dfrac{2+2}{\dfrac{2}{3}}=4\cdot\dfrac{3}{2}=6\)

b: \(B=\dfrac{x^2+1}{x^2-x}-\dfrac{2}{x-1}\)

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

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

\(=\dfrac{\left(x-1\right)^2}{x\left(x-1\right)}=\dfrac{x-1}{x}\)

c: P=A:B

\(=\dfrac{3x+2}{x}:\dfrac{x-1}{x}=\dfrac{3x+2}{x}\cdot\dfrac{x}{x-1}=\dfrac{3x+2}{x-1}\)

Để P là số nguyên thì \(3x+2⋮x-1\)

=>\(3x-3+5⋮x-1\)

=>\(5⋮x-1\)

=>\(x-1\in\left\{1;-1;5;-5\right\}\)

=>\(x\in\left\{2;0;6;-4\right\}\)

Kết hợp ĐKXĐ, ta được: \(x\in\left\{2;6;-4\right\}\)

Thay x=2 vào P, ta được:

\(P=\dfrac{3\cdot2+2}{2-1}=\dfrac{8}{1}=8\)

Thay x=6 vào P, ta được:

\(P=\dfrac{3\cdot6+2}{6-1}=\dfrac{18+2}{5}=\dfrac{20}{5}=4\)

Thay x=-4 vào P, ta được:

\(P=\dfrac{3\cdot\left(-4\right)+2}{-4-1}=\dfrac{-12+2}{-5}=\dfrac{-10}{-5}=2\)

Vì 2<4<8

nên khi x=-4 thì P có giá trị nguyên nhỏ nhất

Để A nguyên thì 3n+3-1 chia hết cho n+1

=>\(n+1\in\left\{1;-1\right\}\)

=>\(n\in\left\{0;-2\right\}\)