Biến đổi \(A=4x^2+9x+29+\frac{4}{x-3}\)

\(\Leftrightarrow A\in Z\)

\(\Leftrightarrow\frac{4}{x-3}\in Z\)

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

Ta có bảng :

\(A=\frac{4x^3-3x^2+2x-83}{x-3}\)

\(=4x^2+9x+29+\frac{4}{x-3}\)

\(\Rightarrow A\in Z\)

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

\(\Rightarrow x-3\inƯ\left(4\right)=\left\{+1;\pm2;\pm4\right\}\)

Chỗ này mình làm tắt luôn nên chịu khó tử hiểu nhé, nếu k đc thì nhắn tin vs mình

\(\Rightarrow x=-1;1;2;4;5;7\)

mọi người giúp em với ạ

a, \(x-1\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)

b, \(2x-1\inƯ\left(-4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

c, \(\dfrac{3\left(x-1\right)+10}{x-1}=3+\dfrac{10}{x-1}\Rightarrow x-1\inƯ\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)

d, \(\dfrac{4\left(x-3\right)+3}{-\left(x-3\right)}=-4-\dfrac{3}{x+3}\Rightarrow x+3\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)

a:

ĐKXĐ: x<>-1/2

Để \(\dfrac{2x^3+x^2+2x+2}{2x+1}\in Z\) thì

\(2x^3+x^2+2x+1+1⋮2x+1\)

=>\(2x+1\inƯ\left(1\right)\)

=>2x+1 thuộc {1;-1}

=>x thuộc {0;-1}

b:

ĐKXĐ: x<>1/3

 \(\dfrac{3x^3-7x^2+11x-1}{3x-1}\in Z\)

=>3x^3-x^2-6x^2+2x+9x-3+2 chia hết cho 3x-1

=>2 chia hết cho 3x-1

=>3x-1 thuộc {1;-1;2;-2}

=>x thuộc {2/3;0;1;-1/3}

mà x nguyên

nên x thuộc {0;1}

c: 

ĐKXĐ: x<>2

\(\dfrac{x^4-16}{x^4-4x^3+8x^2-16x+16}\in Z\)

=>\(\left(x^2-4\right)\left(x^2+4\right)⋮\left(x-2\right)^2\left(x^2+4\right)\)

=>\(x+2⋮x-2\)

=>x-2+4 chia hết cho x-2

=>4 chia hết cho x-2

=>x-2 thuộc {1;-1;2;-2;4;-4}

=>x thuộc {3;1;4;0;6;-2}

\(a,A=\dfrac{x^2-6x+9-x^2+9}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}\\ A=\dfrac{-6x+18}{2\left(x-3\right)\left(x-1\right)}=\dfrac{-6\left(x-3\right)}{2\left(x-3\right)\left(x-1\right)}=\dfrac{-3}{x-1}\\ b,A\in Z\Leftrightarrow x-1\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow x\in\left\{-2;0;2;4\right\}\)

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

a, \(\dfrac{x}{x+2}\) + \(\dfrac{2x}{x-2}\) -\(\dfrac{3x^2-4}{x^2-4}\)

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

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

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

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

Có vài bước mình làm tắc á nha :>

a: \(A=\dfrac{x^2+4x+4+4x^2-x^2+4x-4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x-2}{x\left(x^2+x+2\right)}\)

\(=\dfrac{4x^2+8x}{\left(x+2\right)}\cdot\dfrac{1}{x\left(x^2+x+2\right)}=\dfrac{4}{x^2+x+2}\)

|x+3|=5

=>x=2(loại) hoặc x=-8(nhận)

Khi x=-8 thì \(A=\dfrac{4}{64-8+2}=\dfrac{4}{58}=\dfrac{2}{29}\)

b: A nguyên

=>x^2+x+2 thuộc {1;-1;2;-2;4;-4}

=>x^2+x+2=2 hoặc x^2+x+2=4

=>x^2+x-2=0 hoặc x(x+1)=0

=>\(x\in\left\{1;0;-1\right\}\)