x + 3  chia hết x - 1

x + 3 - ( x - 1 ) chia hết  x - 1

2 chia hết x - 1

Do đó x - 1 thuộc Ư (2) = ( 1,-1,2,-2)

x - 1 = 1 suy ra x = 2

x - 1 = -1 suy ra x = 0

x - 1 = 2 suy ra x = 3

x - 1 = -2 suy ra x = -1

Vậy x = 2, 0, 3, -1

Ta có: x-3/x-1 = x-1-2/x-3 = 1-2/x-3

Để x-3/x-1 có giá trị là số nguyên

suy ra 2 chia hết cho x-3

suy ra x-3 thuộc U(2)={1;2;-1;-2}

suy ra x-3 thuộc {1;2;-1;-2}

suy ra x thuộc {4;5;2;1}

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

Để đạt GT nguyên thì \(\dfrac{4}{x-1}\in Z\)

\(\Rightarrow x-1\inƯ_{\left(4\right)}=\left\{-4;-2;-1;1;2;4\right\}\\ \Rightarrow x\in\left\{-3;-1;0;2;3;5\right\}\)

\(\dfrac{x-1+4}{x-1}=1+\dfrac{4}{x-1}\Rightarrow x-1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

\(A=\frac{2x^3-6x^2+x-8}{x-3}\)

\(=\frac{2x^3-6x^2+x-3-5}{x-3}=2x^2+1-\frac{5}{x-3}\)

Để A là số nguyên thì -5⋮x-3

=>x-3∈{1;-1;5;-5}

=>x∈{4;2;8;-2}

1: Để A là số nguyên thì x+3⋮x-2

=>x-2+5⋮x-2

=>5⋮x-2

=>x-2∈{1;-1;5;-5}

=>x∈{3;1;7;-3}

Để C là số nguyên thì 2x+3⋮x-2

=>2x-4+7⋮x-2

=>7⋮x-2

=>x-2∈{1;-1;7;-7}

=>x∈{3;1;9;-5}

2:

\(\left(x-2\right)^2=25\)

=>\(\left[\begin{array}{l}x-2=5\\ x-2=-5\end{array}\right.\Rightarrow\left[\begin{array}{l}x=7\\ x=-3\end{array}\right.\)

\(C=\dfrac{\left(x^2+3x\right)\left(x^2+2\right)-2}{x^2+2}=x^2+3x-\dfrac{2}{x^2+2}\)

\(C\in Z\Leftrightarrow2⋮\left(x^2+2\right)\)

\(\Leftrightarrow x^2+2=2\Rightarrow x=0\)

ko cần làm câu a nha các bạn

ĐKXĐ: \(x\ne1\)

Ta có: \(B=\dfrac{x^4-2x^3-3x^2+8x-1}{x^2-2x+1}\)

\(=\dfrac{x^4-2x^3+x^2-4x^2+8x-4+3}{x^2-2x+1}\)

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

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

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

Để B nguyên thì \(3⋮\left(x-1\right)^2\)

\(\Leftrightarrow\left(x-1\right)^2\inƯ\left(3\right)\)

\(\Leftrightarrow\left(x-1\right)^2\in\left\{1;3;-1;-3\right\}\)

mà \(\left(x-1\right)^2>0\forall x\) thỏa mãn ĐKXĐ

nên \(\left(x-1\right)^2\in\left\{1;3\right\}\)

\(\Leftrightarrow x-1\in\left\{1;9\right\}\)

hay \(x\in\left\{2;10\right\}\) (nhận)

Vậy: \(x\in\left\{2;10\right\}\)

ĐKXĐ: x<>1

\(B=\frac{x^4-2x^3-3x^2+8x-1}{x^2-2x+1}\)

\(=\frac{x^4-2x^3+x^2-4x^2+8x-4+3}{x^2-2x+1}=\frac{\left(x^2-2x+1\right)\left(x^2-4\right)+3}{x^2-2x+1}\)

\(=x^2-4+\frac{3}{\left(x-1\right)^2}\)

Để B là số nguyên thì 3⋮\(\left(x-1\right)^2\)

=>\(\left(x-1\right)^2\in\left\lbrace1;3\right\rbrace\)

mà x ∈Z

nên \(\left(x-1\right)^2=1\)

=>x-1∈{1;-1}

=>x∈{2;0}

ĐKXĐ: x<>1

\(B=\frac{x^4-2x^3-3x^2+8x-1}{x^2-2x+1}\)

\(=\frac{x^4-2x^3+x^2-4x^2+8x-4+3}{x^2-2x+1}=x^2-4+\frac{3}{\left(x-1\right)^2}\)

Để B là số nguyên thì 3⋮\(\left(x-1\right)^2\)

=>\(\left(x-1\right)^2=1\)

=>x-1∈{1;-1}

=>x∈{2;0}